unable to determine if a string is currently an integer or not

Question

The following funciton drove me nuts. How on earth 100x could be equal to 100 and then 100x is reported as an integer?
For the life of me, I cannot figure it out. You can copy and paste the whole thing and see it for yourself.
I'm missing a simple point somewhere here, help me out guys.

function blp_int($val) {
    $orgval = $val;
    $num = (int)$val;    
    echo "<li><font color=red>val: ". $val . " is being checked to see if it's an integer or not";
    echo "<li><font color=red>orgval: ". $orgval ;
    echo "<li><font color=red>num: ". $num ;
    if ($orgval==$num) {
        echo "<li><font color=red><b>YES IT IS! the [{$orgval}] is equal to [{$num}]</b>";
        return true;
    }
    return false;
}

if (blp_int("100"))
{
    echo "<h1>100 is an integer!</h1>";
}
else
{
    echo "<h1>100 is NOT an integer!</h1>";
}

if (blp_int("100x"))
{
    echo "<h1>100x is an integer!</h1>";
}
else
{
    echo "<h1>100x is NOT an integer!</h1>";
}

the above code, when run returns the following;

val: 100 is being checked to see if it's an integer or not
orgval: 100
num: 100
YES IT IS. the [100] is equal to [100]
100 is an integer!

val: 100x is being checked to see if it's an integer or not
orgval: 100x
num: 100
YES IT IS. the [100x] is equal to [100]
100x is an integer!

I can remedy the situation by adding the following bits

    if (!is_numeric($val))
    {
        return false;
    }

to the top of the blp_int function right off the bat but,.. I'm still super curious to find out why on earth php thinks 100x=100 are equals.

Answer 1

As you can see in this example, casting 100x as an integer converts it to 100. Since you are not using strict comparison, '100x' == 100 is true. PHP removes the x from it to make just 100.

You could use strict comparison (which also compares the types), such that '100x' === 100 would return false. Using it, any time a string was compared to an integer, it would return false.

---

As per your edit: is_numeric may not be the most reliable, as it will return true for numbers formatted as a string, such as '100'. If you want the number to be an integer (and never a string), you could use is_integer instead. I'm not quite sure what exactly you're doing, but i thought I'd add this note.

Answer 2

I think you should use three equal signs in your IF:

if ($orgval===$num) {

Otherwise PHP casts the value 100x to 100 and 100=100.

Documentation: Comparison Operators

Answer 3

What kind of check do you want to do? There are a few ways you could go about it:

if (preg_match('!^[0-9]+$!', $input))

if (intval($input) == $input)

if (intval($input) === $input)

if ('x'.intval($input) === 'x'.$input)

It depends on how closely you want to check if it's an integer. Does it matter if you need to trim() it first?

Answer 4

Either cast it to an int or try http://php.net/manual/en/function.ctype-digit.php. You also need === in your if.