I'm trying to run a method on each element inside a collection. It's an object method residing in the same class:
protected function doSomething()
{
$discoveries = $this->findSomething();
$discoveries->each([$this, 'doSomethingElse']);
}
protected function doSomethingElse($element)
{
$element->bar();
// And some more
}
If I precede the call on Collection::each with the check is_callable([$this, 'doSomethingElse']) it returns true, so apparently it is callable. The call itself however throws an exception:
Type error: Argument 1 passed to Illuminate\Support\Collection::each() must be callable, array given, called in ---.php on line 46
The method trying to be called can be found here.
I'm bypassing this by just passing a closure that itself simply calls that function, but this would definitely a much cleaner solution and I can't find out why it throws the error.
Change the visibility of your callback method to public.
protected function doSomething()
{
$discoveries = $this->findSomething();
$discoveries->each([$this, 'doSomethingElse']);
}
public function doSomethingElse($element)
{
$element->bar();
// And some more
}
Since PHP 7.1 you can leave your function protected. Now you can write:
protected function doSomething()
{
$discoveries = $this->findSomething();
$discoveries->each(\Closure::fromCallable([$this, 'doSomethingElse']));
}
protected function doSomethingElse($element)
{
$element->bar();
// And some more
}
I wasn't able to reproduce your error, but my guess is that you should use $discoveries instead of $this in the callback array, like so:
$discoveries->each([$discoveries, 'doSomethingElse']);
Even though $discoveries and $this are of the same class, and therefore can access each other's protected and private methods, the type-hinting functionality may not check that the object in the callback array is the same class as the current class. However, the is_callable() method will check for this, which may explain why it returns true when you call it from inside the each() method.
There is no type named callable, so when you use it as a type hint, it is referring to a class named callable. See this answer.