使用 laravel builder 同时为子查询、过滤器、计数创建查询答案

【问题标题】：Creating query with laravel builder for Subquery,filter,count at the same使用 laravel builder 同时为子查询、过滤器、计数创建查询
【发布时间】：2018-09-14 11:56:35
【问题描述】：

我有一个这样的查询。我试图将其转换为查询生成器样式。但是因为它太复杂了，所以我遇到了麻烦。

我喜欢使用 whereIn 。但是，如果我使用原始数据库查询，我必须自己编写代码（获取集合放置逗号等），以便我正在寻找一种使其对 laravel 查询构建器友好的方法。

SELECT
    total_amount_until_now/total_orders_until_now as avg_order_value_now,
    total_amount_until_a_month_ago/total_orders_until_a_month_ago as avg_order_value_until_a_month_ago

FROM(
SELECT
  COUNT(DISTINCT a.order_id)
    FILTER (WHERE CURRENT_DATE >= b.order_creation_date AND b.seller_id IN (1, 3))                     AS total_orders_until_now,
  SUM(invoice_amount)
    FILTER (WHERE CURRENT_DATE >= b.order_creation_date AND b.seller_id IN (1, 3))                     AS total_amount_until_now,
  COUNT(DISTINCT a.order_id)
    FILTER (WHERE CURRENT_DATE - INTERVAL '1 month' > b.order_creation_date AND b.seller_id IN (1, 3)) AS total_orders_until_a_month_ago,
  SUM(invoice_amount)
    FILTER (WHERE CURRENT_DATE - INTERVAL '1 month' > b.order_creation_date AND b.seller_id IN (1, 3)) AS total_amount_until_a_month_ago
FROM
  order_items a
  INNER JOIN orders b ON a.order_id = b.order_id)  xyz"

这是我的尝试，我无法成功

DB::select()
        ->selectSub(`total_amount_until_now`, `avg_order_value_now`)
        ->selectSub(`/`, `avg_order_value_now`)
        ->selectSub(`total_orders_until_now`, `avg_order_value_now`)
        ->selectSub(`total_amount_until_a_month_ago`, `avg_order_value_until_a_month_ago`)
        ->selectSub(`/`, `avg_order_value_until_a_month_ago`)
        ->selectSub(`total_orders_until_a_month_ago`, `avg_order_value_until_a_month_ago`)
        ->from(`SELECT COUNT(DISTINCT a.order_id) FILTER ( WHERE CURRENT_DATE >= b.order_creation_date AND b.seller_id IN (1, 3) ) AS total_orders_until_now, SUM(invoice_amount) FILTER ( WHERE CURRENT_DATE >= b.order_creation_date AND b.seller_id IN (1, 3) ) AS total_amount_until_now, COUNT(DISTINCT a.order_id) FILTER ( WHERE CURRENT_DATE - INTERVAL 1 month > b.order_creation_date AND b.seller_id IN (1, 3) ) AS total_orders_until_a_month_ago, SUM(invoice_amount) FILTER ( WHERE CURRENT_DATE - INTERVAL 1 month > b.order_creation_date AND b.seller_id IN (1, 3) ) AS total_amount_until_a_month_ago FROM order_items a INNER JOIN orders b ON a.order_id = b.order_id as xyz`)
        ->get();

【问题讨论】：

抱歉，这太糟糕了，无法阅读。有没有办法只在 laravel 中使用纯 SQL？我保证你的数据库索引运行的速度比 PHP 中运行的任何东西都快。

标签： mysql laravel postgresql

【解决方案1】：

$from = DB::table('order_items AS a')
    ->selectRaw('COUNT(DISTINCT a.order_id) FILTER (WHERE CURRENT_DATE >= b.order_creation_date AND b.seller_id IN (?, ?)) AS total_orders_until_now', [1, 3])
    ->selectRaw('SUM(invoice_amount) FILTER (WHERE CURRENT_DATE >= b.order_creation_date AND b.seller_id IN (?, ?)) AS total_amount_until_now', [1, 3])
    ->selectRaw("COUNT(DISTINCT a.order_id) FILTER (WHERE CURRENT_DATE - INTERVAL '1 month' > b.order_creation_date AND b.seller_id IN (?, ?)) AS total_orders_until_a_month_ago", [1, 3])
    ->selectRaw("SUM(invoice_amount) FILTER (WHERE CURRENT_DATE - INTERVAL '1 month' > b.order_creation_date AND b.seller_id IN (?, ?)) AS total_amount_until_a_month_ago", [1, 3])
    ->join('orders AS b', 'a.order_id', '=', 'b.order_id');
DB::query()
    ->selectRaw('total_amount_until_now/total_orders_until_now as avg_order_value_now')
    ->selectRaw('total_amount_until_a_month_ago/total_orders_until_a_month_ago as avg_order_value_until_a_month_ago')
    ->fromSub($from, 'xyz')
    ->get();

您可能还可以对'1 month' 使用绑定（?），但我不完全确定。你必须尝试一下。

【讨论】：