I am trying to debug a script that I did not write. I have little experience with PHP. The script is working fine but the proper error message is not displayed when I create an error situation. It looks like the author had the same problem and took the easy way out. This snippet establishes the variable value for $serror. It seems like the <br>'s do not belong there but maybe that is permissible. }else if($submit2) { if(!ereg("^[A-Za-z0-9_]{1,16}$",$susername)) { $serror="Invalid username! Use no more than 15 characters and only letters, numbers, and underscores.<br>"; } $this->c=@mysql_query("select username from users where username='$susername'"); $this->d=mysql_fetch_object($this->c); if(is_object($this->d)) { $serror="Username is already in use<br>"; } $this->c=@mysql_query("select username from pending where username='$susername'"); $this->d=mysql_fetch_object($this->c); if(is_object($this->d)) { $serror="Username is already in use<br>"; } echo "<h3>Signup for a New Account Step 3</h3>"; if (!$serror) { PHP: The next snippet is where the variable $serror should be diplayed. Please note that I added the "Hello World" part of this snippet to trace the code. I thought a variable could only be displayed with the echo tag. At any rate this isn't working because what you get on all errors is: $serror Hello World ?> <form method=post> <?=$table2?> <tr align=center> <td colspan=3>$serror <font color=red>Hello World</font><p></td> </tr> PHP: Any insight appreciated. Colbyt