Passing outer function parameter to inner function Laravel

Question

I have the code bellow aimed at retrieving all the service providers using the service ID

 public function getProvidersByPackage($id = null){ $package_id = $id; $providers = ServiceProvider::whereHas('services', function($query) { $query->where('packages.id', 1); })->get(); dd($providers); }

I would like to replace the constant 1 with the variable $id passed to the outer function getProvidersByPackage() my problem is that when I try the following

 public function getProvidersByPackage($id = null){ $package_id = $id; $providers = ServiceProvider::whereHas('services', function($query) { $query->where('packages.id', $id); })->get(); dd($providers); }

I get the error $id is not defined and when I try

 public function getProvidersByPackage($id = null){ $package_id = $id; $providers = ServiceProvider::whereHas('services', function(&$package_id, $query) { $query->where('package.package_id', $package_id); })->get(); }

I get the ArgumentCountError bellow

 Too few arguments to function App\Http\Controllers\ShopController::App\Http\Controllers\{closure}(), 1 passed in /var/www/html/vendor/laravel/framework/src/Illuminate/Database/Eloquent/Builder.php on line 1207 and exactly 2 expected

What could I be doing wrong?

Does this answer your question? In PHP, what is a closure and why does it use the "use" identifier? — Nico Haase
– Nico Haase, Commented Jul 12, 2021 at 11:34

Mohd Abdul Mujib · Accepted Answer · 2021-07-12 11:38:34Z

In PHP, you can pass a variable to a closure using the use syntax

public function getProvidersByPackage($id = null) { $package_id = $id; $providers = ServiceProvider::whereHas('services', function ($query) use ($package_id) { $query->where('package.package_id', $package_id); })->get(); }

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