I am receving an undefined error on a $_POST method [duplicate] - php

Question

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Closed 10 years ago.
Possible Duplicate:
PHP: “Notice: Undefined variable” and “Notice: Undefined index”
This is strange but I am posting an input using the $_POST method which is below:
$c = count($_POST['gridValues']);
But the problem is that I am receiving an error stating:
Notice: Undefined index: gridValues in /web/stud/..../ on line 40
(which is line above)
How come I am receiving this error because the $_POST method is definitly correct?
Below is the whole code:
<?php
ini_set('session.gc_maxlifetime',12*60*60);
ini_set('session.gc_divisor', '1');
ini_set('session.gc_probability', '1');
ini_set('session.cookie_lifetime', '0');
require_once 'init.php';
ini_set('display_errors',1);
error_reporting(E_ALL);
session_start();
?>
$i = 0;
$c = count($_POST['gridValues']);
for($i = 0; $i < $c; $i++ ){
switch ($_POST['gridValues'][$i]){
case "3":
$selected_option = "A-C";
break;
case "4":
$selected_option = "A-D";
break;
case "5":
$selected_option = "A-E";
break;
}
}

Answer 1

You need to check if it is set first:
<?php
if(isset($_POST['gridValues'])) {
$c = count($_POST['gridValues']);
}
?>

Answer 2

There are a few ways to prevent a notice level message about an undefined variable.
Put # in front of the variable. ex: $c = #$_POST['gridValues'] This can be dangerous as you are explicitly telling it to ignore any problems with the variable. See the PHP manual for more info on #: http://php.net/manual/en/language.operators.errorcontrol.php
Turn off (or change) error reporting. ex: ini_set('display_errors', 0);
Check if the variable is defined before using it. ex:
$c = '';
if(!empty($_POST['gridValues'])){
$c = count($_POST['gridValues']);
}