//Check user is logged in in first place if(!isset($_SESSION['user'])) { header("location:******"); }
//Stores mysql login details $host="localhost"; // Host name $sql_username="*****"; // Mysql username $sql_password="*****"; // Mysql password $db_name="*****"; // Database name $tbl_name="admin"; // Table name
//Query to retrieve information $query = "SELECT * FROM admin"; $result = mysql_query($query,$con);
echo "<table border='1'>"; echo "<td><th>Username</th></tr>"; echo '<form action="user_delete.php" method="post">'; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array($result)) { // Print out the contents of each row into a table echo "<tr><td>"; echo $row['id']; echo "</td></tr>"; echo "<tr><td>"; echo $row['username']; echo "</td><td>"; echo '<input type="button" value="" />'; echo "</tr></td>"; }
echo "</form></table>"; ?>
But for some reason it returns an error on the line:
PHP Code:
while($row = mysql_fetch_array($result))
Quote:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/vmrgjdq/public_html/test/users.php on line 29
Can anyone help?
Thanks.
__________________
Last edited by murdocsvan; 05-13-2009 at 01:23 PM.