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PHP - Retrieve From One Table And Inserting This To Another Table.
How would you retrieve information from one table and enter that informatio to another and the same time get other details from form.
Do you retrieve the data and use a form to post this and insert that to another table. Have I confused you??? Similar TutorialsI'm creating a report page and can't figure out how to retrieve multiple rows from a table and display them in html. There are 5 data elements in each row and I'm thinking that using a form in the html might be the best way as I'm totally ignorant about tables in html. I'm a newbie to php and can't figure out how to accomplish this. Here's the code that I have so far:
<?php
$filename = NULL;
session_start();
// start of script every time.
// setup a path for all of your canned php scripts
$php_scripts = '../php/'; // a folder above the web accessible tree
// load the pdo connection module
require $php_scripts . 'PDO_Connection_Select.php';
//*******************************
// Begin the script here
// Connect to the database
if (!$con = PDOConnect("foxclone")):
{
echo "Failed to connect to database" ;
exit;
}
else:
{
$sql = 'SELECT COUNT(IP_ADDRESS) FROM download WHERE FILENAME IS NOT NULL';
$sql1 = 'Update download t2, ip_lookup t1 set t2.country = t1.country, t2.area = t1.area, t2.city = t1.city where ((t2.IP_ADDRESS) = (t1.start_ip) OR (t2.IP_ADDRESS) > (t1.start_ip)) AND ((t2.IP_ADDRESS) = (t1.end_ip) OR (t2.IP_ADDRESS) < (t1.end_ip)) AND (t2.FILENAME is not null and t2.country is null)';
$sql2 = 'SELECT (IP_ADDRESS, FILENAME, country, area, city) from download where FILENAME is not null';
// Update the table
$stmt = $con->prepare($sql1);
$stmt->execute();
// Get count of rows to be displayed in table
$stmt = $con->prepare($sql);
$stmt->execute() ;
$cnt = $stmt->fetch(PDO::FETCH_NUM);
// retrieve one row at a time
$i = 1;
while($i <= $cnt){
$stmt = $con->prepare($sql2);
$row->execute(array('')); // Do I need an array here?
// from here on, I'm lost
$i++;
I'd appreciate any guidance you can provide or understandable tutorials you can point me to. Larry I am not sure if the title is correct; I tried my best.
I'm a PHP/MySQL beginner and I really need some help.
I have a small script that I am using for sending SMS. I recently added a phonebook. The problem with the phonebook right now is that it's available to all users, i.e. they can all update and delete all rows. What I would like to do is make it so that each user can update and delete only their own contacts.
I have a table call contacts. Inside that table there is first name, last name, company and phonenumber.
How can I accomplish this with PHP & MySQL?
CREATE TABLE IF NOT EXISTS `contacts` ( `contact_id` int(10) NOT NULL AUTO_INCREMENT, `firstname` varchar(255) NOT NULL, `lastname` varchar(255) NOT NULL, `company` varchar(255) NOT NULL, `cell_no` text NOT NULL, PRIMARY KEY (`contact_id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ; CREATE TABLE IF NOT EXISTS `users` ( `user_id` int(11) NOT NULL AUTO_INCREMENT, `users_name` varchar(30) NOT NULL, `uname` varchar(30) NOT NULL, `u_pass` varchar(60) NOT NULL, `utype` varchar(30) NOT NULL, `timezone` varchar(30) NOT NULL, `uapi_user` varchar(30) NOT NULL, `uapi_pass` varchar(60) NOT NULL, PRIMARY KEY (`user_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ;
<?php
if (isset($_POST['submit'])){
//form has been submitted1
$firstname = trim($_POST['firstname']);
$lastname = trim($_POST['lastname']);
$company = trim($_POST['company']);
$cellno = trim($_POST['cell_no']);
if($firstname == ''){
echo '<div class="alert alert-danger">First Name is not Valid!</div>';
exit;
}elseif($lastname == ''){
echo '<div class="alert alert-danger">Last Name is not Valid!</div>';
exit;
}elseif($company == ''){
echo '<div class="alert alert-danger">Company is not Valid!</div>';
exit;
}elseif($cellno == ''){
echo '<div class="alert alert-danger">Cellphone Number is not Valid!</div>';
exit;
}else{
$query = "Select cell_no from contacts where cell_no = '".$cellno."' ";
$result = mysql_query($query);
if (!mysql_num_rows($result)) {
$sql = "INSERT INTO contacts(firstname, lastname, company, cell_no)
values('{$firstname}','{$lastname}', '{$company}', '{$cellno}')";
$result = mysql_query($sql);
confirm_query($result);
//echo '<div class="alert alert-success">Successfully added.</div>';
//exit;
?>
<script type="text/javascript">
window.location = "contact_list.php";
</script>
<?php
}
else{
echo '<div class="alert alert-danger">Username. already exist!.</div>';
echo '<p><a href="new_contact.php" class="btn btn-success"> Back </a></p>';
exit;
}}
}else{
$firstname = "";
$lastname = "";
$company = "";
$cellno = "";
}
?>
Hi there reader(s) I've been searching everywhere, and I'm not one to post my issue unless I've searched as hard as I could, I'm not new to PHP, but I am new to MySQL (so go easy on me) and I've got variables all set up, and I've successfully established connection to my Database and my Database's table, but now I cannot find a way to insert the data (variable) into a column of the table. Willing to provide more information! I want to allow users to post entries to a NEWS table and, if they wish, to post an accompanying image to an IMAGES table. Tables are like this: NEWS id // if there'll be an accompanying image, this id to be sent to IMAGES table title subtitle created news_entry category IMAGES image_id f_news_id // the foreign id of the associated post in NEWS table filename caption description So, the user comes to the insert_entry.php page and creates a post. If the user clicks an "Upload accompanying image" link, the news post id must be inserted in the f_news_id field when the image is uploaded. Code excerpt from insert_entry.php: Code: [Select] // Insert the news_entry in the database... // Make the query: $q = "INSERT INTO news (title, subtitle, news_entry, category) VALUES ('$title', '$subtitle', '$news_entry', '$category') "; $r = @mysqli_query ($dbc, $q); // Run the query. if ($r) { // If it ran OK. // Print a message: echo "<h1>Thank you!</h1> <p>You have successfully inserted the News Entry below.</p>"; echo "<h1>" . stripslashes($title) . "</h1><h2>" . stripslashes($subtitle) . "</h2><p>" . stripslashes($news_entry) . "</p>"; // get id of record just created $q = "SELECT id FROM news ORDER BY created DESC LIMIT 1"; $r = mysqli_query($dbc, $q); while ($row = mysqli_fetch_assoc($r)) { // pass the id via GET in the URL echo "<a href='upload_image.php?=" . $row['id'] . "'>Upload image</a>"; } mysqli_close($dbc); ?> Code excerpt from upload_image.php: Code: [Select] // insert news post id into images table if user came via insert_entry.php page // Make the query: require_once ('includes/mysqli_connect.php'); // Connect to the db. $description = mysqli_real_escape_string($dbc, trim($_POST['description'])); $caption = mysqli_real_escape_string($dbc, trim($_POST['caption'])); if (isset($_GET['id'])) { // if there's a NEWS post id $q = "INSERT INTO images (f_news_id, filename, caption, description) VALUES ('$_GET['id']', '{$_FILES['upload']['name']}', '$caption', '$description')"; } else { // if user arrived at upload_image.php otherwise and there's *not* a NEWS post id $q = "INSERT INTO images (filename, caption, description) VALUES ('{$_FILES['upload']['name']}', '$caption', '$description') "; } $r = @mysqli_query ($dbc, $q); // Run the query. if ($r) { // If it ran OK. // Print a message: echo "<p>Info entered in images table.</p>"; Am I going about this the wrong way? Am new to php... so any advice much appreciated... Stuck again on simple code. I am trying to insert some fields extracted from one table into another. I'm using code that worked elsewhere. The SQL statement flies, the script runs, the input array is printed back I get an echo back from the end of the script but nothing is added to the table. Even aded an echo print_r in the conditional and I know the data is getting to the execute command. The script follows with a sample of the input array. I have attached am image of the table I am trying to insert the data into. --Kenoli
The script:
<?php
require '__classes/DB.php';
$sql = "SELECT name, table_id, image_name, description, medium FROM tbl_person_data ";
$stmt = $pdo->query($sql);
$array1 = $stmt->fetchall(PDO::FETCH_ASSOC);
$stmt = $pdo->prepare("INSERT INTO Images (name, person_id, filename, description, medium) VALUES (?,?,?,?,?)");
//$pdo->beginTransaction();
foreach ($array1 as $row)
{
$stmt->execute($row);
}
echo "<pre>";
print_r ($row);
echo "</pre>";
echo '<h4>Got to end of file</h4>';
?>
$array1: The input array
[0] => Array
(
[name] => Carol Lettko
[table_id] => 21
[image_name] => Carol_Lettko-DSC_3022.jpg
[description] => Baby Herons/Brickyard
[medium] => photo
)
[1] => Array
(
[name] =>
[table_id] => 22
[image_name] => Carol_Lettko-DSC_0164.JPG
[description] => Heron/Brickyard
[medium] => photo
)
[2] => Array
(
[name] =>
[table_id] => 23
[image_name] => Carol_Lettko-IMG_5723.jpg
[description] => Kayaker/Brickyard
[medium] => photo
)
i want insert data from text box in html form.there are many text boxes gave name using 2 loops..please tell how can insert data to table?please reply im trying to write a script takes an xml files with tv show info, splits it into the show and the eppisode info and the place it into a table, i have got it to proccess all the info and print it out on a web page, but i cant, for the life of me, get it to insert said data into the table, it seems to be just ignoring the code and prints out the data as if nothing happens, no errors or anything. here is my code (abit messy but im only just starting and its my test.php) Code: [Select] <?php $tvdb_mirror = "http://www.thetvdb.com/api/"; $tvdb_time = "http://www.thetvdb.com/api/Updates.php?type=none"; $dbname = "mediadb"; $dbuser = "root"; $dbpass = ""; $dbserv = "127.0.0.1"; $rss = simplexml_load_file('sample.xml'); $showName = "Show Name = ".$rss->Series->SeriesName; print $showName; print "<br />Show Discription = ".$rss->Series->Overview; print "<br />"; mysql_connect('127.0.0.1', 'root', ''); @mysql_select_db('mediadb') or die("Unable to select database"); foreach ($rss->Episode as $item) { $seasonnum = $item->Combined_season; $EpisodeNumber = $item->EpisodeNumber; if($EpisodeNumber < 10){ $EpisodeNumber = "0".$EpisodeNumber; }; $EpisodeName = $item->EpisodeName; $Overview = $item->Overview; $airdate = $item->FirstAired; $tvdbid = $item ->id; $query = "INSERT INTO eppisodes VALUES('', '1', ".$EpisodeName.", ".$Overview.", ".$airdate.", '1', ".$tvdbid.", '-1', ".$seasonnum.", ".$EpisodeNumber.")"; mysql_query($query); print "<br />".$showName." - ".$seasonnum."x".$EpisodeNumber." - ".$EpisodeName." Overview:<br />".$Overview; } mysql_close(); ?> i am a noob @ php and mysql, but i have doubke and triple checked the names of the db and table. here is an sql dump of my db Code: [Select] -- phpMyAdmin SQL Dump -- version 3.3.5 -- http://www.phpmyadmin.net -- -- Host: 127.0.0.1 -- Generation Time: Nov 13, 2010 at 12:48 PM -- Server version: 5.1.49 -- PHP Version: 5.3.3 SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO"; /*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */; /*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */; /*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */; /*!40101 SET NAMES utf8 */; -- -- Database: `mediadb` -- -- -------------------------------------------------------- -- -- Table structure for table `eppisodes` -- CREATE TABLE IF NOT EXISTS `eppisodes` ( `id` int(11) NOT NULL AUTO_INCREMENT, `showID` int(11) NOT NULL, `eppname` varchar(255) NOT NULL, `eppdesc` longtext NOT NULL, `airdate` date NOT NULL, `format` int(11) NOT NULL, `tvdbid` varchar(20) NOT NULL, `dohave` tinyint(1) NOT NULL, `season` varchar(2) NOT NULL, `eppisode` varchar(3) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ; -- -- Dumping data for table `eppisodes` -- -- -------------------------------------------------------- -- -- Table structure for table `shows` -- CREATE TABLE IF NOT EXISTS `shows` ( `id` int(100) NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, `description` longtext NOT NULL, `TVDBID` int(100) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; -- -- Dumping data for table `shows` -- INSERT INTO `shows` (`id`, `name`, `description`, `TVDBID`) VALUES (1, 'higogo', 'some info', 67546); any help would be very much appericiated. fyi im running win7 with easyPHP 5.3.3 with php 5.3.3, mysql 5.1.49 apache 2.2.16 Hi I'm using this code to insert multiple records. The code executes but nothing is entered into my database. This usually happens when there's a mismatch in data types.
How do I ensure that description goes in as text which in sql is wrapped in single quotes, but also make sure the other variables go in as numeric.
// an array items to insert
$array = array( 'theid' => $theid,
'descr' => $descr,
'costperunit' => $costperunit,
'quantity' => $quantity,
'costperlot' => $costperlot
);
// begin the sql statement
$sql1 = "INSERT INTO descriptions (jobid, description, costperunit, quantity, costperlot) VALUES ";
$it = new ArrayIterator( $array );
// a new caching iterator gives us access to hasNext()
$cit = new CachingIterator( $it );
// loop over the array
foreach ( $cit as $value )
{
// add to query
$sql1 .= "('".$cit->key()."','" .$cit->current()."')";
if( $cit->hasNext() )
{
$sql1 .= ",";
}
}
I have a form that is posting a value from a radio button as well as 16 Hidden values. The form action goes to file2- which is successfully conecting and processing but instead of inserting the correct values I'm just seeing a row of zeros. I don't know where to start looking! This is the code from the posting form : --- <?php // query.php require_once 'login.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); $result = mysql_query("SELECT qid,question,oc1,oc2,oc3,oc4,oc5,oc6,oc7,psc1,psc2,psc3,psc4,psc5,psc6,psc7,psc8,psc9 FROM tblquestions WHERE qid = '1'"); if (!$result) { echo 'Could not run query: ' . mysql_error(); exit; } $row = mysql_fetch_row($result); echo $row[0]; // Question Number echo $row[1]; // Question echo $row[2]; // oc1 echo $row[3]; // oc2 echo $row[4]; // oc3 echo $row[5]; // oc4 echo $row[6]; // oc5 echo $row[7]; // oc6 echo $row[8]; // oc7 echo $row[9]; // pcs1 echo $row[10]; // pcs2 echo $row[11]; // pcs3 echo $row[12]; // pcs4 echo $row[13]; // pcs5 echo $row[14]; // pcs6 echo $row[15]; // pcs7 echo $row[16]; // pcs8 echo $row[17]; // pcs9 ?> <p>Hello </p> <?php echo $row[0]; // Question Number ?> <form id="form1" name="form1" method="post" action="./uinputprocess.php"> <table width="800" border="0"> <tr> <td colspan="6"><p>Question <?php echo $row[0]; // Question Number ?> </p> <?php echo $row[1]; // Question ?> .</td> </tr> <tr> <td width="126"><p align="center"> </p> <p align="center">Strongly Disagree </p></td> <td width="65"><p align="center"> </p> <p align="center">Disagree </p></td> <td width="145"><p align="center"> </p> <p align="center">Unsure but Doubtful </p></td> <td width="152"><p align="center"> </p> <p align="center">Unsure but Probable </p></td> <td width="67"><p align="center"> </p> <p align="center">Agree </p></td> <td width="219"><p align="center"> </p> <p align="center">Strongly Agree </p></td> </tr> <tr> <td><div align="center"> <input type= "radio" name= "ans" value = "1" /> </div></td> <td><div align="center"> <input type= "radio" name= "ans" value = "2" /> </div></td> <td><div align="center"> <input type= "radio" name= "ans" value = "3" /> </div></td> <td><div align="center"> <input type= "radio" name= "ans" value = "4" /> </div></td> <td><div align="center"> <input type= "radio" name= "ans" value = "5" /> </div></td> <td><div align="center"> <input type= "radio" name= "ans" value = "6" /> </div></td> </tr> <tr> <td colspan="6"> </td> </tr> <tr> <td colspan="6"><p>Hidden Fields Here > <input type="hidden" name="oc1" id= <?php "echo $row[2];" ?> /> <input type="hidden" name="oc2" id= <?php "echo $row[3];" ?> /> <input type="hidden" name="oc3" id= <?php "echo $row[4];" ?> /> <input type="hidden" name="oc4" id= <?php "echo $row[5];" ?> /> <input type="hidden" name="oc5" id= <?php "echo $row[6];" ?> /> <input type="hidden" name="oc6" id= <?php "echo $row[7];" ?> /> <input type="hidden" name="oc7" id= <?php "echo $row[8];" ?> /> <input type="hidden" name="psc1" id= <?php "echo $row[9];" ?> /> <input type="hidden" name="psc2" id= <?php "echo $row[10];" ?> /> <input type="hidden" name="psc3" id= <?php "echo $row[11];" ?> /> <input type="hidden" name="psc4" id= <?php "echo $row[12];" ?> /> <input type="hidden" name="psc5" id= <?php "echo $row[13];" ?> /> <input type="hidden" name="psc6" id= <?php "echo $row[14];" ?> /> <input type="hidden" name="psc7" id= <?php "echo $row[15];" ?> /> <input type="hidden" name="psc8" id= <?php "echo $row[16];" ?> /> <input type="hidden" name="psc9" id= <?php "echo $row[17];" ?> /> </p> <p>Pin <label for="textfield"></label> <input name="textfield" type="password" id="textfield" size="20" maxlength="5" /> </p></td> </tr> <tr> <td colspan="6"><p> <input type="submit" name="button" id="button" value="Submit" /> </p> <p> </p></td> </tr> </table> </form> And this is the code from the processing file <?php // query.php require_once 'login.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); $pin =$_POST['pin']; $qid =$_POST['qid']; $oc1 =$_POST['oc1']; $oc2 =$_POST['oc2']; $oc3 =$_POST['oc3']; $oc4 =$_POST['oc4']; $oc5 =$_POST['oc5']; $oc6 =$_POST['oc6']; $oc7 =$_POST['oc7']; $psc1 =$_POST['psc1']; $psc2 =$_POST['psc2']; $psc3 =$_POST['psc3']; $psc4 =$_POST['psc4']; $psc4 =$_POST['psc5']; $psc6 =$_POST['psc6']; $psc7 =$_POST['psc7']; $psc8 =$_POST['psc8']; $psc9 =$_POST['psc9']; $ans =$_POST['ans']; $enter_sql= "INSERT INTO tblresults (pin,qid,oc1,oc2,oc3,oc4,oc5,oc6,oc7,psc1,psc2,psc3,psc4,psc5,psc6,psc7,psc8,psc9) VALUES ('$pin','$qid','$oc1','$oc2','$oc3','$oc4','$oc5','$oc6','$oc7','$psc1','$psc2','$psc3','$psc4','$psc5','$psc6','$psc7','$psc8','$psc9')"; $enter_query =mysql_query($enter_sql) or die (mysql_error()); ?> <body> <p>Thank you - You have successfully entered your answer</p> I'm sorry to be back so soon, but I'm up against another mystery. I'm using the code below to enter a bunch of css data from a spreadsheet into a mysql table. I think the data file is OK. The array created by the script checks out with print_r. (There are many more records than shown. I truncated it to save space.) The problem is that I get this error regarding my sql statement, not the data or anything else: Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'check, name, phone, email, entry_fee, print_fee, image_name, description, med...' at line 1 in /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php:242 Stack trace: #0 /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php(242): PDO->prepare('INSERT INTO tbl...') #1 {main} thrown in /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php on line 242 I've typed it in a dozen times to make sure there are no errors and keep getting the same error. I tried running a test file and gradually increasing the number of placeholders and at some point I always end up getting the same error, I can delete the most recent addition and it works again. Then I can add another placeholder exactly as before and it works the second time. It feels like a ghost in the machine. Any idea what I am doing wrong? An I typing something I don't see?
<?php
require '__classes/Db.php';
$csvData = '1,FALSE,Carol Lettko,,,TRUE,FALSE,Carol_Lettko-DSC_3022.jpg,Baby Herons/Brickyard,photo,,,
,,,925-285-0320,cjl164@aol.com,,,Carol_Lettko-DSC_0164.JPG,Heron/Brickyard,photo,,,
,,,,,,,Carol_Lettko-IMG_5723.jpg,Kayaker/Brickyard,photo,,,
,,,,,,,,,,,,
2,FALSE,Louise Williams,,,TRUE,FALSE,Louise_Williams-BirdsOfAFeatherAOPR.jpg,Alligator with Words,Book Excerpt,,,
,,,510-232-9547,lkw@louisekwilliams.com,,,Louise_Williams-Hope-TheFairyChickenAOPR.jpg,Hope The Fairy Chicken,,,,
,,,The d exatrfrfvct/.*tygrvurr,,,,,,,,,
,,,,,,,,,,,,
3,TRUE,Dorothy Leeland,,lelanddorothy@gmail.com,TRUE,FALSE,DJ_Lee-bridge at dusk 700px width.jpg,Bridge,photo,,,
,,,,,,,DJ_Lee-friends 700px width.jpg,Friends,photo,,,
,,,,,,,DJ_Lee-hybiscus 700 px wide.jpg,Hibiscus,photo,,,
,,,,,,,,,,,,
4,FALSE,Rita Gardner,,,TRUE,FALSE,Rita_Gardner-Explosion - Gardner photo.JPG,Explosion,photo,,,
,,,,tropicrita@msn.com,,,Rita_Gardner-Ferry Point tables and chair - Gardner.JPG,Ferry Point Tables,photo,, ,
,,,,,,,Rita_Gardner-Forks - Gardner photo.JPG,Forks,photo,,,
,,,,,,,,,,,,
';
$lines = explode(PHP_EOL, $csvData);
$array1 = array();
foreach ($lines as $line) {
$array1[] = str_getcsv($line);
}
$stmt = $pdo->prepare("INSERT INTO tbl_person_data (number, check, name, phone, email, entry_fee, print_fee, image_name, description, medium, select, orient, site) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)");
foreach ($array1 as $row)
{
$stmt->execute('$row');
}
echo '<pre>';
print_r($array1);
echo '</pre>';
?>
Hello, I'm having trouble inserting data into a MySQL table. The user has a form which is HEIGHT and WIDTH and NUMBER_OF_OBSERVATIONS. All these values are stored in the same table. NUMBER_OF_OBSERVATIONS does what it needs to and inserts its calculated results into the database, as does DATE_TIME. I've been trying for a couple of days to get this done, and I am having no luck in getting it sorted. Any help is greatly appreciated! Generator.php Code: [Select] $WIDTH = $_POST['WIDTH']; $HEIGHT = $_POST['HEIGHT']; $NUMBER_OF_OBSERVATIONS = $_POST['NUMBER_OF_OBSERVATIONS']; $db1 = new Number_Information(); $db1->openDB(); $OBSERVATION_ID = $db1->insert_Observation(); echo "<br /><br />Success. ID: <strong>$OBSERVATION_ID<strong>"; $db1->closeDB(); Number_Information.php Code: [Select] function insert_Observation() { //$Date_Now = datetime(); $sql = "INSERT INTO Observations (DATE_TIME, HEIGHT, WIDTH) VALUES (NOW(), '{$esc_HEIGHT}','{$esc_WIDTH}' )"; $result = mysql_query($sql, $this->conn); if (!$result) { die("SQL Insertion error: " . mysql_error()); } else { return mysql_insert_id($this->conn); } The table name is Observations Again - Any help is greatly appreciated! Hello, me again. I have created a little forum - very very basic. main_forum.php : Code: [Select] <?php $host="localhost"; // Host name $username="yvonnedp"; // Mysql username $password="yvonne"; // Mysql password $db_name="forum"; // Database name $tbl_name="forum_question"; // Table name // Connect to server and select databse. mysql_connect('localhost', 'yvonnedp', 'yvonne')or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql="SELECT * FROM $tbl_name ORDER BY id"; // OREDER BY id DESC is order result by descending $result=mysql_query($sql); ?> <table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"> <tr> <td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td> <td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td> <td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td> <td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td> <td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ // Start looping table row ?> <tr> <td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td> <td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td> <td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td> <td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td> <td align="center" bgcolor="#FFFFFF"><? echo $rows['datetime']; ?></td> </tr> <?php // Exit looping and close connection } mysql_close(); ?> create_topic.php : Code: [Select] <table width="400" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form id="form1" name="form1" method="post" action="add_topic.php"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3" bgcolor="#E6E6E6"><strong>Create New Topic</strong> </td> </tr> <tr> <td width="14%"><strong>Topic</strong></td> <td width="2%">:</td> <td width="84%"><input name="topic" type="text" id="topic" size="50" /></td> </tr> <tr> <td valign="top"><strong>Detail</strong></td> <td valign="top">:</td> <td><textarea name="detail" cols="50" rows="3" id="detail"></textarea></td> </tr> <tr> <td><strong>Name</strong></td> <td>:</td> <td><input name="name" type="text" id="name" size="50" /></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Submit" /> <input type="reset" name="Submit2" value="Reset" /></td> </tr> </table> </td> </form> </tr> </table> add_topic.php : Code: [Select] <?php $host="localhost"; // Host name $username="yvonnedp"; // Mysql username $password="yvonne"; // Mysql password $db_name="forum"; // Database name $tbl_name="forum_question"; // Table name // Connect to server and select database. mysql_connect('localhost', 'yvonnedp', 'yvonne')or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // get data that sent from form $topic=$_POST['topic']; $detail=$_POST['detail']; $name=$_POST['name']; $datetime=date("d/m/y h:i:s"); //create date time $sql="INSERT INTO $tbl_name(topic, detail, name, datetime)VALUES('$topic', '$detail', '$name', '$datetime')"; $result=mysql_query($sql); if($result){ echo "Successful<BR>"; echo "<a href=main_forum.php>View your topic</a>"; $email_from = "forums@thenewme.co.za"; $email_to = "info@thenewme.co.za"; $email_subject = "The New Me - New Forum Topic"; $email_message = "A new topic has been posted on the Forum!\n"; $email_message .= "Check it out : http://www.thenewme.co.za/forum/main_forum.php"; $headers = 'From: '.$email_from."\r\n"; 'Reply-To: '.$email_from."\r\n" . 'X-Mailer: PHP/' . phpversion(); @mail($email_to, $email_subject, $email_message, $headers); } else { echo "ERROR"; } mysql_close(); ?> add_answer.php : Code: [Select] <?php $host="localhost"; // Host name $username="yvonnedp"; // Mysql username $password="yvonne"; // Mysql password $db_name="forum"; // Database name $tbl_name="forum_answer"; // Table name // Connect to server and select databse. mysql_connect('localhost', 'yvonnedp', 'yvonne')or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // Get value of id that sent from hidden field $id=$_POST['id']; // Find highest answer number. $sql="SELECT MAX(a_id) AS Maxa_id FROM $tbl_name WHERE question_id='$id'"; $result=mysql_query($sql); $rows=mysql_fetch_array($result); // add + 1 to highest answer number and keep it in variable name "$Max_id". if there no answer yet set it = 1 if ($rows) { $Max_id = $rows['Maxa_id']+1; } else { $Max_id = 1; } // get values that sent from form $a_name=$_POST['a_name']; $a_answer=$_POST['a_answer']; $datetime=date("d/m/y H:i:s"); // create date and time // Insert answer $sql2="INSERT INTO $tbl_name(question_id, a_id, a_name, a_answer, a_datetime)VALUES('$id', '$Max_id', '$a_name', '$a_answer', '$datetime')"; $result2=mysql_query($sql2); if($result2){ echo "Successful<BR>"; echo "<a href='view_topic.php?id=".$id."'>View your answer</a>"; // If added new answer, add value +1 in reply column $tbl_name2="forum_question"; $sql3="UPDATE $tbl_name2 SET reply='$Max_id' WHERE id='$id'"; $result3=mysql_query($sql3); $email_from = "forums@thenewme.co.za"; $email_to = "info@thenewme.co.za"; $email_subject = "The New Me - New Forum Answer"; $email_message = "A new answer to a forum topic has been posted on the Forum!\n"; $email_message .= "Check it out : http://www.thenewme.co.za/forum/main_forum.php"; $headers = 'From: '.$email_from."\r\n"; 'Reply-To: '.$email_from."\r\n" . 'X-Mailer: PHP/' . phpversion(); @mail($email_to, $email_subject, $email_message, $headers); } else { echo "ERROR"; } mysql_close(); ?> view_topic : Code: [Select] <?php $tbl_name2="forum_answer"; // Switch to table "forum_answer" $sql2="SELECT * FROM $tbl_name2 WHERE question_id='$id'"; $result2=mysql_query($sql2); while($rows=mysql_fetch_array($result2)){ ?> <table width="400" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <td><table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td bgcolor="#F8F7F1"><strong>ID</strong></td> <td bgcolor="#F8F7F1">:</td> <td bgcolor="#F8F7F1"><? echo $rows['a_id']; ?></td> </tr> <tr> <td width="18%" bgcolor="#F8F7F1"><strong>Name</strong></td> <td width="5%" bgcolor="#F8F7F1">:</td> <td width="77%" bgcolor="#F8F7F1"><? echo $rows['a_name']; ?></td> </tr> <tr> <td bgcolor="#F8F7F1"><strong>Answer</strong></td> <td bgcolor="#F8F7F1">:</td> <td bgcolor="#F8F7F1"><? echo $rows['a_answer']; ?></td> </tr> <tr> <td bgcolor="#F8F7F1"><strong>Date/Time</strong></td> <td bgcolor="#F8F7F1">:</td> <td bgcolor="#F8F7F1"><? echo $rows['a_datetime']; ?></td> </tr> </table></td> </tr> </table><br> <? } $sql3="SELECT view FROM $tbl_name WHERE id='$id'"; $result3=mysql_query($sql3); $rows=mysql_fetch_array($result3); $view=$rows['view']; // if have no counter value set counter = 1 if(empty($view)){ $view=1; $sql4="INSERT INTO $tbl_name(view) VALUES('$view') WHERE id='$id'"; $result4=mysql_query($sql4); } // count more value $addview=$view+1; $sql5="update $tbl_name set view='$addview' WHERE id='$id'"; $result5=mysql_query($sql5); mysql_close(); ?> the problem is. With add_answer, I am not able to insert an answer. This has worked for the very first question, but all subsequent questions, I am unable to add answers. It keeps telling me ERROR. This is where the live forum is : http://www.thenewme.co.za/forum/main_forum.php Can it be an issue with the field that gets autmatically incremented? i have echo'd the data which get entered in when someone types in an answer, and everything is perfect! Can anyone help me? Hi, I'm very new to these forums and php alike. I've been working on this bit of code for weeks now. I've written numerous versions and made endless revisions; trying pieces from many similiar source and well, it's still not quite working out for me. The form I've been trying to finish will have multiple rows. The user will input values specific to each row, but I'm trying to make it so that each row is synonymous with each other in regard to the database table. Like an order form, an inventory form, or a roster. I'd really like to make this code work out because I can think of numerous applications for it's use. I'm sure others could use it for even more. So, as you see; the html code represents the table that contains the form. Currently, I only have five of these rows in the form, but would like for anybody using the code to be able to add in code for as many rows in the table as needed. Code: [Select] <html> <head> <title></title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> </head> <body> <FORM action="post_multiple.php" method="POST"> <table width="80%" border="1"> <tr> <td>Vendor</td> <td>Item Description</td> <td>Quantity</td> <td>Price</td> </tr> <tr> <td><select name="vendor" id="vendor" > <option value="other1">other1</option> <option value="other2">other1</option> <option value="other3">other3</option> <option value="other4">other4</option> <option value="other5">other5</option> </select></td> <td><input name="description" type="text" id="description" size="50" /></td> <td><input name="quantity" type="text" id="quantity" size="50" /></td> <td><input name="price" type="text" id="price" size="50" /></td> </tr> <tr> <td><select name="vendor" id="vendor"> <option value="other1">other1</option> <option value="other2">other1</option> <option value="other3">other3</option> <option value="other4">other4</option> <option value="other5">other5</option> </select></td> <td><input name="description" type="text" id="description" size="50" /></td> <td><input name="quantity" type="text" id="quantity" size="50" /></td> <td><input name="price" type="text" id="price" size="50" /></td> </tr> <tr> <td><select name="vendor" id="vendor"> <option value="other1">other1</option> <option value="other2">other1</option> <option value="other3">other3</option> <option value="other4">other4</option> <option value="other5">other5</option> </select></td> <td><input name="description" type="text" id="description" size="50" /></td> <td><input name="quantity" type="text" id="quantity" size="50" /></td> <td><input name="price" type="text" id="price" size="50" /></td> </tr> <tr> <td><select name="vendor" id="vendor"> <option value="other1">other1</option> <option value="other2">other1</option> <option value="other3">other3</option> <option value="other4">other4</option> <option value="other5">other5</option> </select></td> <td><input name="description" type="text" id="description" size="50" /></td> <td><input name="quantity" type="text" id="quantity" size="50" /></td> <td><input name="price" type="text" id="price" size="50" /></td> </tr> <tr> <td><select name="vendor" id="vendor"> <option value="other1">other1</option> <option value="other2">other1</option> <option value="other3">other3</option> <option value="other4">other4</option> <option value="other5">other5</option> </select></td> <td><input name="description" type="text" id="description" size="50" /></td> <td><input name="quantity" type="text" id="quantity" size="50" /></td> <td><input name="price" type="text" id="price" size="50" /></td> </tr> <tr> <td><select name="vendor" id="vendor"> <option value="other1">other1</option> <option value="other2">other1</option> <option value="other3">other3</option> <option value="other4">other4</option> <option value="other5">other5</option> </select></td> <td><input name="description" type="text" id="description" size="50" /></td> <td><input name="quantity" type="text" id="quantity" size="50" /></td> <td><input name="price" type="text" id="price" size="50" /></td> </tr> </table> <input type="submit" name="Submit" value="Submit" /> </FORM> </body> </html> In the php script I've tried using the count, loop, if, several others functions to try to get it to insert all the completed rows from the form into the table but just can't seem to get it to work out. Anyways, here's a representation of what i'd like to achieve. Not necessarily the code I've written trying to accomplish this, I'd have to upload a book if that were the case. <?php ini_set('display_errors','On'); error_reporting(E_ALL); $host="localhost"; // Host name $username="root"; // Mysql username $password="password"; // Mysql password $db_name="testing"; // Database name $tbl_name="test_multiple"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); //declarations $vendor = $_POST['vendor']; $description = $_POST['description']; $quantity = $_POST['quantity']; $price = $_POST['price']; // Check if button name "Submit" is active, do this $sql1="INSERT INTO $tbl_name (vendor, description, quantity, price) VALUES ('".$vendor."', '".$description."', '".$quantity."', '".$price."')"; $result1=mysql_query($sql1); mysql_close(); ?> If anybody can suggest where I need to go with that, that would be great. Like I said above, I've tried about everything I could find any reason to from forum threads I had found in google searches and whatnot. Also, if there is a good article, link or tutorial that would address this that anybody knows about that would be great. However, I'm not looking for a link to a product or plugin that will do this stuff for me. I'm trying to get halfway decent at php and would like to have working code for myself to build upon, modify, and learn from. Any help is greatly appreciated!!!! I want to take data from one table and insert it into another in the same database, the problem is the two tables have different values so it wouldn't be as simple as using an INSERT INTO script. From table 1 I want to extract a row ID as well as a field called systems. I then want to insert that into the second table which is structured, Id (null) Name (games) Value (the data given from Systems in the other) ID2 (the ID from the other table) Category (news) Any ideas how I can go about this? No idea what I am doing here. I have a joined table which is:
$sql = "SELECT itemnum, image1, title, close, quantity, bidquantity, ". I need to loop through the table on buser and insert select data into two other tables. One of the tables is the control table which will only have one row per buser. The other table I guess you would call the data table of the two and may have more than one row depending. The catch is these will be linked together by an invoice number. Table A, the control table has a field for invoice number, table B does not. Of course once the data is inserted into the new tables they will be joined and from there on it should be easy. I just don't know now how to get there. I've thought about using an array to loop through, but I don't know enough about them to make any sense of this. Not even sure the group by may help me in doing this. Anyone have any ideas on this? I sure do need it and thanks in advance!
<?php
$servername = "xxx";
$username = "xxxx";
$password = "xxxx";
$dbname = "web216-admin-6d5";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$class_id = $_POST['class_id']; // This value is from a dropdown
$checkbox = $_POST['checkbox']; // This is a checkbox next to each record
$user_id = $_POST['user_id']; // this is a hidden value output from the database (in a textbox)
if(isset($_POST['submit']))
{
for($i=0;$i<count($checkbox);$i++){
$query="INSERT INTO Class_List(user_id,class_id)VALUES('".$user_id[$i]."','".$class_id."')";
if (mysqli_query($conn, $query)) {
echo "The selected records have been added successfully !";
echo "<meta http-equiv=\"refresh\" content=\"2;url=search_courses.php\"/>";
} else {
echo "Error: " . $sql . ":-" . mysqli_error($conn);
}
mysqli_close($conn);
}
?>
Hi, I am trying to figure out how to insert the following values into another table if they have a checked box:
$class_id = $_POST['class_id']; // This value is from a dropdown on the top of the page basically, each row in the table has a checkbox If the checkbox is checked then it should insert the user_id field from the checked rows and the value from a drop down (class_id) at the top of a page into another table. I have managed the code up to here but I need a tutorial or guidance to do the rest. At the moment I am getting error HTTP 500 I understand that the code is subject to sql injection, but will sort that out later Edited August 11, 2020 by PythonHelpI am trying to insert a new user into my database from my php code. This is the error message that I am getting from the webpage: Quote Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'order, previousOrder) VALUES ('c_s@gmail.com','test','3','callulm','Smith','17' at line 1 This is the code that I am using: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("deliverpizza", $con); $sql="INSERT INTO customer(userName, password, privilege, firstName, lastName, address, postCode, order, previousOrder) VALUES ('$_POST[username]','$_POST[password]','$_POST[privilege]','$_POST[firstname]','$_POST[lastname]','$_POST[address]','$_POST[postcode]','$_POST[order]','$_POST[previousOrder]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> Hello there! I've been banging my head on this for a while and I just can't seem to get it to work properly. I have a dropdown menu which selects information from table1 using a select statement (this table is called 'lid'). It selects the firstname, lastname and member id from this table and shows it in the dropdown menu. I'm glad I got that part working but the hard thing is inserting the data that the user selects into another table. So when you select the id member from this dropdown menu it only inserts a blank row into table2 (which is called 'teamlid'). Can you guys help me? How can I insert the id member into my table2? What am I doing wrong here? Thanks a million! This is my first post so if I'm doing anything wrong, let me know and I'll fix it asap! My code:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Boast & Drive</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
<style type="text/css">
.wrapper{
width: 650px;
margin: 0 auto;
}
.page-header h2{
margin-top: 0;
}
table tr td:last-child a{
margin-right: 15px;
}
</style>
</head>
<body>
<div class="container-fluid">
<div class="row">
<div class="col-md-12">
<div class="page-header clearfix">
<h2 class="pull-left">Teamleden</h2>
<div class="btn-toolbar">
<a href="read.php" class="btn btn-primary btn-lg pull-right">Terug</a>
</div>
</div>
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
error_reporting(E_ALL);
//Verbinding maken met de database
require_once "login.php";
$sql = "SELECT tl.teamnaam,
tl.tl_ID,
tl.lidnummer,
l.voornaam,
l.achternaam
FROM teamlid tl
JOIN lid l ON tl.lidnummer = l.lidnummer
ORDER BY tl.teamnaam;";
if($result = mysqli_query($conn, $sql)) {
if(mysqli_num_rows($result) > 0) {
echo "<table class='table table-bordered table-striped'>";
echo "<thead>";
echo "<tr>";
echo "<th>Teamnaam</th>";
echo "<th>Tl_ID</th>";
echo "<th>Lidnummer</th>";
echo "<th>Voornaam</th>";
echo "<th>Achternaam</th>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['teamnaam'] . "</td>";
echo "<td>" . $row['tl_ID'] . "</td>";
echo "<td>" . $row['lidnummer'] . "</td>";
echo "<td>" . $row['voornaam'] . "</td>";
echo "<td>" . $row['achternaam'] . "</td>";
echo "<td>";
echo "<a href='update.php?id=". $row['lidnummer'] ."' title='Gegevens wijzigen' data- toggle='tooltip'><span class='glyphicon glyphicon-pencil'></span></a>";
echo "<a href='delete.php?id=". $row['lidnummer'] ."' title='Lid verwijderen' data- toggle='tooltip'><span class='glyphicon glyphicon-trash'></span></a>";
echo "</td>";
echo "</tr>";
}
echo "</tbody>";
echo "</table>";
mysqli_free_result($result);
} else{
echo "<p class='lead'><em>Er zijn geen gegevens om weer te geven.</em></p>";
}
} else{
echo "De volgende fout is gevonden: " . mysqli_error($conn);
}
?>
<form name="dropdown" method="post">
<div class="page-header clearfix">
<h2 class="pull-left">Teamlid toevoegen</h2>
</div>
<p>Selecteer hieronder met behulp van het dropdown menu een lid welke je aan bovenstaand team wilt toevoegen</p>
<div class="container-fluid">
<div class="row">
<?php
// Variabelen aanmaken en tonen met lege waardes
$teamnaam = $lidnummer = '';
// Code voor dropdown. Selecteert voornaam, achternaam en lidnummer van tabel lid)
$sql = "SELECT voornaam, achternaam, lidnummer FROM lid ORDER BY achternaam";
$result = mysqli_query($conn, $sql);
echo "<select id='teamLid' name='teamLid'>";
echo "<option>--Selecteer Lid--</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['lid'] . "'>" . $row['voornaam'] . " " . $row['achternaam'] . " " . $row['lidnummer'] . "</option>";
}
echo "</select>";
if
(isset($_POST["id"]) && !empty($_POST["id"])) {
$id = $_POST["teamLid"];
$stmt = $conn->prepare("INSERT INTO teamlid (teamnaam, lidnummer) VALUES (?,?)");
$stmt->bind_param('si', $param_teamnaam, $param_lidnummer);
$param_teamnaam = $teamnaam;
$param_lidnummer = $lidnummer;
$stmt->execute();
}
// Verbinding sluiten
mysqli_close($conn);
?>
<div>
<input type="hidden" name="id" value="<?php echo $id; ?>" />
<input type="submit" name="submit" class="btn btn-primary" value="Toevoegen">
</div>
</div>
</div>
</form>
</div>
</div>
</body>
</html>
Hi and thank you in advance for helping me. Here is the code fragment: $sql="SELECT UserID,UserName FROM `UserValidation` WHERE UserID='".$_SESSION["UserID"]."'"; $rs=CustomQuery($sql); $RecData=db_fetch_array($rs); global $conn,$strUserValidation; $strSQLSave = "INSERT INTO RecruiterNames (UserID, UserName) values ("; $strSQLSave .= $RecData["UserID"].","; $strSQLSave .= chr(34).$RecData['UserName'].chr(34); $strSQLSave .= ")"; db_exec($strSQLSave,$conn); Here is the error I receive: php error happened Technical information Error type 256 Error description You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '"")' at line 1 URL www.dealwithaces.com/ACESDB/register.php? Error file /home/debra/public_html/dealwithaces.com/ACESDB/include/dbconnection.php Error line 36 SQL query INSERT INTO RecruiterNames (UserID, UserName) values (,"") More info Call stack File: line Function Arguments #0. include/dbconnection.php:36 db_query 1. INSERT INTO RecruiterNames (UserID, UserName) values (,""); 2. Resource id #10; #1. include/dbconnection.php:47 db_exec 1. INSERT INTO RecruiterNames (UserID, UserName) values (,""); 2. Resource id #10; #2. include/events.php:25 AfterSuccessfulRegistration 1. Array ( [UserName] => bsbnick [Password] => kaos55 [UserEmail] => mike@world-class-multimedia.com [GroupID] => 1 [UserActive] => 0 ) ; #3. register.php:298 Global scope N/A I need two fields from UserValidation table (UserID and UserName) to be inserted into table RecruiterNames (UserID and UserName), but ONLY when GroupID = 3 Thank you again for your help. Mike i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
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