How to check whether there are variables at once in one if?

Question

PHP

There is such a code. It works but...

if(!isset($_GET['id']))
  {
    exit('Отсутствует обязательный параметр "ID"');
  }

When I add another one, the script execution stops. Where is the mistake?

if(!isset($_GET['id'] or $_GET['id2']))
  {
    exit('Отсутствует обязательный параметр "ID"');
  }

Answer 1

This is the correct way to check two variables:

if (!isset($_GET['id']) || !isset($_GET['id2'])) {
    exit('Отсутствует обязательный параметр "ID"');
}

Answer 2

Enable error reporting and be more careful.
You do this: take the value of $_GET['id']. Does it not exist? Checking id2. There is? So it returns one. We check with isset if a unit exists.
And for you you need:
if (!isset($_GET['id']) && $_GET['id2']) ...
(if both values ​​don't exist). If you want an error to pop up if at least one value does not exist, you need if (!isset($_GET['id']) || $_GET['id2']) ...
Also, !empty is better for your purpose , since for isset "0" is a completely valid value, and for empty it will return false