Laravel Eloquent Model 依赖注入，使用预先格式化的 Id 来查找（）

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;
   }
}