是否可以在 Laravel 的 map 函数中应用条件？

问题描述

我想在 foodOrders 中应用一些条件。是否可以在 map 函数中应用条件作为 where $foodOrder->foodOrders->where('invoice_id',null)->get();

 public function getTableList(Request $request){
    $skip =$request->skip;
    $limit=$request->limit;
    $totaltable = Table::get()->count();

    $table = Table::skip($skip)->take($limit)->orderBy('id','DESC')->get();
    $table->map(function($foodOrder){
        $foodOrder->foodOrders;
    });
}

以下是此查询返回的输出。但是我只想要invoice_id为空的数据

{
  "success": true,"message": "Lists of Table.","data": [
    {
      "id": 2,"table_number": "TN02","food_orders": [
        {
          "id": 16,"food_items_id": 1,"table_id": 2,"invoice_id": null,"quantity": 2,"price": "2000.00"
         },{
          "id": 17,"food_items_id": 2,"quantity": 3,"price": "150.00"
        }
      ]
    },{
      "id": 1,"table_number": "TN01","created_at": "2020-10-25 10:44:31","updated_at": "2020-10-25 10:44:31","food_orders": [
        {
          "id": 14,"table_id": 1,"invoice_id": 39,"quantity": 1,"price": "2000.00"
        }
      ]
    }
  ]
}

解决方法

不是最好的方法，但这应该可以

$table->map(function($foodOrder){
    $foodOrder->foodOrders=$foodOrder->foodOrders()->where('invoice_id',null)->get();
});

注意范围，您需要它们来应用 where 条件。我们说 = 以便传播进一步的更改。

,

我认为您需要使用的是 filter 而不是 map。还是我错了？

,

您可以过滤掉 food_orders 的白名单记录。

$table->map(function($record){
    $record->food_orders = collect($record->food_orders)
        ->filter(fn($foodOrder) => is_null($foodOrder->invoice_id));

   return $record;
});

从上面显示的输出数据来看，food_orders 似乎是模型上的 hasMany 关系，因此在急切加载时很容易过滤掉该关系

假设 Table 模型有一个为 food_orders 定义的关系

class Table extends Model
{
    public function food_orders()
    {
        return $this->hasMany(FoodOrder::class);
    }

    //... rest of the class code
}

可以在预先加载时约束相关模型

$table = Table::with([
        'food_orders' => fn($query) => $query->whereNull('invoice_id')
    ])
    ->skip($skip)
    ->take($limit)
    ->orderBy('id','desc')
    ->get();