Laravel 5.4：無法從關系中檢索數據

Question

我試圖通過將每個名稱存儲在兩個UNION'd表（訪問和報告）的數組中來顯示每個故障單的受理人的姓名（來自Users表的外鍵），但它給了我這個錯誤。 ErrorException未定義屬性：stdClass :: $ assignee。

//HomeController
    $accesses = DB::table('accesses')
                ->select(array('id', 'fullname','emp_id','shift','state','resolved_at', 'closed_at','assigned_to'))
                ->where('state','=','Assigned');

    $all = DB::table('reports')
                ->select(array('id', 'fullname','emp_id','shift','state','resolved_at', 'closed_at','assigned_to'))
                ->union($accesses)
                ->where('state', '=', 'Assigned')
                ->get();

    $names[] = array();

    foreach ($all as $one)//store in array to display in a chart
    {
      $names[] = $one->assignee->name; //error here
    }




   //Report Model
   public function assignee()
   {
      return $this->belongsTo(User::class, 'assigned_to');
   }

   //Access Model
   public function assignee()
   {
      return $this->belongsTo(User::class, 'assigned_to');
   }

  //Report Migration
  public function up()
  {
    Schema::create('reports', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('user_id')->nullable();
        $table->string('fullname');
        $table->string('emp_id');
        $table->string('shift');
        $table->longText('report');
        $table->string('status')->default('Pending'); //Pending, Approved
        $table->string('state')->default('Open'); //Open, Assigned, Resolved, Closed
        $table->date('resolved_at')->nullable();
        $table->date('closed_at')->nullable();
        $table->integer('assigned_to')->nullable();
        $table->longText('action')->nullable();
        $table->timestamps();
    });
   }

   //Access Migration
   public function up()
   {
    Schema::create('accesses', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('user_id')->nullable();
        $table->string('fullname');
        $table->string('emp_id');
        $table->string('shift');
        $table->string('request');
        $table->longText('note')->nullable();
        $table->string('status')->default('Pending'); //Pending, Approved
        $table->string('state')->default('Open'); //Open, Assigned, Resolved, Closed
        $table->date('resolved_at')->nullable();
        $table->date('closed_at')->nullable();
        $table->integer('assigned_to')->nullable();
        $table->longText('action')->nullable();
        $table->timestamps();
    });
   }

它給了我這個錯誤

結果應該是這樣的

Answer 1

你應該使用集合的merge方法：

$accesses = Access::select(array('id', 'fullname','emp_id','shift','state','resolved_at', 'closed_at','assigned_to'))
            ->where('state','=','Assigned')
            ->get();

$reports = Report::select(array('id', 'fullname','emp_id','shift','state','resolved_at', 'closed_at','assigned_to'))
            ->where('state', '=', 'Assigned')
            ->get();

$all = $accesses->merge($reports);

Answer 2

您的問題並不完全清楚，但似乎您實際上只需要用戶名。 這是未經測試的，因為我沒有要測試的數據集。 但它應該找到具有指定的訪問權限或已分配狀態的所有用戶的名稱。

$names = DB::table('users')
    ->select('users.name')
    ->leftJoin('accesses', function ($join) {
        return $join->on('users.id', '=', 'accesses.assigned_to')
            ->where('accesses.state', '=', 'Assigned');
    })
    ->leftJoin('reports', function ($join) {
        return $join->on('users.id', '=', 'reports.assigned_to')
            ->where('reports.state', '=', 'Assigned');
    })
    ->where(function ($query) {
        return $query->whereNotNull('accesses.id')
            ->orWhereNotNull('reports.id');
    })
    ->groupBy('users.id');