PHP: is_int return wrong result

Question

I am posting numeric value by a form and in php file use var_dump(is_int($my_var)); but it return bool(false) though i am posting 1 and if I use var_dump(is_int(1)); it is fine and return bool(true) what is wrong....?

check $my_var type, if is_int($my_var) returns false, $my_var must be something else than an integer — Ben
– Ben, Commented Jul 9, 2012 at 12:04
if your variable $my_var contains input from a form (as your first sentence suggests), it will contain data of type string - php doesn't do any automatic conversion of form data. — codeling
– codeling, Commented Jul 9, 2012 at 12:05

Floern · Accepted Answer · 2012-07-09 12:07:18Z

8

Variables transmitted by a POST request are strings, so you're calling is_int() on a string which returns false.

You may use is_numeric() or filter_var() instead or simply cast your variable to an integer.

answered Jul 9, 2012 at 12:07

Floern

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Frankline · Accepted Answer · 2012-07-09 12:10:46Z

2

// First check if it's a numeric value as either a string or number
    if(is_numeric($int) === TRUE){

        // It's a number, but it has to be an integer
        if((int)$int == $int){

            return TRUE;

        // It's a number, but not an integer, so we fail
        }else{

            return FALSE;
        }

    // Not a number
    }else{

        return FALSE;
    }

Also, instead of getting the variable as

$my_var = $_POST["value"];

try this instead to see if the value is really passed.

$my_var = $_REQUEST["value"];

answered Jul 9, 2012 at 12:10

Frankline

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1 Comment

KingCrunch Over a year ago

You can reduce your whole code snippet to return (int)$int == $int;

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