Test for NULL value in a variable that may not be set

Question

Considering that:

• The isset() construct returns TRUE if a variable is set and not NULL
• The is_null() function throws a warning if the variable is not set

Is there a way to test whether a variable exists, no matter it's NULL or not, without using the @ operator to suppress the notice?

---

EDIT

Together with your first replies, I've been thinking about this and I'm getting the conclusion that inspecting get_defined_vars() is the only way to distinguish between a variable set to NULL and an unset variable. PHP seems to make little distinctions:

<?php

$exists_and_is_null = NULL;

// All these are TRUE

@var_dump(is_null($exists_and_is_null));
@var_dump(is_null($does_not_exist));

@var_dump($exists_and_is_null===NULL);
@var_dump($does_not_exist===NULL);

@var_dump(gettype($exists_and_is_null)=='NULL');
@var_dump(gettype($does_not_exist)=='NULL');

?>

Answer 1

if (isset($var) && (is_null($var)) {
    print "\$var is null";
}

This should do the trick.

Answer 2

$result = array_key_exists('varname', get_defined_vars());

Answer 3

As you already found out, you cannot :

• rely on isset, as it return false for a variable that's null.
• use $not_exists===null, as it'll raise a notice.

But you could be able to use a combinaison of :

• get_defined_vars to get the list of existing variables, including those which are null,
• and array_key_exists to find out if an entry exists in that list.

For instance :

$exists_and_null = null;
$exists_and_not_null = 10;

$defined_vars = get_defined_vars();

// true
var_dump(array_key_exists('exists_and_null', $defined_vars) 
    && $defined_vars['exists_and_null']===null);

// false
var_dump(array_key_exists('exists_and_not_null', $defined_vars) 
    && $defined_vars['exists_and_not_null']===null);

// false
var_dump(array_key_exists('not_exists', $defined_vars) 
    && $defined_vars['not_exists']===null);

A couple of notes :

• In the first case, the variable exists => there is an entry in the list returned by get_defined_vars, so the second part of the condition is evaluated
  • and both parts of the condition are true
• In the second case, the variable exists too, but is null
  • which means the first part of the condition is true, but the second one is false,
  • so the whole expression is false.
• In the third case, the variable doesn't exist,
  • which means the first part of the condition is false,
  • and the second part of the condition is not evaluated -- which means it doesn't raise a notice.

But note this is probably not that a good idea, if you care about performances : isset is a language construct, and is fast -- while calling get_defined_vars is probably much slower ^^

Answer 4

I used a self created function to check this easily, keep in mind it will fire off a PHP warning (I only monitor E_ERROR when I develop).

/**
 * isNullOrEmpty()
 * 
 * @global Returns a boolean letting you know if a value is null or empty
 * @param mixed $arg
 * @return
 */
    function isNullOrEmpty( $arg )
    {
        if ( !is_array( $arg ) )
        {
            $arg = array( $arg );
        }

        foreach ( $arg as $key => $value )
        {
            $value = trim($value);
            if( $value == "" || $value == null )
            {
                return true;
            }
        }
        return false;
    }

Answer 5

I would argue here that any code requiring such a comparison would have gotten its semantics wrong; NULL is an unset value in a language that has no straightforward way of distinguishing between the two.