PHP Error : Undefined Index ($_POST Variable)

Question

I'm trying to do a PHP connection script, with form and data processing in the same page, but when I try my code I always get 2 errors :

Notice: Undefined index: password in C:\Users\Simon\Documents\Site FTM\connexion.php on line >7

Fatal error: Call to a member function fetch() on a non-object in C:\Users\Simon\Documents\Site FTM\connexion.php on line 18

I understand that PHP don't find the index called 'password' in $_POST, but I'm sure that the form is leading to the same page (#)... Can somebosy help me ? Here is the whole code, on PasteBin : http://pastebin.com/A9z1HHvf

Sorry, the error messages are in french, but all the other variable names are in english. You just need to know that $errDisplay is use to store all my custom error messages and display them at the bottom ofmy page.

And what does your code look like? Especially line 18...

Broxzier
– Broxzier

2013-09-29 17:36:01 +00:00
Commented Sep 29, 2013 at 17:36 — Broxzier
– Broxzier, Commented Sep 29, 2013 at 17:36

raina77ow · Accepted Answer · 2013-09-29 17:19:44Z

This line is problematic:

$isFormOK = isset($_POST['username']) AND isset($_POST['password']);

AND has lower precedence than =, that's why this line reads like this:

( $isFormOK = isset($_POST['username']) )
  AND isset($_POST['password']);

In other words, whether or not $_POST['password'] is in the script, it works the same way - setting $isFormOk value based only on $_POST['username'] existence.

What you should do instead is either use && form of boolean and (it has higher precedence than =):

$isFormOK = isset($_POST['username']) && isset($_POST['password']);

... or, even better, just use the fact that isset can take more than one argument:

$isFormOK = isset($_POST['username'], $_POST['password']);

Quoting the doc:

If multiple parameters are supplied then isset() will return TRUE only if all of the parameters are set. Evaluation goes from left to right and stops as soon as an unset variable is encountered.

Moeed Farooqui · Accepted Answer · 2013-09-29 17:06:38Z

-1

if(isset($_REQUEST['password'])){
$pass = $_REQUEST['password'];
}

Now You can use $pass as it contains the values which is being posted.

answered Sep 29, 2013 at 17:06

Moeed Farooqui

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1 Comment

Adam Hopkinson Over a year ago

The OP's code already references $_POST['password'] - all $_REQUEST does is take both GET and POST

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