dalty52 Posted July 18, 2006 Share Posted July 18, 2006 Anyone have any ideas why the DELETE query at the bottom of this script is not working? I just want to delete one row in the database that corresponds to the $id variable. Thanks![code]<?php require ('../mysqlconnect.php');$id = $_GET['id'];function form() { global $id; $q = "SELECT * FROM gigs WHERE id=$id"; $result = @mysql_query ($q); if ($result) { while ($part = mysql_fetch_array ($result, MYSQL_ASSOC)) { print '<h1>Do you want to delete the gig: ' . $part['location'] . '?</h1>'; print '<form method="post" action="' . $_SERVER['PHP_SELF'] . '">'; print '<input type="radio" name="delete" value="yes"/>Yes'; print '<input type="hidden" name="check" value="1">'; print '<input type="submit" name="submit" value="Submit">'; }}else { print "could not retrieve data"; } }if ($_POST['check']) { process(); }else { form(); } function process() { global $id; if ($_POST['delete']) { $id = $_GET['id']; $q = "DELETE FROM gigs WHERE id=$id"; $result = @mysql_query($q); if ($result) { print 'Deleted!<br><a href="addgig.php">Return</a>'; } else { print 'Could not connect to the database.'; $e = mysql_error(); print "$e"; } }else { print 'Not Deleted';} }?>[/code] Quote Link to comment Share on other sites More sharing options...
AndyB Posted July 18, 2006 Share Posted July 18, 2006 No, but you could find out by changing this:[code]$result = @mysql_query($q);[/code]to this:[code]$result = mysql_query($q) or die("Error ". mysql_error(). " with query ". $query);[/code] Quote Link to comment Share on other sites More sharing options...
dalty52 Posted July 18, 2006 Author Share Posted July 18, 2006 The error returned this: 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 '' at line 1The version on the server is 3.23.49, and I checked the manual but could not find any answers. I'm sure it's something simple I am overlooking. Thanks for the help! Quote Link to comment Share on other sites More sharing options...
dalty52 Posted July 18, 2006 Author Share Posted July 18, 2006 Ok, figured it out. I was sending the form, so the $id variable lost it's value. I created a hidden form input that spit out the id value and then posted it into the function I was having problems with. Worked like a charm. Thanks for the help. Quote Link to comment Share on other sites More sharing options...
fenway Posted July 18, 2006 Share Posted July 18, 2006 It doesn't look like you're validating that UID at all... does that mean with one line of JS I could delete a random record in this table? Quote Link to comment Share on other sites More sharing options...
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