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Question

PHP if-statement using $_POST variable doesn't seem to work. Why?

Answer 1

+3 A:

You probably don't have register_globals set. This is depreciated and will be removed in 6.x. So for good programming you should instead of $fname try $_POST['fname'] on your second page.

Pete 2009-11-22 02:59:24

Answer 2

+5 A:

On your second page, instead of checking for $fname, check for $_POST['fname'] instead. I think that is all you are missing.

darthnosaj 2009-11-22 02:59:40

Thank you!!! It works now!!!

brilliant 2009-11-22 03:53:18

You're very welcome. I hope your future php endeavors go well!

darthnosaj 2009-11-22 03:58:02

Answer 3

+1 A:

It might help to set the post values as variables and work with that. Something like this:

foreach($_POST as $key => $value) { $$key = $value; }

Then whatever is posted will be available rather than using $_POST['xxxxx'] in your logic.

chris 2009-11-22 03:36:03

This is an even worse idea than register_globals, as it can overwrite any variables you have used in your script previously. And you'll still get loads of warnings for variables that aren't passed in.

bobince 2009-11-22 04:07:01

Yes, it can overwrite variables previously set, but you can do that anywhere. Naming is one of the most difficult aspects of programming! However, I would rather not be repeating myself with $_POST['key'] throughout my logic.

chris 2009-12-22 18:43:45

Answer 4

+1 A:

pass.php needs to look like this

<html>
<head>
<title>Secon Page</title>
</head>
<body>
<?php
if ($_POST['fname'] =="Jack")
  echo "You are Jack!";
else
  echo "You are not Jack!";
?>
</body>
</html>

Ben Shelock 2009-11-22 03:41:27

Thank you very much.

brilliant 2009-11-22 14:34:45

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PHP if-statement using $_POST variable doesn't seem to work. Why?

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