jwwceo Posted August 27, 2006 Share Posted August 27, 2006 I just posted a few minutes ago about a GET problem...It turns out that is wasn't a problem, just a warning...It is actually working fine...but my code is still not working...the following snippet is supposed to update or delete a row in a MySQL database....when I press either submit or delete I get the correct response, that is, is says "your shirt has been deleted", but nothing actually happens in the database...any ideas....Thanks!!!<?phpinclude 'library/config.php';include 'library/opendb.php';error_reporting(0);$shirt_id = $_GET['shirt_id'];if(isset($_POST['delete'])) {$query = "DELETE from shirts WHERE shirt_id='$shirt_id'";mysql_query($query) or die("Error: ". mysql_error(). " with query ". $query); echo "Your shirt has been deleted!";}elseif(isset($_POST['submit'])) { $name=$_POST['name']; $comment=$_POST['comment']; $link=$_POST['link']; $site=$_POST['site']; $image=$_POST['image']; $active=$_POST['active']; $commission=$_POST['commission']; $query="UPDATE shirts SET name='$name', comment='$comment', link='$link', site ='$site', image ='$image', active ='$active', commission ='$commission' WHERE shirt_id ='$shirt_id'"; mysql_query($query) or die("Error: ". mysql_error(). " with query ". $query); echo "Your shirt has been updated!";} else{$data = mysql_query("SELECT * FROM shirts WHERE shirt_id ='$shirt_id'") or die(mysql_error());$info = mysql_fetch_array( $data );echo "<a href='admin.php'><img src='images/logo.gif' border='0'></a><br><form method='post' action='edit.php'><TABLE><TR> <TD>Name:</TD> <TD><INPUT TYPE='TEXT' NAME='name' size=50 value='".$info['name']."'></TD></TR><TR> <TD>Comment</TD> <TD><textarea rows='5' cols='50' name='comment'>".$info['comment']."</textarea></TD></TR><TR> <TD>Link</TD> <TD><INPUT TYPE='TEXT' NAME='link' size=50 value='".$info['link']."'></TD></TR><TR> <TD>Site</TD> <TD><INPUT TYPE='TEXT' NAME='site' size=50 value='".$info['site']."'></TD></TR><TR> <TD>Image</TD> <TD><INPUT TYPE='TEXT' NAME='image' size=50 value='".$info['image']."'></TD></TR><TR> <TD>Active</TD>"; if($info['image'] == 'true') { echo "<TD><input type='radio' name='active' value='true' CHECKED> Active<br><input type='radio' name='active' value='false'> Inactive </TD>"; } else { echo "<TD><input type='radio' name='active' value='true' > Active<br><input type='radio' name='active' value='false' CHECKED> Inactive </TD>"; }echo"</TR><TR> <TD>Commission</TD> <TD><INPUT TYPE='TEXT' NAME='commission' size=50 value='".$info['commission']."'></TD></TR><TR> <TD></TD> <TD><INPUT TYPE='submit' name='submit' value='Edit this Tee'> <INPUT TYPE='submit' name='delete' value=' Delete this Tee'></TD></TR></TABLE></form>";}?></body></html> Link to comment https://forums.phpfreaks.com/topic/18834-new-problem/ Share on other sites More sharing options...
Orio Posted August 27, 2006 Share Posted August 27, 2006 Try:[code]<?phpinclude 'library/config.php';include 'library/opendb.php';@$shirt_id = $_GET['shirt_id'];if(isset($_POST['delete'])) {$query = "DELETE from shirts WHERE shirt_id='$shirt_id'";mysql_query($query) or die("Error: ". mysql_error(). " with query ". $query); echo "Your shirt has been deleted!";}?>[/code]And see if you get an error.Orio. Link to comment https://forums.phpfreaks.com/topic/18834-new-problem/#findComment-81275 Share on other sites More sharing options...
litebearer Posted August 27, 2006 Share Posted August 27, 2006 Perhaps I am missing the boat (I think it sailed 50 plus years ago), but nowhere in your example have you set a value to shirt_id.Lite... Link to comment https://forums.phpfreaks.com/topic/18834-new-problem/#findComment-81344 Share on other sites More sharing options...
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