Notice: Undefined index [duplicate]

Answer 1

Try this:

$thread_id = isset($_GET['id']) ? intval($_GET['id']) : 0;

Which reads as "if $_GET['id'] is set, set $thread_id to intval($_GET['id']), if not, set to 0"

Answer 2

You can get rid of all notices by simply using error_reporting(4);. Another, maybe crude but working, method to suppress the notice and sanitize the variable could be:

$thread_id = (int)@$_GET["id"];

If you do a lot of these things, you maybe want to take a look at filter_var() and put all the functionality into a separate function (eg. get_int("id")).

Answer 3

isset() returns a boolean value: TRUE / FALSE. If you want to determine whether $_GET["id"] is set, you need to do so in an IF-Statement, or via a ternary operator.

if (isset($_GET["id"])) {
  $thread_id = $_GET["id"]; // Bad idea taking values in from $_GET directly
} else {
  die("Thread id not set");
}

Answer 4

This error is dependent on error/warning level set in php.ini. Its good practice to use...

array_key_exists('id', $_GET);

http://us2.php.net/manual/en/function.array-key-exists.php

also: make sure you sanitize that input from the $_GET before you do that. You're leaving yourself open to mysql injection.