Best way to test for a variable's existence in PHP; isset() is clearly broken

chazomaticus_30497 0
Viewed 149450

From the isset() docs:

isset() will return FALSE if testing a variable that has been set to NULL.

Basically, isset() doesn't check for whether the variable is set at all, but whether it's set to anything but NULL.

Given that, what's the best way to actually check for the existence of a variable? I tried something like:

if(isset($v) || @is_null($v))

(the @ is necessary to avoid the warning when $v is not set) but is_null() has a similar problem to isset(): it returns TRUE on unset variables! It also appears that:

@($v === NULL)

works exactly like @is_null($v), so that's out, too.

How are we supposed to reliably check for the existence of a variable in PHP?

Edit: there is clearly a difference in PHP between variables that are not set, and variables that are set to NULL:

<?php
$a = array('b' => NULL);
var_dump($a);

PHP shows that $a['b'] exists, and has a NULL value. If you add:

var_dump(isset($a['b']));
var_dump(isset($a['c']));

you can see the ambiguity I'm talking about with the isset() function. Here's the output of all three of these var_dump()s:

array(1) {
  ["b"]=>
  NULL
}
bool(false)
bool(false)

Further edit: two things.

One, a use case. An array being turned into the data of an SQL UPDATE statement, where the array's keys are the table's columns, and the array's values are the values to be applied to each column. Any of the table's columns can hold a NULL value, signified by passing a NULL value in the array. You need a way to differentiate between an array key not existing, and an array's value being set to NULL; that's the difference between not updating the column's value and updating the column's value to NULL.

Second, Zoredache's answer, array_key_exists() works correctly, for my above use case and for any global variables:

<?php
$a = NULL;
var_dump(array_key_exists('a', $GLOBALS));
var_dump(array_key_exists('b', $GLOBALS));

outputs:

bool(true)
bool(false)

Since that properly handles just about everywhere I can see there being any ambiguity between variables that don't exist and variables that are set to NULL, I'm calling array_key_exists() the official easiest way in PHP to truly check for the existence of a variable.

(Only other case I can think of is for class properties, for which there's property_exists(), which, according to its docs, works similarly to array_key_exists() in that it properly distinguishes between not being set and being set to NULL.)

isset variables php
17 Answers

If the variable you are checking would be in the global scope you could do:

array_key_exists('v', $GLOBALS)
Zoredache_202670

You can use the compact language construct to test for the existence of a null variable. Variables that do not exist will not turn up in the result, while null values will show.

$x = null;
$y = 'y';

$r = compact('x', 'y', 'z');
print_r($r);

// Output:
// Array ( 
//  [x] => 
//  [y] => y 
// )

In the case of your example:

if (compact('v')) {
   // True if $v exists, even when null. 
   // False on var $v; without assignment and when $v does not exist.
}

Of course for variables in global scope you can also use array_key_exists().

B.t.w. personally I would avoid situations like the plague where there is a semantic difference between a variable not existing and the variable having a null value. PHP and most other languages just does not think there is.

user_1204480

Object properties can be checked for existence by property_exists

Example from a unit test:

function testPropertiesExist()
{
    $sl =& $this->system_log;
    $props = array('log_id',
                   'type',
                   'message',
                   'username',
                   'ip_address',
                   'date_added');

    foreach($props as $prop) {
        $this->assertTrue(property_exists($sl, $prop),
                           "Property <{$prop}> exists");
    }
}
Admin1

If I run the following:

echo '<?php echo $foo; ?>' | php

I get an error:

PHP Notice:  Undefined variable: foo in /home/altern8/- on line 1

If I run the following:

echo '<?php if ( isset($foo) ) { echo $foo; } ?>' | php

I do not get the error.

If I have a variable that should be set, I usually do something like the following.

$foo = isset($foo) ? $foo : null;

or

if ( ! isset($foo) ) $foo = null;

That way, later in the script, I can safely use $foo and know that it "is set", and that it defaults to null. Later I can if ( is_null($foo) ) { /* ... */ } if I need to and know for certain that the variable exists, even if it is null.

The full isset documentation reads a little more than just what was initially pasted. Yes, it returns false for a variable that was previously set but is now null, but it also returns false if a variable has not yet been set (ever) and for any variable that has been marked as unset. It also notes that the NULL byte ("\0") is not considered null and will return true.

Determine whether a variable is set.

If a variable has been unset with unset(), it will no longer be set. isset() will return FALSE if testing a variable that has been set to NULL. Also note that a NULL byte ("\0") is not equivalent to the PHP NULL constant.

user_504530

I have to say in all my years of PHP programming, I have never encountered a problem with isset() returning false on a null variable. OTOH, I have encountered problems with isset() failing on a null array entry - but array_key_exists() works correctly in that case.

For some comparison, Icon explicitly defines an unused variable as returning &null so you use the is-null test in Icon to also check for an unset variable. This does make things easier. On the other hand, Visual BASIC has multiple states for a variable that doesn't have a value (Null, Empty, Nothing, ...), and you often have to check for more than one of them. This is known to be a source of bugs.

staticsan_288320

Try using

unset($v)

It seems the only time a variable is not set is when it is specifically unset($v). It sounds like your meaning of 'existence' is different than PHP's definition. NULL is certainly existing, it is NULL.

user_204710

I don't agree with your reasoning about NULL, and saying that you need to change your mindset about NULL is just weird.

I think isset() was not designed correctly, isset() should tell you if the variable has been set and it should not be concerned with the actual value of the variable.

What if you are checking values returned from a database and one of the columns have a NULL value, you still want to know if it exists even if the value is NULL...nope dont trust isset() here.

likewise 

$a = array ('test' => 1, 'hello' => NULL);

var_dump(isset($a['test']));   // TRUE
var_dump(isset($a['foo']));    // FALSE
var_dump(isset($a['hello']));  // FALSE

isset() should have been designed to work like this:

if(isset($var) && $var===NULL){....

this way we leave it up to the programmer to check types and not leave it up to isset() to assume its not there because the value is NULL - its just stupid design

user_5037730
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