PHP: issue with switch statement (wrong return)

Question

I have this method:

private function convertStatusStringToIntZeroOrOne(string $status)
{

    $status = strtolower($status);

    switch ($status) {

        case "off":
        case "0":
        case 0:
            $int_status = 0;
            break;

        case "on":
        case "1":
        case 1:
            $int_status = 1;
            break;

        default:
            $int_status = 1;
            break;


    }

    return $int_status;
}

The $status parameter, when is the string "On" (with O letter capitalize), return 0 (zero).

Of course, I need return as 1.

Thank you

Because you're doing a type juggle from string to int when testing for 0. And 'on' == 0 is true. — Jonnix
– Jonnix, Commented Jan 24, 2019 at 10:19

Nigel Ren · Accepted Answer · 2019-01-24 10:25:03Z

6

As you had numeric 0 and 1 in the options of the switch it was using a numeric comparison - "on" to a number is 0 and so it matched against 0.

As you have the parameter as type string a number would be converted to a string, so remove the numeric comparisons...

function convertStatusStringToIntZeroOrOne(string $status)
{
    $status = strtolower($status);

    switch ($status) {
        case "off":
        case "0":
            $int_status = 0;
            break;
        case "on":
        case "1":
            $int_status = 1;
            break;
        default:
            $int_status = 1;
            break;
    }

    return $int_status;
}
echo convertStatusStringToIntZeroOrOne("On");

Although you could reduce the function to...

function convertStatusStringToIntZeroOrOne(string $status)
{
    $status = strtolower($status);
    return ($status == "off" || $status == 0)?0:1;
}

answered Jan 24, 2019 at 10:25

Nigel Ren

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1 Comment

Bananaapple Over a year ago

Could be reduced even further to return (int) ! in_array(strtolower($status), ['off', '0'], true); - but that's less clear :-)

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