Checking NULL against database record -> isset vs. is_null

Question

I provided code, where first argument is path to image or default NULL in MySQL database,

function formatImage($img = NULL, $alt = NULL)
    {
        // if(isset($img)) DOESN'T WORK
        if(!$img == NULL)
        {
            return '<img src="' . $img . '" alt="' . $alt .'" />';
        }
        else
        {
            return NULL;
        }
    }

Why isset doesn't work? Instead I have to check if is (!$img == NULL).

In case of using checking argument with isset, and that $img is NULL in database. as output im getting whole HTML IMG tag with empty src attibutes, and alt attribute which is actaully title fetched from database. Setting display property to none to images without src attribute isn't acceptable.

Perhaps $img ends up containing a string 'NULL' - so his approach to using isset() didn't work out that well. — SquareCat
I'm pretty sure a var_dump($img) just before the if will shed some light here — Mosty Mostacho

SquareCat SquareCat · Accepted Answer · 2013-12-03T01:14:43

Try this:

if($img != NULL) {
    return '<img src="' . $img . '" alt="' . $alt .'" />';
}

Explanation:

If you compare !$img against NULL you basically do a boolean reversal on $img and then seeing if that compares to NULL - which is not what you want.

Also, please note that you should use

if($img !== NULL)

instead. != will compare the values of the variables provided, while !== will compare the values AND the datatype (which in case of NULL might be your best interest).

Checking NULL against database record -> isset vs. is_null

2 Answers