PHP | Switch statement crashes execution

Question

I have this statement that includes in a php-file. And with it execution crashes. Nothing executes before and after the statement.

    switch ($type) {
    case "contact_page" :
        $subject = "Вопрос со страницы Контакты";
        $user_email = filter_var($_POST["email"], FILTER_SANITIZE_EMAIL);
        break;
    case "call_ask" :
        $subject = "Просьба позвонить";
        break;
    case "calculator" :
        $subject = "Калькулятор";
        $user_email = filter_var($_POST["email"], FILTER_SANITIZE_EMAIL);
        break;
    default :
        $output = json_encode(['type' => 'error', 'text' => 'Запрос из неверного типа формы.']);
        die($output);
}

When I replace this with if...else statement all executes good. Can't find a mistake.

so start commenting out each of those lines and figure out which one causes the crash. there's nothing that looks wrong. — Marc B, Jun 27 '16 at 16:10
@KA_lin: pointless+useless. default's the last in that switch, so there's no need for a break, and break is not REQUIRED anywhere, so the presence/absence of a break cannot cause a crash. — Marc B, Jun 27 '16 at 16:10
how you are calling this function? and what is the output of `var_dump($type);`? — Alive to die - Anant, Jun 27 '16 at 16:10
Looks like the problem is some crazy character is hiding somewhere. — Praveen Kumar Purushothaman, Jun 27 '16 at 16:11
@MarcB I tried this before writing. Compiling stops in the first _case_ or _default_ , even when I leave only _default_ it isn't work. Or changing to _case_ steps to numbers. When I leave empty _switch_ without nothing, then execution. — Paul Burilichev, Jun 27 '16 at 16:18
@Anant even with constant value or only _default_ nothing works. And _if...else_ works, but it is not an issue. — Paul Burilichev, Jun 27 '16 at 16:22
@PraveenKumar no, Russian is only the value. And this language exists later too. Also _if...else_ works. — Paul Burilichev, Jun 27 '16 at 16:23
@PaulBurilichev Grrr... True that. Do one thing. Wrap the Russian content with some English text before and after and see? Just check? — Praveen Kumar Purushothaman, Jun 27 '16 at 16:24

score 2 · Accepted Answer · answered Jun 27 '16 at 16:24

2

The php version does not support the new array notation (5.4+ from what I know).

Use in the default case:

 $output = json_encode(array('type' => 'error', 'text' => 'Запрос из неверного типа формы.'));

answered Jun 27 '16 at 16:24

ka_lin

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Yes, it is the problem. Thank you for attention. Really I forget this. – Paul Burilichev Jun 27 '16 at 16:29
That's funny that people still in old versions of PHP. For example I use always latest release of PHP (now v. 7) and I don't care if there will be exceptions, I'll fix it. – num8er Jun 27 '16 at 16:30

score 0 · Answer 2 · answered Jun 27 '16 at 16:27

0

Upgrade PHP to >5.4

or update code from:

json_encode(['type' => 'error', 'text' => 'Запрос из неверного типа формы.']);

to

json_encode(array('type' => 'error', 'text' => 'Запрос из неверного типа формы.'));

answered Jun 27 '16 at 16:27

num8er

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Unfortunately, I can't upgrade php and haven't permission. And, yes, this is the issue. – Paul Burilichev Jun 27 '16 at 16:32
oh... those admins... (: – num8er Jun 27 '16 at 16:32

PHP | Switch statement crashes execution

2 Answers2