Laravel Eloquent Model Dependency Injection with preformated Id to find() - php

Question

Normally we can simplify finding User by id logic in controller by injecting the User Class in parameter. Like this:
class UserController extends Controller
{
public function show(User $id)
{
return $user;
}
}
But now I must treat the Id to find like this:
<?php
namespace App\Http\Controllers;
class UserController extends Controller
{
public function show(User $id)
{
$preformattedId = '98'.$id;
$user = User::find($preformattedId );
return $user;
}
}
My basic question is: how I can achieved that same trick to my preformatted id in below code like the above code?
Note: I have to use the Id this way because i work with legacy database that actually adding that '98' prefix in every Id, despite that we only use characters after that prefix.

Answer 1

You can use Inversion of Control by using explicit binding on your router.
In your RouteServiceProvider
/**
* Bootstrap any application services.
*
* #return void
*/
public function boot()
{
parent::boot();
Route::bind('user', function ($value) {
return User::find('98'.$value);
});
}
Or in your User model
/**
* Retrieve the model for a bound value.
*
* #param mixed $value
* #param string|null $field
* #return \Illuminate\Database\Eloquent\Model|null
*/
public function resolveRouteBinding($value, $field = null)
{
return $this->find('98'.$value);
}
https://laravel.com/docs/7.x/routing#explicit-binding

Answer 2

You can share your route file ?
But if your file is
Route::get('user/{id}', 'UserController#show');
When you use
class UserController extends Controller
{
public function show(User $id)
{
// you don't need use find function, it is make automatic by laravel
$user = $id;
return $user;
}
}
if you want to get id, just remove User type inside show parameter
class UserController extends Controller
{
public function show($id)
{
$preformattedId = '98'.$id;
$user = User::find($preformattedId );
return $user;
}
}