Gem Posted May 17, 2010 Share Posted May 17, 2010 Sorry guys - This is driving me insane! Can anyone tell me why this is not working?? <?php include ("includes/dbconnect120-gem.php"); if(isset($_POST['submit1'])) { $sqlalbum="INSERT INTO album (album_name) VALUES ('$newalbumname')"; mysql_query($sqlalbum) or die ('Error, album_query failed : ' . mysql_error()); } ?> <!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=utf-8" /> <title>Untitled Document</title> </head> <body> <form name="newalbum" method="post" action=""> <input name="newalbumname" type="text" size="30" maxlength="30" value="<?php echo $newalbumname; ?>" /> <input type="submit" name="submit1" value="Create Album" /> </form> </body> </html> Table info (album) album_id int(3) album_name varchar(255) latin1_swedish_ci upload_cat_id int(3) I dont get any errors, but it doesnt work either. It creates a new row - but all it contains is the id ... I dont understand why the album_name isnt coming up in the db ?! Hope you can help Thanks Gem Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted May 17, 2010 Share Posted May 17, 2010 Where in your code, before it is being used in the INSERT query, are you setting $newalbumname from the corresponding $_POST variable? Quote Link to comment Share on other sites More sharing options...
Gem Posted May 17, 2010 Author Share Posted May 17, 2010 Not sure what you mean buddy - but thats the whole code ^ I thought ( ) that $newalbumname would be taken from the form? Quote Link to comment Share on other sites More sharing options...
Gem Posted May 17, 2010 Author Share Posted May 17, 2010 Oh - im an idiot. $sqlalbum="INSERT INTO album (album_name) VALUES ('$_POST[newalbumname]')"; Fixed it - thats anyway :S Quote Link to comment Share on other sites More sharing options...
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