Compare with int and string in IF condition

Question

Code in PHP :

<?php
    $key = 0;
    if($key == 'monty' && $key == 'anil'){
        echo 'Mackraja';
    }else{
        echo "Nothing";
    }
?>

OR

<?php
    $key = 2;
    if($key == '2abc' || $key == 'abc2'){
        echo 'Mackraja';
    }else{
        echo "Nothing";
    }
?>

OUTPUT: Mackraja

Answer 1

If you compare a number with a string or the comparison involves numerical strings, then each string is converted to a number and the comparison performed numerically. These rules also apply to the switch statement. - Source

This means that

<?php
    var_dump(0 == "a"); // 0 == 0 -> true
    var_dump("1" == "01"); // 1 == 1 -> true
    var_dump("10" == "1e1"); // 10 == 10 -> true
    var_dump(100 == "1e2"); // 100 == 100 -> true

So when you're comparing $key (with value 0) to the strings, the strings have value 0 and they both return true. This way, it will always output "Mackraja".

Changing the value of $key to any other integer will return false.

---

Note that

The type conversion does not take place when the comparison is === or !== as this involves comparing the type as well as the value. - Source

This means that changing the comparison operator to === will mean that the values have to match completely - i.e., a string 1 will not be equal to an integer 1: they have to match formats:

echo $key == 'monty' && $key == 'anil' ? 'Mackraja' : 'Nothing';
//will echo "Mackraja"

echo $key === 'monty' && $key === 'anil' ? 'Mackraja' : 'Nothing';
//will echo "Nothing"