Why does casting a string as an integer not work?

Question

// EDIT: The end goal is to detect whether user-submitted value is a string, or an integer. Because the value is obtained from form input, it is cast as string even when the value provided is an integer number.

I am performing a simple comparison to detect if a string variable is equal to its integer value. The value is passed from an HTML form field.

When the string value is a single letter, it evaluates to true, when it should not.

if ( ($val == (int)$val ) && strlen( $val ) == strlen( (int)$val ) ){
}

I've also tried using intval() rather than casting int variable type.

It is always evaluating as true. But r does not == 0, so how can this be?

$val = 'r'; 
echo 'Does $val = (int)$val? ' . ($val == (int)$val);

echo '<br/><br/>';
echo '$val was: ' . $val . '<br/>';
echo '(int)$val was: ' . (int)$val;

Output:

Does $val = (int)$val? 1

$val was: r
(int)$val was: 0

when you compare with ==, values are implicitely cast. it's the same as (int)$val == (int)$val. thing is: what are you even trying to achieve with your code? — Franz Gleichmann
@FranzGleichmann thanks for this. User can pass a record ID (integer) or a name (string). I build a MySQL query depending on whether it's integer (user_id = $val) or whether it's string (user_name LIKE '%val%'`) — Patrick Moore
Yeah, shouldn't ($val == (int)$val) be ($val = (int)$val) ? or are you trying to use a ternary operator here? — Funk Forty Niner
I use if (is_numeric($searchval)) when I do a string or ID entry search. — IncredibleHat

Franz Gleichmann Franz Gleichmann · Accepted Answer · 2018-02-21T17:43:17

To check if your string is actually a number (which is what you want according to your comments), you don't have to hack together some custom test. PHP provides a built-in function for this:

if(is_numeric($val)) {
    //do integer stuff
} else {
    //do string stuff
}

Why does casting a string as an integer not work?

2 Answers