PHP cannot automatically use function result as array (in PHP 5.3)

Question

When I try to run something like

getParent($child)[0]->user

I get the following error:

PHP Parse error:  syntax error, unexpected '[', expecting ')' in /.....

The problem can be avoided if I do something like this:

$get_parent = getParent($child);
$parent = $get_parent[0]->user;

Is there a better way to do in php 5.3

Answer 1

No, before PHP5.4, you have to do that.

Array dereferencing comes from PHP 5.4.

But if getParent return an object which implement the ArrayAccess interface, you could chain it with:

$parent = getParent($child)->offsetGet(0)->user;

If it just return an array, then a temp variable is necessary.

Answer 2

Actually, there is a way to achieve that without temporary variable. But I definitely wouldn't recommend to use it (because, yes, it's one-liner and it's not using temporary variable, but no - it's not readable):

function getParent($child=null)
{
   //mock:
   return array(
      (object)(array('user'=>'foo', 'data'=>'fee')),
      (object)(array('user'=>'bar', 'data'=>'bee')),
   );
};

//array(null) will have 1 key, 0;
//however, to get another offset N, use array(N => null) instead
$result = array_shift(array_intersect_key(getParent('baz'), array(null)))->user;

-so use temporary variable if your version in <5.4

Where it may be helpful - is in debugger, where you're forced to use "one-liners" to check some expression(s)

Answer 3

There is no better way to do it when using PHP version < 5.4.

If you still want to use one line instead of 2 and you always want to get the first element, you can try the following

echo reset(getParent($child))->user;

This reset the array pointer to 0 and returns the value.