skipskap Posted April 25, 2007 Share Posted April 25, 2007 Hey all, I have been trying to get this to work all day ! Basically, I have a database called pink, table pink, with 3 fields - userID, name, colour. I have a 'remove_user' page that will show a drop down box and contain all the users and their userID number, but my code won't work! Here it is: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Delete Customer</title> </head> <body> <?php // connect to database files include 'config.php'; include 'connect.php'; if(isset($_POST['add'])) { $delete=$_POST["delete"]; // Remove customer } else { ?> <form method="post"> <table width="500" border="2" cellspacing="0" cellpadding="0"> <tr> <td colspan="2">Please fill out the form below:</td> </tr> <tr> <td width="198">Name:</td> <td width="294"> <?php $sql = "SELECT * FROM pink "; $result = mysql_query ( $sql, $conn, $result ); while ( $row = mysql_fetch_array ( $result ) ) { echo "<option value=\"{$row['userID']}\">{$row['name']}</option>"; } ?> </td> </tr> <tr> <td>Submit:</td> <td> <input name="add" type="submit" id="add" value="Delete User"> </td> </tr> </table> </form> <?php // close database } include 'close.php'; echo "Record Deleted"; ?> </body> </html> Any ideas anyone? Link to comment https://forums.phpfreaks.com/topic/48627-reading-fields-from-database-into-a-drop-down-box/ Share on other sites More sharing options...
fenway Posted April 26, 2007 Share Posted April 26, 2007 Why the 3rd parameter to mysql_query? Link to comment https://forums.phpfreaks.com/topic/48627-reading-fields-from-database-into-a-drop-down-box/#findComment-238649 Share on other sites More sharing options...
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