Is someone can explain me this..:
$bob = $_POST['foo'] ;
is_int($bob) FAIL but is_numeric($bob) is OK .
So i know i cannot use is_int directly on $_POST but here i transfert the post value in another variable before..
Whats wrong please ?!
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albator
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Why you make a question that does not matter with the assumption? You write about `$_POST` but in your example you use `is_int` and `is_numeric` with `$bob` – Luca Rainone Dec 06 '13 at 15:48
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so theres no ways to test is_int for a data coming to $_POST ?? – albator Dec 06 '13 at 15:49
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2Use `is_numeric` and if that is true, cast to an int. – PKeidel Dec 06 '13 at 15:50
1 Answers
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$_POST values are strings, no matter whether they contain a numeric value or not. Just transferring them to a different variable doesn't change that.
You'd have to typecast the variable:
$bob = (int) $_POST['foo'];
but note that non-integer values are cast to 0 in this case.
Pekka
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thanks, so if i really need to test is_int , i need to do $bob = (int) $_POST['foo']; THEN if(is_int($bob)) , thats right ?! – albator Dec 06 '13 at 15:53
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@albatros that would not make sense: `(int)` will always create an integer. The problem is that it will convert any non-integer to 0. Do as PKeidel says above, test whether it is numeric first, then convert it into an int. – Pekka Dec 06 '13 at 15:54