Is there a better PHP way for getting default value by key from array (dictionary)?

In Python one can do:

foo = {}
assert foo.get('bar', 'baz') == 'baz'

In PHP one can go for a trinary operator as in:

$foo = array();
assert( (isset($foo['bar'])) ? $foo['bar'] : 'baz' == 'baz' );

I am looking for a golf version. Can I do it shorter/better in PHP?

UPDATE [March 2020]:

assert($foo['bar'] ?? 'baz' == 'baz');

It seems that Null coalescing operator ?? is worth checking out today.

found in the comments below (+1)

Solution 1:

Time passes and PHP is evolving. PHP 7 now supports the null coalescing operator, ??:

// Fetches the value of $_GET['user'] and returns 'nobody'
// if it does not exist.
$username = $_GET['user'] ?? 'nobody';
// This is equivalent to:
$username = isset($_GET['user']) ? $_GET['user'] : 'nobody';

// Coalescing can be chained: this will return the first
// defined value out of $_GET['user'], $_POST['user'], and
// 'nobody'.
$username = $_GET['user'] ?? $_POST['user'] ?? 'nobody';

Solution 2:

I just came up with this little helper function:

function get(&$var, $default=null) {
    return isset($var) ? $var : $default;
}

Not only does this work for dictionaries, but for all kind of variables:

$test = array('foo'=>'bar');
get($test['foo'],'nope'); // bar
get($test['baz'],'nope'); // nope
get($test['spam']['eggs'],'nope'); // nope
get($undefined,'nope'); // nope

Passing a previously undefined variable per reference doesn't cause a NOTICE error. Instead, passing $var by reference will define it and set it to null. The default value will also be returned if the passed variable is null. Also note the implicitly generated array in the spam/eggs example:

json_encode($test); // {"foo":"bar","baz":null,"spam":{"eggs":null}}
$undefined===null; // true (got defined by passing it to get)
isset($undefined) // false
get($undefined,'nope'); // nope

Note that even though $var is passed by reference, the result of get($var) will be a copy of $var, not a reference. I hope this helps!

Solution 3:

Use the error control operator @ with the PHP 5.3 shortcut version of the ternary operator:

$bar = @$foo['bar'] ?: 'defaultvalue';