Warning: array_filter() expects parameter 2 to be a valid callback, function 'empty' not found or invalid function name....
Why is empty considered a invalid callback?
$arr = array_filter($arr, 'empty');
This works: if(empty($arr['foo'])) die();
empty() is not a function but a language construct and array_filter() can only accept a function as its callback.
This is given as a small note on the manual page:
Note: Because this is a language construct and not a function, it cannot be called using variable functions
To work around this you can wrap empty in another function for example:
function empty_test($val) {
return empty($val);
}
And then call it like so:
$arr = array_filter($arr, 'empty_test');
$arr = array_diff($arr, array_filter($arr));
See the documentation page on empty():
Note: Because this is a language construct and not a function, it cannot be called using variable functions
So basically empty() is not a function, and because callback must be a function, empty() can not be passed as callback.
But you can create callback that may use empty(). The following should work in PHP > 5.3:
$arr = array_filter($arr, function($var){
return empty($var);
});
In PHP < 5.3 you will need to create similar function first and then pass it to the array_filter().
Did it help?
empty() is a language construct, and not a true function in terms of PHP, so you can't pass its name as an argument to functions like array_filter() and call_user_func_array().
From the manual:
Note: Because this is a language construct and not a function, it cannot be called using variable functions
For a workaround, just wrap it in another user-defined function; see Treffynnon's answer.
You can use just array_filter() function without callback:
Remove empty array elements in PHP
$arr = array("PHP", "HTML", "CSS", "", "JavaScript", null, 0);
print_r(array_filter($arr)); // removing blank, null, false, 0 (zero) values
Result:
Array
(
[0] => PHP
[1] => HTML
[2] => CSS
[4] => JavaScript
)
