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PHP - Using A Variable To Name A Database
Hi,
I'm trying to use a variable- "$newdbname" in a script that sets up a new database. I'm using this variable because I have an include file that sets all login info and the database name as variables. here's the code. or at lest the relevant part. <?php require 'dbinfo.inc.php'; $db = mysql_connect($servname,$dbusername,$dbpassword) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db($database, $db) or die(mysql_error($db)); // create the user table $query = 'CREATE TABLE ($newdbname)( user_id int(30) NOT NULL AUTO_INCREMENT, username varchar(20) NOT NULL, password varchar(41) NOT NULL, PRIMARY KEY (user_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); if I run the script like this, I get the following 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 '($newdbname)( user_id int(30) NOT NULL AUTO_INCREMENT, usernam' at line 1 if I run the code as this, without parentheses around the variable, it creates a table named "$newdbname" <?php require 'dbinfo.inc.php'; $db = mysql_connect($servname,$dbusername,$dbpassword) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db($database, $db) or die(mysql_error($db)); // create the user table $query = 'CREATE TABLE $newdbname ( user_id int(30) NOT NULL AUTO_INCREMENT, username varchar(20) NOT NULL, password varchar(41) NOT NULL, PRIMARY KEY (user_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); what am I doing wrong? Similar TutorialsQuote i need to store a variable from database like if i have "copies" in one of my column in my database then i have to store a particular value for copies store it to $copies here i want that i can store value of copies into $copies $update_book="update book set copies=copies-1 where bookid='$bookid'"; $result=mysql_query($update_book,$linkID1); if($result) { print "<html><body background=\"header.jpg\"> <p>book successfully subtracted from database</p></body></html>"; } else { print "<html><body background=\"header.jpg\"> <p>problem occured</p></body></html>"; } } I'm stumped on what's causing this problem. $descriptor2 is entering "0". But $_POST['States'] does have a value. I put $testable in as a variable to test that in place of descriptor2 and it works just fine. Code: [Select] $descriptor1 = isset($_GET['id']) ? (int)$_GET['id'] : null; $descriptor2 = isset($_POST['States']) ? (int)$_POST['state'] : null; $descriptor3 = isset($_POST['continent_regions']) ? (int)$_POST['continent_regions'] : null; $descriptor4 = isset($_POST['Continents']) ? (int)$_POST['Continents'] : null; $descriptor5 = isset($_POST['Country']) ? (int)$_POST['Country'] : null; $testable = $_POST['States']; // my test variable $query = "INSERT INTO plant_locations_link(plant_id,state,continent_regions,continents,country) VALUES ($descriptor1,$descriptor2,$descriptor3,$descriptor4,$descriptor5)"; // if I change descriptor2 to testable, it inputs the number I need. This is driving me crazy .... Need to use a variable for the database name on this code ... Code: [Select] <!DOCTYPE html> <html> <head> <title>Table Definition's Tool</title> <style type="text/css"> th { font-size: 110%; border-bottom: 2px solid black; } td { padding: 3px; border-bottom: 1px solid #aaa } </style> </head> <body> <h1>ITG's table's definition</h1> <table> <?php error_reporting (E_ALL ^ E_NOTICE); //Variables $instance=$_REQUEST['instance']; $database_name= $_REQUEST['database_name']; $table=$_REQUEST['table']; require 'utils.php'; // Connect via Windows authentication $server = $instance; $connectionInfo = array( 'CharacterSet' => 'UTF-8' ); $db = sqlsrv_connect($server, $connectionInfo); if ($db === false) { exitWithSQLError('Database connection failed'); } /* Set up and execute the query. */ $query = "SELECT COLUMN_NAME, ORDINAL_POSITION, COLUMN_DEFAULT, DATA_TYPE, CHARACTER_MAXIMUM_LENGTH FROM $database_name.INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME='$table'"; // Run query $qresult = sqlsrv_query($db, $query); if ($qresult === false) { exitWithSQLError('Query of product data failed.'); } echo '<tr><th>COLUMN NAME</th><th>POSITION</th><th>DEFAULT</th><th>TYPE</th><th>LENGHT</th></tr>'; // Retrieve individual rows from the result while ($row = sqlsrv_fetch_array($qresult)) { echo '<tr><td>', htmlspecialchars($row['COLUMN_NAME']), '</td><td>', htmlspecialchars($row['ORDINAL_POSITION']), '</td><td>', htmlspecialchars($row['COLUMN_DEFAULT']), '</td><td>', htmlspecialchars($row['DATA_TYPE']), '</td><td>', htmlspecialchars($row['CHARACTER_MAXIMUM_LENGTH']), "</td></tr>\n"; } // null == no further rows, false == error if ($row === false) { exitWithSQLError('Retrieving schema failed.'); } // Share Release result liststatement resource and close connection sqlsrv_free_stmt($qresult); sqlsrv_close($db); ?> </table> </body> </html> Is a simple code where the main MS-SQL query is Code: [Select] SELECT COLUMN_NAME, ORDINAL_POSITION, COLUMN_DEFAULT, DATA_TYPE, CHARACTER_MAXIMUM_LENGTH FROM $database_name.INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME='$table' The user should be able to introduce the database and table's name and above query will display the table's schema. However, does not work. I am able to use a variable for the table, but the program does nothing when I use $database_name inside the query. What am I doing wrong? Help, please .... I am trying to take a variable from a database. But it keeps equaling nothing. Ive tried using mysql_error() and sI get nothing. I also keep tryed mysql_num_rows() and I get 3(just what Id exspect). Here is the code. Code: [Select] <?php if($userb) { $q1 = " SELECT sub FROM login_info WHERE user='$user_log' LIMIT 1 "; $s1 = mysql_query($q1); $scribe = mysql_fetch_assoc($s1); $sub = explode('![sep]!' ,$scribe['sub']); foreach($sub as $key => $name) { $q2 = " SELECT * FROM subscribe WHERE user = '$name' "; echo 'Llama'; $s2 = mysql_query($q2); $num1 = mysql_num_rows($s2); echo '<h3>' .$name .' Posted ' .$num1 .' Storie(s)! </h3>'; while($stor = mysql_fetch_assoc($s2)); { $story = $stor['stor']; $sid = $stor['sid']; $q3 = " SELECT * FROM story_info WHERE id = $sid "; echo $q3 .'<br />'; $s3 = mysql_query($q3) or die(mysql_error()); $rows = mysql_fetch_assoc($s3); $viewsdb = $rows['views']; $titledb = $rows['title']; $userdb = $rows['user']; $catdb = $rows['cat']; $ratdb = $rows['rating']; $id_db = $rows['story_id']; $sumdb = shorten($rows['sum']); echo "<h3><a href='?p=page&id=$id_db'> $titledb </a> </h3>"; echo "<div id='fun_info'>"; echo "$sumdb <br />"; echo "By <a href='?p=profile&user=$userdb'> $userdb </a> <br /> "; echo "$viewsdb Views | Rated: $ratdb | Catagory: <a href='?p=cat_view&gen=$catdb'> $catdb </a> </div>"; } } } else { login('?p=sub'); } Hey all Im working on an assignment for school and currently I am trying to inser the variable $uid which currently = 2.. But for someone reason when the post happens it inserts a 0 instead of a 2. Here is my insert Code: [Select] mysql_query( "INSERT INTO blog_posts (title, post, author_id, date_posted) ". "VALUES ('$btitle', '$bpost', '$uid', CURDATE())" ); I have a table that has 5 columns Quote player_id fname lname team I'm trying to get all the values from that table using this sql command Code: [Select] include_once('../database_connection.php'); $player_id = addslashes($_REQUEST['edit_player']); $sql_query = "Select * from nba_info where player_id = '$player_id'"; $result = MYSQL_QUERY($sql_query); "edit_player" above is came from a different page. They I'm fetching the data using while Code: [Select] while($row = mysql_fetch_array($result)) { $row['player_id']; $row['fname']; $row['lname']; $row['team']; $row['email']; } Then I'm trying to pass that values to a variable here Code: [Select] $id = $row['player_id']; $fname = $row['fname']; $fname = $row['lname']; $lname = $row['team']; $email = $row['email']; When I'm trying to put that variables in a textbox value, they are not showing up Here Code: [Select] <table> <form name="edit_player_form" method="post" action=""> <tr> <td>Players ID</td> <td><input type="text" name="playerid" size="20" value="<?php echo $id; ?>" /></td> </tr> <tr> <td>First Name</td> <td><input type="text" name="fname" size="20" value="<?php echo $fname; ?>" /></td> </tr> <tr> <td>Last Name</td> <td><input type="text" name="lname" size="20" value="<?php echo $lname; ?>" /></td> </tr> <tr> <td>Team</td> <td><input type="text" name="team" size="20" value="<?php echo $team; ?>" /></td> </tr> <tr> <td>Email</td> <td><input type="text" name="email" size="20" value="<?php echo $email; ?>" /></td> </tr> <tr> <td><div align="center"> <input type="submit" name="Submit" value="Edit This Player"> </div></td> </tr> </form> </table> I wonder why it ain't showing up on the textbox value? tried almost everything... Anyone? Hello Guys, I have a column named message_status it can only have the values "read" or "unread" in it. I would like to show an image in the column next to it. If the message is "read" then I would like to display a green dot, and if it is unread I would like to display a red dot. I would also like to be able to have anything in the row that's "unread" to appear in bold. I completely stumped by this! I have written the following code, which doesn't seem to work, presumably because it is run after the table has been displayed? I am very new to this so please be gentle! Regards, AJLX $result = mysql_query("SELECT * FROM messages where username ='$username'ORDER BY {$_GET['orderbycol']} $sort"); echo "<table border='1'> <tr width='200'> <th>View</th> // code to display table removed </tr>"; $green_dot = $message_status; $red_dot = $message_status; $green = ""; $red = ""; $bold = ""; while ($row = mysql_fetch_array($result)) { echo "<tr>"; $id = $row['ID']; echo "<td><a href='/view_message.php/?view_message=$id'>View</a></td>"; echo "<td>" .$green . $red."</td>"; echo "<td>" .$bold . $message_status = $row['message_status'] . "</td>"; echo "<td>" .$bold. $row['date_sent'] . "</td>"; echo "<td>" .$bold. $row['contact_name'] . "</td>"; echo "<td>" .$bold. $row['subject'] . "</td>"; echo "</tr>"; } if ($green_dot = 'read'){ $green ='<img src=\"assets/red.jpg\"/>'; echo $green; } else { $red ='<img src=\"assets/red.jpg\"/>'; $bold = "<b>"; echo $red; Quote i want to store a value from a database and use it as variable in php code can anyone help me out in this code i want to store value of copies in a php variable and want that it should be more than 0 (zero) for furthur calculations $update_book="update book set copies=copies-1 where bookid='$bookid'"; $result=mysql_query($update_book,$linkID1); if($result) { print "<html><body background=\"header.jpg\"> <p>book successfully subtracted from database</p></body></html>"; } else { print "<html><body background=\"header.jpg\"> <p>problem occured</p></body></html>"; } Hello Guys, I want to insert an array variable into the database e.g Code: [Select] foreach($html->find('div[class=PostContent]') as $element) { echo $element; $sq = "INSERT INTO articles(original_text) VALUES ('$element') WHERE article_link='$item_url'"; $result = mysql_query($sql1) or die('Query failed: ' . mysql_error()); } i want to insert the variable $element into the database but i'm not able to do so for some reason! $element contains only a paragraph of text. how can i insert the variable $element? Hello all! I am somewhat new to PHP and SQL, but I think I have a basic understanding of how the language works. I have read through a million different websites and forums, and have tried a million different ways but am just not getting anywhere! This is what I am trying to do... I have a text link that I want to click through to an outside URL from my website. This outside URL was already entered into an SQL database, and I just cant seem to get the text link to bring up the outside URL from the database. This is what I have so far... 1.) My webpage that I am working on is www.mywebsite.com/articles.php 2.) I want the text "Read more..." on this webpage to click through to an outside URL, like "http://www.google.com", which was already entered into an SQL database under the title of "article_link" in a new window 3.) So I would like for "Read more..." to click through the outside URL that was entered into "article_link" in the database Code: [Select] <?php echo <p><a href="\' . $article_link . '\" target="_blank">Read more...</a></p>; ?> Nothing I have tried has worked so far, nothing clicks through to the outside URL in the "article_link" entry from the database. Sorry if I am being redundant, I have been at this all day! Please help and thank you in advance!! Hello guys,
I do not know if anyone work or have worked with the "elFinder (file manager)" .. I incorporate "elFinder" to my platform .. I have the following question .. Have its connector:
<?php
error_reporting(0); // Set E_ALL for debuging
include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderConnector.class.php';
include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinder.class.php';
include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeDriver.class.php';
include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeLocalFileSystem.class.php';
// Required for MySQL storage connector
include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeMySQL.class.php';
// Required for FTP connector support
include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeFTP.class.php';
/**
* Simple function to demonstrate how to control file access using "accessControl" callback.
* This method will disable accessing files/folders starting from '.' (dot)
*
* @param string $attr attribute name (read|write|locked|hidden)
* @param string $path file path relative to volume root directory started with directory separator
* @return bool|null
**/
function access($attr, $path, $data, $volume) {
return strpos(basename($path), '.') === 0 // if file/folder begins with '.' (dot)
? !($attr == 'read' || $attr == 'write') // set read+write to false, other (locked+hidden) set to true
: null; // else elFinder decide it itself
}
$opts = array(
// 'debug' => true,
'roots' => array(
array(
'driver' => 'LocalFileSystem', // driver for accessing file system (REQUIRED)
'path' => '/home/', // path to files (REQUIRED)
'URL' => dirname($_SERVER['PHP_SELF']) . '/home/', // URL to files (REQUIRED)
'accessControl' => 'access' // disable and hide dot starting files (OPTIONAL)
)
)
);
// run elFinder
$connector = new elFinderConnector(new elFinder($opts));
$connector->run();
I want him to make the call the "$screen" (folder will be created automatically by the user) of each user .. and not the "/home/" (as an example of put) .. the "$screen" comes from my platform, and I add the "elFinder" to it ..Imagine that each user creates "1/2/3 or 4, etc .." folders .. Hello all, I can't get the data store in the $ART_ID variable to pass into the database. The original $Artisan variable is set up like this: 1. Artisan Name. So the explode is taking just the number. If I put an echo after the explode and the $ART_ID variable it outputs the correct information but it doesn't store in the database as that data. The query is in correct order too. Thanks in advance. $CType_Type = $_REQUEST["CTYPE_Type"]; $Artisan = $_REQUEST['Artisan']; $Quantity = $_POST['Quantity']; $HAnswer1 = $_POST["HAnswer1"]; $HAnswer2 = $_POST["HAnswer2"]; $HAnswer3 = $_POST["HAnswer3"]; $HAnswer4 = $_POST["HAnswer4"]; $break = explode(".", $Artisan); $ART_ID = $break[0]; if(!$Quantity) { die('Quantity field is empty. Please enter the quantity of handicrafts made.'); } else { $ctypeQuery = mysql_query("SELECT CTYPE_ID FROM CraftType WHERE CTYPE_Type = '".$CType_Type."'"); while($row = mysql_fetch_array($ctypeQuery)) { $CTYPE_ID = $row["CTYPE_ID"]; $sql = ("INSERT INTO Handicraft VALUES (`HANDI_ID`, '".$Quantity."', 'NULL', '".$CTYPE_ID."', '".$ART_ID."', 'NULL', '1')"); if(!mysql_query($sql)) { die('Error inserting Handicraft Type into table: ' . mysql_error()); } else { -----data in this section doesn't affect the rest of the code---- } } } Hello Everyone, I am pretty new to the forums and was curious if i could get some help here. Basically, in a nutshell, i have PayPal integrated into my website. I will use this to collect money from clients. when a client logs into his/her account they see their balance (which is pulled from the database to correspond with the user that's logged-in). Now, everytime a payment is submitted a notify_url is contacted after payment has been verified, that notify_url is the code written below. What I am trying to execute here is when this notify_url is called the current balance is reduced from the amount paid through paypal. In the second If condition, you will see that the word success is being entered into the paypal.txt file, which is working perfectly fine. Now, you will also see the variable $update_balance; which is suppose to update the original balance with the balance paid through PayPal BUT IT'S NOT!! WHY?? LOL Thank You in advance! <?php ob_start(); session_start(); include_once ('/home/rdewebde/public_html/includes/paypal.php'); $myPaypal = new Paypal(); $myPaypal->ipnLog = TRUE; include_once "/home/rdewebde/public_html/includes/_config.php"; $username = "".$_SESSION['username'].""; $users_data = mysql_query("SELECT * FROM `members` WHERE `username`='".$username."'"); $user_info = mysql_fetch_array($users_data); $current_amount = $user_info['balance']; $deduct_amount = $myPaypal->ipnData['payment_gross']; $new_amount = $current_amount - $deduct_amount; $update_balance = mysql_query("UPDATE `members` SET `balance` = '$new_amount' WHERE `username` = '".$username."'"); if ($myPaypal->validateIpn()) { if ($myPaypal->ipnData['payment_status'] == 'Completed') { $update_balance; file_put_contents('/home/rdewebde/public_html/lounge/paypal.txt', 'SUCCESS'); } else { file_put_contents('/home/rdewebde/public_html/lounge/paypal.txt', "FAILURE\n\n" . $myPaypal->ipnData); } } ?> Hi all, Thanks for reading. I'm hella frustrated at this script I wrote: for some reason, it will not work correctly. Basically, it works. The first 4 names in the table on the database show up when searched. But, anything past these four names in the database will not show up as a result when searched! I'm pulling my hair out here! It's really simple - take a gander: Code: [Select] if (isset($_POST['submit'])) { $search = $_POST['search']; $searchQuery = mysql_query("SELECT * FROM Accounts WHERE FullName='$search'"); if (mysql_num_rows($searchQuery) == 0) { $result = "Your search returned no results. Please try again."; } else { $results = 1; while ($getSearchResults = mysql_fetch_array($searchQuery)) { $fullName = $getSearchResults['FullName']; $result = "Name: ".$fullName.""; } } } ?> ...and the HTML form... Code: [Select] <form action="search.php" method="post"> <p>Search: <input type="text" name="search" size="35" maxlength="100" /></p> <p><input type="submit" value="Search" name="submit" /></p> <?php echo $result; ?> </form> Does anyone have any ideas? I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here. It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere. <?PHP require('connect.php'); ?> <form action ='' method='post'> <select name="id"> <?php $extract = mysql_query("SELECT * FROM cars"); while($row=mysql_fetch_assoc($extract)){ $id = $row['id']; $make= $row['make']; $model= $row['model']; $year= $row['year']; $color= $row['color']; echo "<option value=$id>$color $year $make $model</option> ";}?> </select> Which attribute would you like to change?<br /> <input type="radio" name="getchanged" value="make"/>Make<br /> <input type="radio" name="getchanged" value="model"/>Model<br /> <input type="radio" name="getchanged" value="year" />Year<br /> <input type="radio" name="getchanged" value="color" />Color<br /><br /> <br /><input type='text' value='' name='tochange'> <input type='submit' value='Change' name='submit'> </form> //This is where I need help... <?PHP if(isset($_POST['submit'])&&($_POST['tochange'])){ mysql_query(" UPDATE cars SET '$_POST[getchanged]'='$_POST[tochange]' where id = '$_POST[id]' ");}?> Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment. My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted." The program should not accept any numbers greater than 100 or any characters. Once I do this, I must take a second number and do a similar thing. Finally, I must have a statement show up at the bottom stating which number is greater. Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters. Any help you can provide will be greatly appreciated! I have just re-installed Xampp and suddenly my sites are now displaying lots of: Notice: Use of undefined constant name - assumed 'name' in ... Notice: Use of undefined constant price - assumed 'price' in ... this is an example of the line its refering too: $defineProducts[1001] = array(name=>'This is a product', price=>123); My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user and $priv are undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working. The code I'm using:
<?php
session_start();
function verify() {
//verify that the user is logged in via the login page. Session_start has already been called.
if (!isset($_SESSION['loggedin'])) {
header('Location: /index.html');
exit;
}
//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users.
$user === $_SESSION['name'];
//Connect to Databse
$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
//SQL Statement to lookup privlege information.
if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
//LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
while ($row = $result->fetch_assoc()) {
$priv === $row["su"];
}
}
// close SQL connection.
mysqli_close($link);
// Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators.
if ($priv !== 1) {
echo $_SESSION['name'];
echo "you have privlege level of $priv";
echo "<br>";
echo 'Your account does not have the privleges necessary to view this page';
exit;
}
}
verify();
?>
after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks |
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