The error complains about an unexpected "!", right ?
What if you do
if (!isset($title)) {
That's the correct syntax, you know ?...
Luis
----- Original Message -----
From: "jh" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, July 09, 2003 9:03 AM
Subject: [PHP-WIN] Re: Forms-PHP-and-errors
Hi,
Thanks for reply.
I have wrapped your code in <?php ?> like so:-
<?php
if !isset($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
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