PHP - Create Table Using Id As Name
I have a table in my database for users. On the registration page I want to create another table with the id of the user as the table name.
$sql = "CREATE TABLE IF NOT EXISTS `id_prod` ( ) how do i modify this line so that it takes the id from the user table and creates a new table with id as name followed by prod. Similar TutorialsFor the life of me I can not figure out why this is not creating the table in the database. Second set of eyes would be great.
function install_kudos() { global $wpdb; $table_name = $wpdb->prefix . 'acikudosnew'; if ($wpdb->get_var('SHOW TABLES LIKE ' .$table_name) != $table_name) { $sql = "CREATE TABLE $table_name ( kudoid int(9) NOT NULL AUTO_INCREMENT, kudomsg text NOT NULL, kudoagent text NON NULL, kudocust text NOT NULL, kudoacct int(16) NOT NULL, kudoclient varchar(100) NOT NULL, kudoloc text NOT NULL, kudoentry TIMESTAMP DEFAULT CURRENT_TIMESTAMP, kudoadmin text NOT NULL, kudopic varchar(55) DEFAULT '' NOT NULL, PRIMARY KEY (kudoid) )"; require_once( ABSPATH . 'wp-admin/includes/upgrade.php' ); dbDelta( $sql ); } } register_activation_hook(__FILE__, 'install_kudos');
I want to show a table with a max of three columns. I am using an if statement to count the rows and then start a new row (after 3 cols), I cant see my error. The output just gice one col with all the data in it. Any help appreciated. Code: [Select] <form action="<?php $_SERVER['PHP_SELF'] ?>" method="POST" id="cover" enctype="multipart/form-data"> <table width="700"><?php /* display picture and radio button */ //add counter for more than 3 images $i=0; $size=3; while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $coverID=$row['coverID']; echo"<tr>"; echo "<td>"; echo"<img src=\"/wordpress_3/wp-content/plugins/Authors2/jackets/{$row['pix']}\" />"; echo "<input type = 'radio' name ='cover' value= '$coverID'/>"; ?> <br /> <input type="submit" value="submit"> <?php echo"</td>"; $i++; if($i==$size) { echo "</tr><tr>"; $i=0; } } ?></table> </form><?php Hello, please let me know how to write xml file, i read somewhere we can use psql.....anyone knows please help! Thanks Code: [Select] <? $link = mysql_connect($dbhost,$dbuser,$dbpass) or die ('Could not connect: ' . mysql_error()); mysql_select_db($dbname) or die ('Error connecting to database'); $sql = "SELECT * FROM papers WHERE ctitle='May 2012'"; $result = mysql_query($sql); $num = mysql_num_rows($result); //number of entry $data = mysql_query("select * from papers where ctitle='May 2012' ORDER BY id ASC"); $Wrd = new COM("Word.Application"); $Wrd->Application->Visible = False; $DocName = "MyDoc/MyWord.doc"; $WrdDoc = $Wrd->Documents->Add(); $WTable = $WrdDoc->Tables->Add($Wrd->Selection->Range, 2, $num); // Colums, Rows while($info=mysql_fetch_array($data)) { $i=1; $WTable->Cell($i,1)->Range->Text = $info['title']; $WTable->Cell($i,2)->Range->Text = $info['name'] $info['surname'] $info['institution'] $info['country']; $i++; } $Wrd->ActiveDocument->SaveAs(realpath($DocName)); $Wrd->Application->Quit; $Wrd = null; ?> Word Created <a href="<?=$DocName?>">Click here</a> to Download. php script is not working. Can someone please help me to detect the errors? i want create a table in ms word file for all entry in my database... "2 columns, number of rows" Cell(1,1) Cell(1,2) Cell(2,1) Cell(2,2) etc... Cell(3,1) Cell(3,2) i'm trying to achive this results layout: example: catagory comment comment comment catagory catagory catagory comment what is being pulled now is if their is 2 comments two 1 catagory then 2 catagorys are being returned with the same id. what i want is to return one cataogry with many comments. thanks in advance for your help. Code: [Select] <?php require_once('Connections/Del_Comments.php'); ?> <?php $all_ids = array(); $str_ids = ""; // first parent query $maxRows_sourceType = 10; $pageNum_sourceType = 0; if (isset($_GET['pageNum_sourceType'])) { $pageNum_sourceType = $_GET['pageNum_sourceType']; } $startRow_sourceType = $pageNum_sourceType * $maxRows_sourceType; mysql_select_db($database_Del_Comments, $Del_Comments); $query_sourceType = "SELECT a.Id, a.Type, a.Dates, a.UIdFk, b.Id as Did, b.comment, b.dates as Day, b.sfk as Sfk , c.sfk as sfk1, d.Memo as memo FROM asstatusupdate as a left join asstatusdata as b on a.id = b.sfk left join asmanystatusupdate as c on b.sfk = c.sfk left join ascomments as d on d.id = c.cfk where a.uidfk='1' order by Dates asc"; $query_limit_sourceType = sprintf("%s LIMIT %d, %d", $query_sourceType, $startRow_sourceType, $maxRows_sourceType); $sourceType = mysql_query($query_limit_sourceType, $Del_Comments) or die(mysql_error()); $row_sourceType = mysql_fetch_assoc($sourceType); if (isset($_GET['totalRows_sourceType'])) { $totalRows_sourceType = $_GET['totalRows_sourceType']; } else { $all_sourceType = mysql_query($query_sourceType); $totalRows_sourceType = mysql_num_rows($all_sourceType); } $totalPages_sourceType = ceil($totalRows_sourceType/$maxRows_sourceType)-1; // add the array while ($row_source = mysql_fetch_assoc($sourceType)) { $all_ids[] = $row_source['Sfk']; $str_ids .= $row_source['Sfk'].','; } // remove the array $str_ids = (substr($str_ids,-1) == ',') ? substr($str_ids, 0, -1) : $str_ids; echo $str_ids; //echo $all_ids; //second child query $maxRows_sourceComments = 10; $pageNum_sourceComments = 0; if (isset($_GET['pageNum_sourceComments'])) { $pageNum_sourceComments = $_GET['pageNum_sourceComments']; } $startRow_sourceComments = $pageNum_sourceComments * $maxRows_sourceComments; mysql_select_db($database_Del_Comments, $Del_Comments); $query_sourceComments = "SELECT c.sfk as sfk1, d.Memo as memo FROM asmanystatusupdate as c left join ascomments as d on d.id = c.cfk where c.uidfk0='1' and c.sfk in ($str_ids)"; $query_limit_sourceComments = sprintf("%s LIMIT %d, %d", $query_sourceComments, $startRow_sourceComments, $maxRows_sourceComments); $sourceComments = mysql_query($query_limit_sourceComments, $Del_Comments) or die(mysql_error()); $row_sourceComments = mysql_fetch_assoc($sourceComments); if (isset($_GET['totalRows_sourceComments'])) { $totalRows_sourceComments = $_GET['totalRows_sourceComments']; } else { $all_sourceComments = mysql_query($query_sourceComments); $totalRows_sourceComments = mysql_num_rows($all_sourceComments); } $totalPages_sourceComments = ceil($totalRows_sourceComments/$maxRows_sourceComments)-1; $resultComments = @mysql_query($query_sourceComments, $Del_Comments); $numComments = @mysql_num_rows($resultComments); $result = @mysql_query($query_sourceType, $Del_Comments); $num = @mysql_num_rows($result); // column count for parent $thumbcols = 1; // column count for parent query $thumbrows = 1+ round($num / $thumbcols); // column count for child query $thumbrowsComments = 1+ round($numComments/$thumbcols); // header print '<br />'; print '<table align="center" width="500" border="3" cellpadding="0" cellspacing="0">'; if (!empty($num)) { print '<tr><td colspan="3" align="center"><strong>Returned Num of Record Sets: ' .$num. ' and comments count' .$numComments. '</strong></td></tr>'; } // table layout for parent query function display_table() { // global variables global $num, $result, $thumbrows, $thumbcols, $resultComments, $numComments, $thumbrowsComments ; // row count for parent for ($r=1; $r<=$thumbrows; $r++) { // print table row print '<tr>'; //format the columns for ($c=1; $c<=$thumbcols; $c++) { print '<td align="center" valign="top">'; $row = @mysql_fetch_array($result); $row1 = @mysql_fetch_array($resultComments); $Id = $row['Id']; $Type = $row['Type']; $Dates = $row['Dates']; $Comment = $row['Comment']; $Sfk1 = $row['sfk1']; $Memo = $row['memo']; // test if not empty show record sets if (!empty($Id)) { // print output from parent query // grab type dates comment and format for there column echo '<td valign="top" align="center">'; echo "$Type"; echo ' '; echo "$Dates"; echo '<br />';echo '<br />';echo '<br />'; echo "$Comment"; echo '</td>'; echo '<td>'; echo "$Id"; echo '</td>'; echo '<td>'; echo "$str_ids"; echo '</td>'; echo '<tr>'; echo '<td>'; echo "$Sfk1"; echo '</td>'; echo '<td>'; echo "$Memo"; echo '</td>'; echo '</tr>'; } // closing the $id loop else { print ' '; } print '</td>'; //closing the table data } //closing the rows print '</tr>'; } // closing the outter loop //} //closing fk id loop print '</td>'; //closing the table data } //closing the rows print '</tr>'; //} // closing the main loop // call the main table display_table() ; print '</table>'; ?> <?php var_dump(substr('a', 1)); // bool(false) ?> Had some great help yesterday, but I'm stuck using <?php ?> on every line, thanks to Joomla Anyway, I'm trying to create a table, and it doesn't seem to be working, to create it, what is the correct format? <?php <table width='400px' border ='1'>; ?> ?? output will be put out on a per line within a if statement, which works, just the framing of the table and tr and td seem to be eluding me. Thanks Im messing around with functions and arrays but cant seem to get this to work. It basically creates a simple table with the parameters you specify. The array however doesnt go into the table properly. function asf_create_table($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. Hi, I need a quick tip. I want this script to produce an output of "Table created" if it succesfully creates a table. <?php $con = mysql_connect("localhost","****","****"); if (!$con) { die('Could not connect: ' . mysql_error()); } // Create table mysql_select_db("my_db", $con); $sql = "CREATE TABLE People ( FirstName varchar(15), LastName varchar(15), Age int )"; // Execute query mysql_query($sql,$con); mysql_close($con); ?> I tried to do it on my own, but I guess I'm still completely new to php I tried: <?php $con = mysql_connect("localhost","****","****"); if (!$con) { die('Could not connect: ' . mysql_error()); } // Create table if (mysql_select_db("my_db", $con)) { echo "Table created"; } else { echo "Error creating Table: " . mysql_error(); } $sql = "CREATE TABLE People ( FirstName varchar(15), LastName varchar(15), Age int )" // Execute query mysql_query($sql,$con); mysql_close($con); ?> It gave me an error with the last two lines mysql_query($sql,$con); mysql_close($con); I removed them and it really did give the output "Table created" yet it didn't create the table in the database. I'm having a feeling that I need to add an "if" statement, something of the sort. if (sql(CREATE TABLE Persons)) But I'm also getting the feeling something there is wrong, I still haven't tried it. Help? i m tryin to create table from mysql coding and when i run the script its runs without any errors and just show that its created table but whn i chk into phpmyadmin i cnt find my table there and also when i use some copy paste from some website its just do the same thing but this time table is created but i cant find any difference between my table and the website table but still my script wont work please help me where is the mistake $connection=mysql_connect("127.0.0.1","root",""); $db=mysql_select_db("addressbook",$connection); if($db){ $SQL="CREATE TABLE address_data_new { ID int(7) NOT NULL auto_increment, First_name varchar(50) NOT NULL, ################## MY SCRIPT################ Last_name varchar(50) NOT NULL, email varchar(50) NOT NULL, PRIMARY KEY (ID), UNIQUE Id (ID) }"; /* $SQL="create table address_data { ID int(7) NOT NULL auto_increment, ############# working script############ First_Name varchar(50) NOT NULL, Surname varchar(50) NOT NULL, email varchar(50), PRIMARY KEY (ID), UNIQUE id (ID) )"; */ mysql_query($SQL); mysql_close($connection); echo "table added successfully"; } else { echo "database not found"; } ?> Hey yall! I'm working on a new site idea and I've run across a problem that I know is simple enough but I'm stumped. It's in the signup form What I want to do is insert the new user into the 'Login' table and get the user id that was just created and use that to create a table with the user id in the name. Here is what I have: // now we insert user into 'Login' table mysql_real_escape_string($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 What I mean by dynamic is say I created a function to generate a random code. I put it in a variable named $key. I make a form with a hidden input type with the name and value of $key. When the form posts, it generates the table name to whatever $key is. Because I tried something like this & its not querying. Heres my code: <?php $dbc = mysqli_connect('localhost', 'cpacrop_todo', 'pass', 'cpacrop_todo') or die('Failed to connect to database!'); function make_key($num_chars) { if ((is_numeric($num_chars)) && ($num_chars > 0) && (! is_null($num_chars))) { $key = ''; $accepted_chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890'; //seed srand(((int)((double)microtime()*1000003)) ); for ($i=0; $i<=$num_chars; $i++) { $random_number = rand(0, (strlen($accepted_chars) -1)); $key .= $accepted_chars[$random_number] ; } return $key; } } $key = make_key(6); ?> <form action="" method="post"> <input type="hidden" name="<?php echo "$key"; ?>" /> <input type="submit" name="submit" value="Submit" /> </form> <?php if ($_POST['submit'] == "Submit") { $query = "CREATE TABLE `$key` ( `id` INT AUTO_INCREMENT, `title` VARCHAR(35), `description` VARCHAR(365) );"; mysqli_query($dbc,$query) or die('Unable to query database.'); echo "Success"; } ?> If this is possible, what am I doing wrong? Thanks! Dear All, As I would like to create json as below format { "data": [ { "title": "After getting a COVID-19 Vaccine", "content": [ { "list": "You may have some side effects, which are normal signs that your body is building protection." }, { "list": "Following symptoms can take place, and it usually resolves within 2 to 3 days." }, { "list": "Common Side Effects pain, redness, swelling, warmth, nausea, muscle pain, tiredness, headache" } ] }, { "title": "After getting a COVID-19 Vaccine", "content": [ { "list": "You may have some side effects, which are normal signs that your body is building protection." }, { "list": "Following symptoms can take place, and it usually resolves within 2 to 3 days." }, { "list": "Common Side Effects pain, redness, swelling, warmth, nausea, muscle pain, tiredness, headache" } ] } ] } The information is from mysql select command: SELECT (tb_helpinfo.title_eng) as title,(tb_helpdetail.content_eng) as content FROM tb_helpinfo INNER JOIN tb_helpdetail ON tb_helpinfo.id = tb_helpdetail.helpinfo_id
This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=355885.0 Hello
I have a mysql table like this:
id | user | car 1 | 1 | fiat 500 2 | 2 | vw polo 3 | 2 | vw golf 4 | 3 | renault clio 5 | 2 | fiat panda 6 | 3 | seat ibiza From this table how can i get a query that the result be like: user 2 - 3 cars user 3 - 2 cars user 1 - 1 cars Thanks Have any of you ever run into this one. Can't quite figure this one out.
Error Code: 1064. 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 'DECLARE cUserID INT; DECLARE done INT DEFAULT FALSE; DECLARE CUR1 CURSOR FOR ' at line 9
Hello everyone! I am trying to insert a student into a table (with TIMESTAMP; works with VARCHAR, not TIMESTAMP). Can anyone help?
Variable
$time_stamp = date("D M j G:i:s T Y");Populate DB Query ("DROP TABLE IF EXISTS enrolled") || !$link->query("CREATE TABLE enrolled(course_id VARCHAR(50), student_id VARCHAR(50), user_ip VARCHAR(50), time_stamp TIMESTAMP(6))Insert Query INSERT INTO enrolled(course_id,student_id,user_ip,time_stamp) VALUES('$course','$number','$user_ip','$time_stamp') Edited by MatthewPatten, 12 December 2014 - 08:32 AM. In this output table, I would like to be able to click on any number in the ama row, add .html to that value and then execute the link. Example: If the ama number in row 3 is 890543, add .html to the number = 890543.html, then click on it to link to that web page. <?php // Connect to database ========================================= include("connect_db.php"); $table1='passwords'; $table2='airplanes'; // send query =================================================== $result = mysql_query("SELECT * FROM $table2") or die(mysql_error()); if (!$result) { die("Query to show fields from table failed"); } echo "<table border='10' cellpadding='3' cellspacing='2'>"; echo "<p>View All Airplanes</p>"; echo "<tr> <th>ID</th> <th>AMA #</th> <th>Model Name</th> <th>Model MFG</th><th>Wingspan</th><th>Engine</th><th>Decibels</th></tr>"; // keeps getting the next row until there are no more to get ================ while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table ========================== echo "<tr><td>"; echo $row['id']; echo "</td><td>"; echo $row['ama']; // Append .html to this number echo "</td><td>"; echo $row['model_name']; echo "</td><td>"; echo $row['model_mfg']; echo "</td><td>"; echo $row['wingspan']; echo "</td><td>"; echo $row['engine']; echo "</td><td>"; echo $row['decibels']; echo "</td></tr>"; } echo "</table>"; ?> How would I create a prepared mysql temp table using PHP? I currently write my prepared statements like the example below. How would I create a "TEMP" table? Thanks!
$connection8y = new mysqli("host", "xxx", "xxx", "db"); This topic has been moved to Other Libraries and Frameworks. http://www.phpfreaks.com/forums/index.php?topic=355753.0 I Have Taken The User_name In Session and i want to Make a Table as User_name as Prefix and When The User Logs in i want to show tht Table To the User !!!! |