PHP的IF条件中的一些困惑

Question

This is the code

$a = 'Rs 15.25';
if ( $a != '' && $a! = 0 ) {
    echo "Inside If";
} else {
    echo "Outside If";
}

actually I want to Print "Inside If" so that's why I put $a='Some String Value'. But it always prints "Outside If". Then I changed my code to

$a = 'Rs 15.25';
if ( $a != '' && $a != '0' ) {
    echo "Inside If";
} else {
    echo "Outside If";
}

I have just added single quotes to 0. Then i got the exact output as i want. But I didn't understand why this happens. So please help me with this.

Answer 1

PHP does weak type comparison, that is, it converts both operands to the same type before doing the actual comparison.

If one of the operands is a number, the other one is converted to a number as well. If the second operand is a string and contains no digits, it is silently converted to the number 0.

To avoid this whole issue, use string type checking with the operator !== (=== for equality).

if($a !== '' && $a !== 0) {
    echo "Inside If";
} else {
    echo "Outside If";
}