就像在用户模型laravel 8上一样

Question

嘿，伙计们，我想在laravel 8上添加一个类似和不喜欢的用户模型，而且只有当用户喜欢的是经过身份验证的用户，并且有人可以帮助我如何使它与所有用户一起成为可能时，它才能工作。

用户模型：

这是用户模型：

 class User extends Authenticatable
    {
        use HasApiTokens, HasFactory, Notifiable;

    /**
     * The attributes that are mass assignable.
     *
     * @var array<int, string>
     */
    protected $fillable = [
        'name',
        'first_name',
        'image',
        'cover_image',
        'country_id',
        'email',
        'password',
    ];

    /**
     * The attributes that should be hidden for serialization.
     *
     * @var array<int, string>
     */
    protected $hidden = [
        'password',
        'remember_token',
    ];

    /**
     * The attributes that should be cast.
     *
     * @var array<string, string>
     */
    protected $casts = [
        'email_verified_at' => 'datetime',
    ];



    public function likes()
    {
        return $this->hasOne(UserLike::class,);
    }


}

这是类似的模式：

namespace App\Models;

use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;
use App\Models\User;

class UserLike extends Model
{
    use HasFactory;
    protected $fillable = [
        'user_id',
        'liked_id',
    ];

    public function user()
    {
        return $this->belongsTo(User::class);
    }
}

这是类似的迁移：

use Illuminate\Database\Migrations\Migration;
use Illuminate\Database\Schema\Blueprint;
use Illuminate\Support\Facades\Schema;

class CreateUserLike extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {
        Schema::create('user_likes', function (Blueprint $table) {
            $table->id();
            $table->integer('user_id');
            $table->integer('liked_id');
            $table->timestamps();
        });
    }

    /**
     * Reverse the migrations.
     *
     * @return void
     */
    public function down()
    {
        Schema::dropIfExists('user_like');
    }
}

控制器:类UserLikeController扩展控制器{ //公共函数类似于不喜欢的用户($id){

        $user = User::find($id);
        if(!$user)
        {
            return response([
                'message' => 'User not found.'
            ], 403);
        }

        $userlike = $user->likes()->where('user_id', auth()->user()->id)->first();

        // if not liked then like
        if(!$userlike)
        {
            UserLike::create([
                'user_id' => auth()->user()->id,
                'liked_id' => $id,
            ]);

            return response([
                'message' => 'Liked'
            ], 200);
        }
        // else dislike it
        $userlike->delete();

        return response([
            'message' => 'Disliked'
        ], 200);
    }
}

Answer 1

您不需要在控制器中使用Auth用户，只需要让所有用户访问该函数并检查您的路由，它不应该有Auth中间件，您的构造函数也不应该有中间件(Auth)。