I need to check if value is defined as anything, including null. isset treats null values as undefined and returns false. Take the following as an
If you are dealing with object properties whcih might have a value of NULL you can use: property_exists() instead of isset()
<?php
class myClass {
public $mine;
private $xpto;
static protected $test;
function test() {
var_dump(property_exists($this, 'xpto')); //true
}
}
var_dump(property_exists('myClass', 'mine')); //true
var_dump(property_exists(new myClass, 'mine')); //true
var_dump(property_exists('myClass', 'xpto')); //true, as of PHP 5.3.0
var_dump(property_exists('myClass', 'bar')); //false
var_dump(property_exists('myClass', 'test')); //true, as of PHP 5.3.0
myClass::test();
?>
As opposed with isset(), property_exists() returns TRUE even if the property has the value NULL.
Here some silly workaround using xdebug. ;-)
function is_declared($name) {
ob_start();
xdebug_debug_zval($name);
$content = ob_get_clean();
return !empty($content);
}
$foo = null;
var_dump(is_declared('foo')); // -> true
$bla = 'bla';
var_dump(is_declared('bla')); // -> true
var_dump(is_declared('bar')); // -> false
See Best way to test for a variable's existence in PHP; isset() is clearly broken
if( array_key_exists('foo', $GLOBALS) && is_null($foo)) // true & true => true
if( array_key_exists('bar', $GLOBALS) && is_null($bar)) // false & => false
is_null($bar) returns true, since it has no values at all. Alternatively, you can use:
if(isset($bar) && is_null($bar)) // returns false
to check if $bar is defined and will only return true if:
$bar = null;
if(isset($bar) && is_null($bar)) // returns true
IIRC, you can use get_defined_vars() for this:
$foo = NULL;
$vars = get_defined_vars();
if (array_key_exists('bar', $vars)) {}; // Should evaluate to FALSE
if (array_key_exists('foo', $vars)) {}; // Should evaluate to TRUE
You could use is_null and empty instead of isset(). Empty doesn't print an error message if the variable doesn't exist.