Question

我只需要获取以下查询返回的roomnumber数组：

$roomnumbers = Room::with(['floorroomcount' => function($query){
                $query->with('roomnumber')->get();
        }])->where('roomtype_id', $roomtype_id)->get();

尝试：以下采摘返回floorroomcount

$roomnumbers->pluck('floorroomcount');

但是我需要roomnumber数组，我怎么能得到？

Answer 1

这会在一个集合中为您提供所有roomnumber个结果：

$roomnumbers->pluck('floorroomcount')->collapse()->pluck('roomnumber')->collapse();

Answer 2

这是有效的，但有很多循环和直接回声，如果有什么可以简化，请告诉我：

   $roomnumbers = Room::with(['floorroomcount.roomnumber'])->where('roomtype_id', $roomtype_id)->get();

    $floorroomcounts =  $roomnumbers->pluck('floorroomcount');

    $records =  $floorroomcounts->map(function($floorroomcount, $value){

                    return $floorroomcount->pluck('roomnumber')->flatten();

                })->values()->all();        

    foreach($records as $record){

        foreach($record as $row){
            echo '<option value='.$row->id.'>'.$row->roomnumber.'</option>';
        }

    }

    //return response()->json($roomnumbers);

Answer 3

您可以这样缩短@Jonas Staudenmeir的答案：

$roomnumbers->pluck('floorroomcount.*.roomnumber.*')->collapse();

pluck('*')在此特定情况下与collapse()基本相同。

Answer 4

尝试，

$roomnumbers = Room::with(['floorroomcount' => function($query){
    $query->with('roomnumber')->get();
}])
->where('roomtype_id', $roomtype_id)
->get();

$records = $roomnumbers->map(function($element, $value){
    return $element->map(function($e, $v){
        return $e->roomnumber;
    });
})->values()->all();

map()是一个Laravel集合方法，因此您需要在控制器顶部导入集合外观，如：use Illuminate\Support\Collection;

Laravel从嵌套关系中拔出一个数组

4 个答案: