Notice: Undefined index: intakeid [duplicate] - php

Question

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
I am trying since last hour to update my database using php but unable to update.
I get the error
Notice: Undefined index: intakeid in C:\wamp\www\Multi_Edit\edit_save.php on line 3.
<?php
include('dbcon.php');
$intakeid=$_POST['intakeid'];
$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$password=$_POST['password'];
$confirmpassword=$_POST['confirmpassword'];
$homeaddress=$_POST['homeaddress'];
$email=$_POST['email'];
$N = count($intakeid);
for($i=0; $i < $N; $i++)
{
$result = mysql_query("UPDATE registration SET firstname='$firstname[$i]',
lastname='$lastname[$i]',
password='$password[$i]',
confirmpassword='$confirmpassword[$i]',
homeaddress='$homeaddress[$i]',
email='$email[$i]'
WHERE intakeid='$intakeid[$i]`enter code here`'")or die(mysql_error());
}
error_reporting(E_ALL);
ini_set("display_errors", 5);
?>

Answer 1

Its always best method to check $_POST['value'] variable has value set or not using isset() function , otherwise you will get NOTICES as you have it already. check this

Answer 2

It happens simply there is no variable called "intakeid" in $_POST array. If you still need to get it (whereas it is set) you can do something like follows.
$intakeid=(isset($_POST['intakeid']) ? $_POST['intakeid'] : NULL ;
This will put $intakeid to null if the post variable not have the value and if it has it will put the posted value