Question

我想总结一个嵌套的集合。

我有以下表格：

场地
优惠
订单

venue可以包含多个offers，offer可以有多个orders。

示例数据可能是：

场地

id        |  name
==========================
5         |  Pizza 4 Less
10        |  Poundland

优惠

id | venue_id | name
==================================
17 | 5        | Buy 1 get one free 
24 | 5        | 30% off pizza
32 | 10       | 50% off
50 | 10       | 20% off

订单

id | offer_id | bill | paid
===========================
85 | 17       | 15   | true
86 | 17       | 20   | true
87 | 17       | 90   | true
88 | 24       | 14   | true
89 | 32       | 15   | true
90 | 32       | 65   | true
91 | 50       | 24   | true
92 | 50       | 1000 | false

我想使用Laravel Elqouent模型来获取每个场地的总金额。因此，对于上述数据，我希望得到以下结果：

id | name          | total_paid
===============================
5  | Pizza 4 Less  | 139
10 | Poundland     | 104

请注意，总计不包括未支付的订单（即订单92）

我目前的做法如下：

$venues = Venue::with(['offers.orders' => function ($query) {
        $query->where('paid', '=', true);
    }])
    ->get();

$totals = [];
foreach ($venues as $venue) {
    $totalPaid = 0;
    foreach ($venue->offers as $offer) {
        $totalPaid += $offer->orders->sum('bill');
    }
    $totals[$venue->name] = $totalPaid;
}

如您所见，上面的代码效率低且时间长。

有更好的方法吗？

Answer 1

清理版本，但效率不高

// In your Venue Model
public function getTotalPaidAttribute()
{
    return $this->offers->sum('TotalPaid');
}

// In your Offer Model
protected $appends = ['totalPaid'];

public function getTotalPaidAttribute()
{
    return $this->orders->sum('paid');
}

// use it : 
foreach($venues as $venue){
     //do what you want with it
     $venue->totalPaid;
}

（编辑）的

正如评论中所说，这种方法可能更清洁，但效率不高：

高效的方式：

// In your Venue Model
public function orders(){
    return $this->hasManyThrough(Order::class, Offer::class)
        ->selectRaw('sum(paid) as aggregate, venue_id')
        ->groupBy('venue_id');
}

public function totalPaid(){
    if ( ! array_key_exists('orders', $this->relations)) $this->load('orders');
    $relation = $this->getRelation('orders');

     return ($relation) ? $relation->aggregate : 0;
}

public function getTotalPaidAttribute()
{
    return $this->totalPaid();
}

也许我搞砸了你的钥匙，你可能不得不使用关系的完整声明：

return $this->hasManyThrough(
        Order::class, Offer::class
        'venue_id', 'offer_id', 'id'
    );

但我刚刚完成了我的项目，它就像一个魅力

Laravel：用嵌套关系

1 个答案: