拉拉维尔等效于用户：：所有（） 函数

Question

伙计们我有问题。Laravelapi

在我的控制器中，当我使用时，一切都很好，但我需要过滤数据，所以当我使用收集时，它会给我错误。错误是collection$teacher$teachersCall to undefined method Illuminate\\Database\\Query\\Builder::mapInto();

我想要过滤后的数据

    $teachers = DB::table('users')->where('role', '=', 1);

    $teacher = Users::all();

    return UserResource::collection($teachers);

我试过这个

$teachers = DB::table('users')->where('role', '=', 1)->get();

$teachers = DB::table('users')->where('role', '=', 1)->get()->toArray();

所以我需要解决方案

这是我的资源

<?php

namespace App\Http\Resources;

use Illuminate\Http\Resources\Json\JsonResource;

class User extends JsonResource

{

    /**

     * Transform the resource into an array.

     *

     * @param  \Illuminate\Http\Request  $request

     * @return array

     */

    public function toArray($request)

    {

        return parent::toArray($request);

    }

}

Answer 1

试试这个$teachers = DB::table('users')->where('role', '=', 1);$teacher = Users::all();return new UserResource($teachers);

Answer 2

我不知道你的，但在我的环境中，这些代码行工作得很好。尝试使用命令更新您的拉拉维尔。composer update

Answer 3

return UserResource::collection(User::where('role', '=', 1)->get());这应该对你有用，你能试试吗？更新您仍然可以将表与 连接，但如果您真的想使用eloquentDB::Table()然后你必须像这样指定你的方法。toArray()&nbsp; &nbsp; public function toArray($request)&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; // return parent::toArray($request);&nbsp; &nbsp; &nbsp; &nbsp; return [&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 'role' => $this->role,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 'id' => $this->id,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 'name' => $this->name,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 'email' => $this->email,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 'created_at' => $this->created_at,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 'updated_at' => $this->updated_at,&nbsp; &nbsp; &nbsp; &nbsp; ];&nbsp; &nbsp; }并会给你结果。return UserResource::collection(DB::table('users')->where('role', '=', 1)->get());