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PHP - How Can Create Insert Function?
HI All,
I would like create query when I add some one is look what the last number in field and add one ? example : you can see the img in attach IDT is primary key + auto increment Customer_id is number field customer _name is name how can create this login IDT Customer_ID Customer_name 112 5 bbbb 113 6 ccccc 114 7 eeeee Also I need the inster function and is look the last field in customer_ID and Increase one ( +1) Any help please/ THanks Similar TutorialsHey yall! I'm working on a new site idea and I've run across a problem that I know is simple enough but I'm stumped. It's in the signup form What I want to do is insert the new user into the 'Login' table and get the user id that was just created and use that to create a table with the user id in the name. Here is what I have: // now we insert user into 'Login' table mysql_real_escape_string($insert = "INSERT INTO `Login` (`UID`, `pass`, `HR`, `mail`, `FullName`) VALUES ('{$_POST['username']}', '{$_POST['pass']}', '{$_POST['pass2']}', '{$_POST['e-mail']}', '{$_POST['FullName']}')"); mysql_query($insert) or die( 'Query string: ' . $insert . '<br />Produced an error: ' . mysql_error() . '<br />' ); $error="Thank you, you have been registered."; setcookie('Errors', $error, time()+20); // Get user ID mysql_real_escape_string($checkID = "SELECT * FROM Login WHERE `mail` = '{$_POST['e-mail']}'"); while ($checkIDdata = mysql_fetch_assoc($checkID)) { $userID = $checkIDdata; // now we create table 'Transactions" for the user mysql_real_escape_string($create = "CREATE TABLE `financewatsonn`.`Transactions_{$userID}` ( `ID` INT( 20 ) NOT NULL AUTO_INCREMENT COMMENT 'Transaction ID', `name` VARCHAR( 50 ) NOT NULL COMMENT 'Name/Location', `amount` VARCHAR( 50 ) NOT NULL COMMENT 'Amount', `date` VARCHAR( 50 ) NOT NULL COMMENT 'Date', `category` VARCHAR( 50 ) DEFAULT NULL COMMENT 'Category', `delete` INT( 1 ) NOT NULL DEFAULT '0', UNIQUE KEY `ID` ( `ID` ) ) ENGINE = MYISAM DEFAULT CHARSET = utf8 COMMENT = 'User ID {$userID}'"); mysql_query($create) or die( 'Query string: ' . $create . '<br />Produced an error: ' . mysql_error() . '<br />' ); } I know in 'Get user ID' that I need to get the ID but I'm not sure how to get that information. i get the error Quote Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource at 166 which is the while line under Get user ID Hi to everybody I need ur help because I’m trying to make a script in php to write the same data but with different date in MySQL depending on the splitting ($data06).... like a schedule...
For example if $data06 = annual and $data03 = “2019/01/01” programm must create in MySQL : 2019/01/01 2020/01/01 2021/01/01 2022/01/01 2023/01/01
if $data06=“half year” will create 10 date , increasing 6 moths ...
But I have a problem at finish of code switch ($data06) { case 'annual': $numrate = 5; $aumdata = "+12 months"; break; case 'half year': $numrate = 10; $aumdata = "+6 month"; break;
default: echo "error"; $sql = "INSERT into $nometb ( name, scadenza ) values ( '$data01', '$data03' )"; if ($conna->query($sql) === TRUE) {} else {die('ERROR'. $conna->error); }
//IMPORT NEXT AND NEW DATE $newDate = date_create($data03); for ($mul = 2; $mul <= $numrate; ++$mul) {
$datanuova = date_create($data03); $datanuova->modify($aumdata); $datanuova->format('yy/m/d'); $newDate = $datanuova->format('yy/m/d');
$sql = "INSERT into $nometb ( name, scadenza ) values ( '$data01', '$newDate' )"; if ($conna->query($sql) === TRUE) {} else {die('ERROR'. $conna->error); }
//HERE THERE IS ERROR $data03 = $newDate;
if ($conna->query($sql) === TRUE) {} else {die('ERRORE NELL\'IMPORTAZIONE'. $conna->error); } }
} }
}
How I can resolve ?
many thanks Francesco I'm trying to after submission
1. create a csv
2. insert record into db
3. send email that's created in form submission - $msg.
4. send email with attachment to only the email I specify (if possible)
if not attach it to the email that's created at submission.
I've search and search and found different methods but doesn't work with my code.
I kept 3 lines at the top but can't get them to work ... either I don't get an email after submission or don't get an attachement.
Can some one help?
<?php
$random_hash = md5(date('r', time()));
$csvString = "..."; // your entire csv as a string
$attachment = chunk_split(base64_encode($csvString));
$to = "email@email.com";
if(isset($_POST['submit']))
{
// VALIDATION
if(empty($_POST['firstName']))
{
"First Name Required";
}
if(empty($_POST['lastName']))
{
"Last Name Required";
}
if(empty($error))
{
$to = "$to";
$subject = 'The Form';
$headers = "MIME-Version: 1.0 \r\n";
$headers .= "Content-Type: text/html; \r\n" ;
$msg .="<html>
<head></head>
<body>
<table width='100%' cellspacing='0' border='0' cellpadding='0'>
<tr><td>
<table width='100%' cellspacing='0' border='0' cellpadding='0'>
<tr><td>This is the email sent.</td></tr>
</table>
</body>
</html>";
include('con.php');
$con = mysqli_connect($host,$user,$pass,$dbName);
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"thetable");
$firstName = mysqli_real_escape_string($con, $_POST['firstName']);
$lastName = mysqli_real_escape_string($con, $_POST['lastName']);
$sql = "SELECT * FROM thetable WHERE `firstName` = '{$firstName}' OR `lastName` = '{$lastName}'";
$result = mysqli_query($con,$sql);
if(($result->num_rows)>= 1)
{
$theerror = "You exist";
}
else
{
$sql="INSERT INTO thetable(firstName, lastName) VALUES ('$_POST[firstName]','$_POST[lastName]'";
$success = "Sent ... Insert it!!!";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
$result = @mail($to, $subject, $msg, $headers);
}
mysqli_close($con);
{
}
}
}
?>
Edited by barkly, 27 October 2014 - 02:59 PM. Im messing around with functions and arrays but cant seem to get this to work. It basically creates a simple table with the parameters you specify. The array however doesnt go into the table properly. function asf_create_table($rows, $cols, $border=1, $padding=5, $td_border=1, $contents) { $table = "<table style=\"border: {$border}px solid; padding:{$padding}px;\">"; for ($t_rows=0; $t_rows<$rows; $t_rows++) { $table .= "<tr>"; } for ($t_cols=0; $t_cols<$cols; $t_cols++) { for ($i=0; $i<$cols; $i++) { $table .= "<td style=\"border: {$td_border}px solid;\">"; $table .= $contents[$i]; $table .= "</td>"; } } for ($t_rows=0; $t_rows<$rows; $t_rows++) { $table .= "</tr>"; } $table .= "</table>"; echo $table; } $t_contents = array("Cell 1", "Cell 2", "Cell 3", "Cell 4"); asf_create_table("4", "4", "1", "5", "1", $t_contents); instead of 4 cells each with Cell # in them i get 16 cells with the cell #. the 4 displayed 4 times. I have a class built for an INSERT query but it is passing two sets of records into the database rather than one. Code: [Select] class DatabaseInsert { function DatabaseConnectionRequire() { include("../scrips/php/database.connection.class.php"); include("../scrips/php/database.settings.php"); include("../scrips/php/database.connection.class.invoke.php"); } function ArticleInsert($values,$fields,$table) { $values_imploded = implode(" ",$values); $fields_imploded = implode(" ",$fields); $i = "INSERT INTO $table ($fields_imploded) VALUES ($values_imploded)"; mysql_query($i) or die(mysql_error()); if (!mysql_query($i)) { echo "Sorry, something whent wrong there..."; } else { echo "<strong><p style='color:green;'>Content added sucessfully!!!</p></strong>"; } } } HI,i am using java script to create a add row function in the php .but when the first row data can insert into database ,the 2nd row data cannot insert into database ,can help me to check my coding? thanks a lot Code: [Select] <SCRIPT language="javascript"> function addRow(tableID) { var table = document.getElementById(tableID); var rowCount = table.rows.length; var row = table.insertRow(rowCount); var cell1 = row.insertCell(0); var element1 = document.createElement("input"); element1.type = "checkbox"; cell1.appendChild(element1); var cell2 = row.insertCell(1); var element2 = document.createElement("input"); element2.type = "text"; cell2.appendChild(element2); var cell3 = row.insertCell(2); var element3 = document.createElement("input"); element3.type = "text"; cell3.appendChild(element3); var cell4 = row.insertCell(3); var element4 = document.createElement("input"); element4.type = "text"; cell4.appendChild(element4); var cell5 = row.insertCell(4); var element5 = document.createElement("input"); element5.type = "text"; cell5.appendChild(element5); } function deleteRow(tableID) { try { var table = document.getElementById(tableID); var rowCount = table.rows.length; for(var i=0; i<rowCount; i++) { var row = table.rows[i]; var chkbox = row.cells[0].childNodes[0]; if(null != chkbox && true == chkbox.checked) { table.deleteRow(i); rowCount--; i--; } } } catch(e) { alert(e); } } </SCRIPT> THE PHP CODE Code: [Select] <?php require_once ('../../../Connections/admin_db.php'); mysql_select_db("admin_db"); if ((isset($_POST["Submit"])) && ($_POST["Submit"] == "Submit")) { $i=0; foreach($_POST['abc'] as $value ) { $abc = $_POST['abc'][$i]; $level = $_POST['level'][$i]; $level_desc = $_POST['level_desc'][$i]; $pc_desc = $_POST['pc_desc'][$i]; //Insert Data into Instructor Profile Info $q = "INSERT INTO plo_pc(p_name,plo_id,plo_criteria,plo_level,level_dec,plo_desc) VALUES ('$list','$plo_id','$abc','$level','$level_desc','$pc_desc') " ; mysql_query($q) or die(mysql_error()) ; $i=$i+1; } } ?> Hello all, so I created an insert function and it seems no matter what I try that it won't add values using the query function inside a table from the respective variables, I would like to know why is this happening? Here is the code can you tell me why it doesn't insert anything in the database? It shows no errors when it runs but then again when I check the tables they're empty!
function insert(){
$user = $_POST['user'];
$pass = md5($_POST['pass']);
$priv = "User";
$mail = $_POST['mail'];
$avatar = $_FILES['avatar']['name'];
$date="now()";
$submit = $_POST['submit'];
$query = "INSERT INTO user(user,pass,priv,mail,avatar,date) VALUES(`$user`,`$pass`,`$priv`,`$mail`,`$avatar`,`$date`);";
if($submit){
$res = mysqli_query($con,$query) or die(mysqli_error($con));
}
}
Looking for help writing a function that does the following: Look for http:// or https:// in a string and replaces it with an <a href> tag. For example: Here is my link http://www.mydomain.com some further text here. Becomes: Here is my link <a href="http://www.mydomain.com">http://www.mydomain.com</a> some further text here. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=308768.0 Is there any way to tell my php ajax file to run the update query if the data already exist and if not, then create the row in the database? I have both the update and the insert functions created, but was just wondering if I could tell php which one to use without passing through a parameter. I am simply trying to use stripslashes for my mysqli insert statement, and errors are driving me nuts.. I've tried several variation and pattern with apostrophes and quotes to no avail. Should I even be using stripslashes to clean my data? Or is there a better function? Notice: Use of undefined constant title - assumed 'title' in C:\wamp\www\php\simple_classifieds\add_posting.php on line 57 $query = "INSERT INTO Postings (id, city_id, title, description) VALUES ('','$_POST[city]','" . stripslashes($_POST[title]) . "','$_POST[description]')" or mysqli_error(); i have been trying to insert date with when users post there comment but when i echo the date() with the comments..it just display 0000.00.00.00 just like that and when i checked my DB it was like that too..please what can i do.this is my code Thankd in advance Code: [Select] <?php include"header.php"; if(isset($_POST['submit'])) { $postdate=mktime(0,0,0,date("m"),date("d")+1,date("y")); $comment=mysql_real_escape_string($_POST['comment']); if($comment!=='') { $ins="INSERT INTO post(post_content,post_date)VALUES('$comment','$postdate')"; mysql_query($ins) or die(mysql_error()); } else { echo"You can not post an empty page"; } } Not sure if this is right. I can't get it to insert my record. Can someone please tell me if I'm doing it right?
cus_functions.php
function dbRowInsert($table_name, $form_data)
{
global $conn;
$fields = array_keys($form_data);
$sql = "INSERT INTO ".$table_name."
(`".implode('`,`', $fields)."`)
VALUES('".implode("','", $form_data)."')";
return mysqli_query($sql);
}
process.php
include("new_db.php");
include("cus_functions.php");
$do=$_GET['do'];
if($do=='addpro'){
if(isset($_POST['submit'])){
$input = $_POST['title'];
$comp = '0';
$form_data = array('title' => $input, 'completed' => $comp);
dbRowInsert('projects', $form_data);
}}
how i can make a insert using this fuctions I m learning php, as using this functions (mysqli abstract) but after update wont work any more.
/** insert data array */
public function insert(array $arr)
{
if ($arr)
{
$q = $this->make_insert_query($arr);
$return = $this->modifying_query($q);
$this->autoreset();
return $return;
}
else
{
$this->autoreset();
return false;
}
}
complement
/** insert query constructor */
protected function make_insert_query($data)
{
$this->get_table_info();
$this->set_field_types();
if (!is_array(reset($data)))
{
$data = array($data);
}
$keys = array();
$values = array();
$keys_set = false;
foreach ($data as $data_key => $data_item)
{
$values[$data_key] = array();
$fdata = $this->parse_field_names($data);
foreach ($fdata as $key => $val)
{
if (!$keys_set)
{
if (isset($this->field_type[$key]))
{
$keys[] = '`' . $val['table'] . '`.`' . $val['field'] . '`';
}
else
{
$keys[] = '`' . $val['field'] . '`';
}
}
$values[$data_key][] = $this->escape($val['value'], $this->is_noquotes($key), $this->field_type($key), $this->is_null($key),
$this->is_bit($key));
}
$keys_set = true;
$values[$data_key] = '(' . implode(',', $values[$data_key]) . ')';
}
$ignore = $this->ignore ? ' IGNORE' : '';
$delayed = $this->delayed ? ' DELAYED' : '';
$query = 'INSERT' . $ignore . $delayed . ' INTO `' . $this->table . '` (' . implode(',', $keys) . ') VALUES ' . implode(',',
$values);
return $query;
}
before update this class i used to insert data like this
$db = Sdba::table('users');
$data = array('name'=>'adam');
$db->insert($data);
this method of insert dont works on new class. if i try like this i got empty columns and empty values.
thanks for any help complete class download http://goo.gl/GK3s4E
Here is function that backup mysql database. It's working fine, but I want to download file instead of creating it. /* backup the db OR just a table */ function backup_tables($host,$user,$pass,$name,$tables = '*') { $link = mysql_connect($host,$user,$pass); mysql_select_db($name,$link); //get all of the tables if($tables == '*') { $tables = array(); $result = mysql_query('SHOW TABLES'); while($row = mysql_fetch_row($result)) { $tables[] = $row[0]; } } else { $tables = is_array($tables) ? $tables : explode(',',$tables); } //cycle through foreach($tables as $table) { $result = mysql_query('SELECT * FROM '.$table); $num_fields = mysql_num_fields($result); $return.= 'DROP TABLE '.$table.';'; $row2 = mysql_fetch_row(mysql_query('SHOW CREATE TABLE '.$table)); $return.= "\n\n".$row2[1].";\n\n"; for ($i = 0; $i < $num_fields; $i++) { while($row = mysql_fetch_row($result)) { $return.= 'INSERT INTO '.$table.' VALUES('; for($j=0; $j<$num_fields; $j++) { $row[$j] = addslashes($row[$j]); $row[$j] = str_replace("\n","\\n",$row[$j]); if (isset($row[$j])) { $return.= '"'.$row[$j].'"' ; } else { $return.= '""'; } if ($j<($num_fields-1)) { $return.= ','; } } $return.= ");\n"; } } $return.="\n\n\n"; } //save file $date= date("d-M-Y_h-i"); $handle = fopen('Backup-'.$date.'.sql','w+'); fwrite($handle,$return); fclose($handle); } Usage: Quote backup_tables('HOST','USER','PASS','TABLE'); But this will create backup file on server. I also tried to add following code in function (at the end): echo $return; header('Content-type: text/plain'); header("Content-Disposition: attachment; filename=\"$filename\""); How to just download file, without showing it on screen instead of using fwrite/fclose function? hi there..im new to php mysql and im having trouble inserting a string data to mysql from a php date() function. here's my code: Code: [Select] $year = date('Y'); echo $year; $insertSQL = sprintf("INSERT INTO tbl_elections (election_id=$year)"); mysql_select_db($database_organizazone_db, $organizazone_db); $Result1 = mysql_query($insertSQL, $organizazone_db) or die(mysql_error()); when i try to output the $year variable on a webpage, it returns "2012" but when i try to insert this data into my database table, it returns an error like this: check the manual that corresponds to your MySQL server version for the right syntax to use near '=2012)' is there a way to convert "2012" into a normal string data type? Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1 query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '') Code: [Select] if (empty($errors)) { require_once ('dbconnectionfile.php'); $query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8', '$description9', '$description10', '$description11')"; $result = @mysql_query ($query); if ($result) { $who_donated=mysql_insert_id(); $query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why) VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')"; $result2 = @mysql_query ($query2); if ($result2) {echo "Info was added to both tables! yay!";} echo "table one filled. Table two was not."; echo $who_donated; //header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id"); exit(); } else { echo 'system error. No donation added'; Hello everyone, I am working on a form that is similar to a shopping cart system and I am thinking of creating a button that submits the checked value and saves them to a $_SESSION variable. And also a link that links to a cart.html that takes the values of a $_SESSION variable. I am have trouble figuring what tag/attribute should I use in order to achieve that.
Right now my code attached below submits the checked values to cart.html directly. However I want my submit button to save the checked box to a $_SESSION variable and STAY on the same page. And then I will implement a <a> to link to the cart.php.
I researched a little bit about this subject and I know it's somewhat related to ajax/jquery. I just wanted to know more about it from you guys. I appreciate your attention for reading the post and Thanks!
Below is the form that I currently have:
<form name= "finalForm" method="POST" action="cart.php">
<input type="Submit" name="finalSelected"/>
<?php foreach($FinalName as $key => $item) {?>
<tr>
<td><input type="checkbox" name="fSelected[]" value="<?php echo htmlspecialchars($FinalID[$key])?>" />
<?php echo "$FinalID[$key] & $item";?>
</td>
</tr>
<?php } ;?>
Below is the code for cart.php
<?php
require ('connect_db.php');
if(isset($_POST['finalSelected']))
{
if(!empty($_POST['fSelected']))
{
$chosen = $_POST['fSelected'];
foreach ($chosen as $item)
echo "aID selected: $item </br>";
$delimit = implode(", ", $chosen);
print_r($delimit);
}
}
if(isset($delimit))
{
$cartSQL = "SELECT * from article where aID in ($delimit)";
$cartQuery = mysqli_query($dbc, $cartSQL) or die (mysqli_error($dbc));
while($row = mysqli_fetch_array($cartQuery, MYSQLI_BOTH))
{
$aTitle[] = $row[ 'name' ];
}
}
?>
<table>
<?php if(isset($delimit)) {
$c=0; foreach($aTitle as $item) {?>
<tr>
<td>
<?php echo $aTitle[$c]; $c++;?>
</td>
</tr>
<?php }}?>
</table>
Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated I am trying to use text boxes to insert numbers into the database based on what is inputed. If I have a string, like this for example: $variable = 09385493; And I want to insert it into the database like this: mysql_query("INSERT INTO integers(number) VALUES ('$variable')"); When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as 9385493 Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too. Thanks |
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