haakuturi Posted July 7, 2007 Share Posted July 7, 2007 hey people - first post and I'm making a fool of myself already lol here's my code: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>CDDB@HOME :: Add New Artist</title> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1" /> </head> <body> <?php if (isset($_POST['ArtistName'])): // A new author has been entered // using the form below. $dbcnx = @mysql_connect('xxxxx', 'xxxxx', 'xxxxx'); if (!$dbcnx) { exit('<p>Unable to connect to the ' . 'database server at this time.</p>'); } if (!@mysql_select_db('cddb')) { exit('<p>Unable to locate the ' . 'database at this time.</p>'); } $name = $_POST['ArtistName']; $sql = "INSERT INTO TblArtist SET ArtistName='$ArtistName'"; if (@mysql_query($sql)) { echo '<p>New artist added</p>'; } else { echo '<p>Error adding new artist: ' . mysql_error() . '</p>'; } ?> <p><a href="<?php echo $_SERVER['PHP_SELF']; ?>">Add another artist</a></p> <p><a href="manageartists.php">Return to artist list</a></p> <?php else: // Allow the user to enter a new author ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <p>Enter the new artist:</p> <label>Name: <input type="text" name="ArtistName" /></label><br /> <input type="submit" value="SUBMIT" /> </form> <?php endif; ?> </body> </html> Now what's happening is that when I add a new artist, I'm told that it was successfully added, but when I view the artists, there's a blank space! I haven't got a screenshot of it, but here's the general idea: Manage Artists Sarah McLachlan Edit Delete Placebo Edit Delete Deep Forest Edit Delete Edit Delete Enigma Edit Delete Natalie Merchant Edit Delete Edit Delete Edit Delete Any clues? Please? I just know it's something really really obvious that I just can't see Link to comment https://forums.phpfreaks.com/topic/58823-solved-im-being-stupid-and-ive-overlooked-something/ Share on other sites More sharing options...
sasa Posted July 7, 2007 Share Posted July 7, 2007 change $name = $_POST['ArtistName']; $sql = "INSERT INTO TblArtist SET ArtistName='$ArtistName'"; to $name = $_POST['ArtistName']; $sql = "INSERT INTO TblArtist SET ArtistName='$name'"; Link to comment https://forums.phpfreaks.com/topic/58823-solved-im-being-stupid-and-ive-overlooked-something/#findComment-291826 Share on other sites More sharing options...
haakuturi Posted July 8, 2007 Author Share Posted July 8, 2007 lol thanks Link to comment https://forums.phpfreaks.com/topic/58823-solved-im-being-stupid-and-ive-overlooked-something/#findComment-292390 Share on other sites More sharing options...
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