PHP switch statement "echoing" the wrong case

Question

$year = readline('Type your year of birth: ');

$age = 2023 - $year;

switch ($age) {
    case ($age < 0):
    echo 'I don't see in the future.';
    break;
    case ($age >= 0) && ($age <= 3):
    echo 'Congrats. A newborn capable of using the PC.';
    break;
    case ($age > 3):
    echo 'Our system calculated that you are:' . ' ' . $age . ' ' . 'years old';
    break;
}

So this is from my first lesson of PHP and the statement "echoes" the first case if I input 2023 but it should echo the second one. Any idea why this happens?

Easiest way to start debugging is to echo $age before your switch statement and see what that variable is — Kosy Onyenso
– Kosy Onyenso, Commented Jan 28, 2023 at 18:29
@KosyOnyenso I've already done that. It's 0 and it should echo 'Congrats.. etc' — user21100580
– user21100580, Commented Jan 28, 2023 at 18:32
There's a syntax error in this code, not sure how you're getting any output. — Alex Howansky
– Alex Howansky, Commented Jan 28, 2023 at 18:32
Assuming the quote syntax error is a typo (and fixed), then as to why when $age == 0 falls into the case $age < 0 it is due to loose comparisons between 0 and false which the switch statement employs. — Computable
– Computable, Commented Jan 28, 2023 at 18:59

Kosy Onyenso · Accepted Answer · 2023-01-28 18:33:55Z

1

Change the switch ($age) to switch (true). Try this:

switch (true) {
    case ($age < 0):
    echo "I don't see in the future.";
    break;
    case ($age >= 0) && ($age <= 3):
    echo "Congrats. A newborn capable of using the PC.";
    break;
    case ($age > 3):
    echo "Our system calculated that you are:" . " " . $age . " " . "years old";
    break;
    default:
    break;
}

answered Jan 28, 2023 at 18:33

Kosy Onyenso

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Comments

Mureinik · Accepted Answer · 2023-01-28 18:36:31Z

0

cases are values, not expressions to evaluate. It looks like you meant to use an if-else:

if ($age < 0) {
    echo 'I don\'t see in the future.';
} elseif ($age >= 0) && ($age <= 3) {
    echo 'Congrats. A newborn capable of using the PC.';
} else if ($age > 3) { # Could have also been an "else", without the condition
    echo 'Our system calculated that you are:' . ' ' . $age . ' ' . 'years old';
}

answered Jan 28, 2023 at 18:36

Mureinik

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