我是一名 JS 開(kāi)發(fā)人員，但試圖用一些 Laravel 幫助我的團(tuán)隊(duì)。我有他的查詢?nèi)缦拢?customers = ShopUser::selectRaw('shop_users.name, shop_users.email, shop_users.unique_id, shop_users.created_at, SUM(orders.total) AS total_spent, MIN(orders.created_at) AS first_purchase, MAX(orders.created_at) AS last_purchase, count(orders.id) AS total_orders')
->leftJoin('orders', 'orders.customer_id', 'shop_users.id')
->groupBy('shop_users.id')
->get();然后我想把花費(fèi)超過(guò) 1,000 美元unique_id的這個(gè)查詢的用戶total放在一個(gè)數(shù)組中。有沒(méi)有辦法在上面的查詢中做到這一點(diǎn)，還是應(yīng)該在此之后進(jìn)行單獨(dú)的迭代來(lái)排序？
2 回答
慕碼人2483693
TA貢獻(xiàn)1860條經(jīng)驗(yàn) 獲得超9個(gè)贊
我想你可以使用 havingRaw
ShopUser::selectRaw('shop_users.name, shop_users.email, shop_users.unique_id, shop_users.created_at, SUM(orders.total) AS total_spent, MIN(orders.created_at) AS first_purchase, MAX(orders.created_at) AS last_purchase, count(orders.id) AS total_orders')
->leftJoin('orders', 'orders.customer_id', 'shop_users.id')
->havingRaw('total_spent > ?', [1000])
->groupBy('shop_users.id')
->get();如果要將結(jié)果作為數(shù)組返回，可以使用 pluck
ShopUser::selectRaw('shop_users.name, shop_users.email, shop_users.unique_id, shop_users.created_at, SUM(orders.total) AS total_spent, MIN(orders.created_at) AS first_purchase, MAX(orders.created_at) AS last_purchase, count(orders.id) AS total_orders')
->leftJoin('orders', 'orders.customer_id', 'shop_users.id')
->havingRaw('total_spent > ?', [1000])
->groupBy('shop_users.id')
->get()
->pluck('unique_id')
->all()我還沒(méi)有測(cè)試過(guò)，但我希望這會(huì)有所幫助
慕蓋茨4494581
TA貢獻(xiàn)1850條經(jīng)驗(yàn) 獲得超11個(gè)贊
由于在 Laravel 中有很多方法可以實(shí)現(xiàn)這個(gè)目標(biāo)，我更喜歡 Eloquent Query Builder方法而不是Row Queries。
我假設(shè)您需要的只是一個(gè)總數(shù)超過(guò) 1,000 的唯一 ID 列表。例如：[233123、434341、35545123]。
因此，您在代碼中編寫(xiě)的任何其他選擇、分組和過(guò)濾器都將被忽略。
&shopUserIds = ShopUser::whereHas('orders', function ($query) {
$query->where(DB::row('SUM(total)'), '>', '1000');
})
->get()
->map->id
->toArray();- 2 回答
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