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'])) ? $foo['bar'] : 'baz' == 'baz' );
I am looking for a golf version. Can I do it shorter/better in PHP?
assert($foo['bar'] ?? 'baz' == 'baz');
It seems that Null coalescing operator ??
is worth checking out today.
found in the comments below (+1)
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!