Hi,
     Thanks for reply.
        I have wrapped your code in <?php    ?> like so:-
<?php
if !isset($title) {
  // Title contains no valid data
  $title = ""
}

if !isset($firstname) {
  // Title contains no valid data
  $firstname = ""
}

if !isset($lastname) {
  // Title contains no valid data
  $lastname = ""
}
>
?>

called it p-6-2.php and this is the error I get:-
Parse error: parse error, unexpected '!', expecting '(' in D:\web\myforms\p-6-2.php on 
line 4

I have used my own form which uses javascript to validate the fname and surname 
fields, but it still outputs the same error messages!
------------------------------------------------------------------------
Notice: Undefined variable: title in D:\web\myforms\p-6-2.php on line 11

Title is required.
Notice: Undefined variable: fname in D:\web\myforms\p-6-2.php on line 17

First name is required.
Notice: Undefined variable: surname in D:\web\myforms\p-6-2.php on line 23

Last name is required.

Please use your browser's back button to return to the form, correct the errors, and 
re-submit the form.

-----------------------------------------------------
If I send the output of my form to phpinfo() , I get all my form fields listed in ' 
PHP Variables ' in ' get ' and ' post ' and the data entered into my form is there 
also.
In php.ini, globles is set off, like so:- register_globals=Off
Like I say, I am new to php, so I haven't a clue what to do!

Thanks
jh


--
PHP Windows Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

Reply via email to