Just curious why this would trigger a warning.
Note: query string param "test" is not included in the URL
//Notice: Undefined index: test
if($_GET['test'] === 'somestring' && !empty($_GET['test'])) {
}
//Valid - No Warning. Why is this Valid? Param 'test' was never given
if(!empty($_GET['test']) && $_GET['test'] === 'somestring') {
}
Is this because PHP evaluates from LEFT to RIGHT? And not the whole condition?
if(!empty($_GET['test']) && $_GET['test'] === 'somestring') {
Because you're using &&, if any condition is false, further conditions are not checked. Conditions are checked in the order they're written. Since here, first condition !empty($_GET['test']) is false, the next one does not evaluate hence no warning.
In the first example:
//Notice: Undefined index: test
if($_GET['test'] === 'somestring' && !empty($_GET['test'])) {
}
you are trying to access the value of the variable first, before checking if it exists, which is what the second example is basically doing. After checking that it does not exist, it exits the conditional without testing the second part.
Simply put, it comes down to operator precedence.
The documentation shows that && is evaluated left-to-right, which means the left part of the clause is evaluated first. In your case, that clause results in a warning. If you reversed your clause order, the empty() check would return false, then the logic would short circuit and stop evaluating.
&&statements will stop at the firstfalseand||statements will stop at the firsttrue. In PHP you can use this to write otherwise-unsafe statements, like your second example.