Question

So here is the code –

    $info = 0;
  switch ( $info ) {
    case ( $info < 11 ):
     $ts = "zero";
     break;
    case ( $info <= 44 && $info >= 11):
     $ts = "two";
     break;
    case ( $info > 44 ):
     $ts = "three";
     break;
   }

echo $ts;

It doesn’t give out zero, bur it gives out “two”, maybe you have an idea why? It only happens if $info = 0.

Answer 1

It’s because case expects a value, not a condition, and the conditions you’re providing are being converted into values before comparison with the switch argument.

With $info being 0:

$info < 11 is true, hence gives 1.

$info <= 44 && $info >= 11 is false, hence gives 0.

$info > 44 is false, hence gives 0.

So it’s matching the first case where the $info value of 0 is equal to the condition converted to a number, which is the second one. That’s why you’re seeing "two".

I would suggest changing it into an if/else variant:

if ($info < 11) {
    $ts = "zero";
} elseif ($info >= 11 && $info <= 44) {
    $ts = "two";
} elseif ($info > 44) {
    $ts = "three";
}

or possibly better, since it removes extraneous checks and ensures that some value is always assigned:

if ($info < 11) {
    $ts = "zero";
} elseif ($info < 45) {
    $ts = "two";
} else {
    $ts = "three";
}