Question

我有这样的疑问:(对于复杂性道歉，我不确定在不影响问题的情况下我可以删除的内容）

SELECT COUNT(*) AS total,
SUM(o.total) AS total_loss,
SUM((SELECT SUM(cost_price) FROM `orders_items` WHERE orders_id = o.orders_id)) AS cost_total ,
SUM((SELECT COUNT(*) FROM refunds AS r1 WHERE r1.order_id = r.order_id AND NOT r.reason IS NULL)) AS refund_count ,
SUM((SELECT COUNT(*) FROM exchanges AS e1 WHERE e1.order_id = e.order_id AND e.type = :countResend AND NOT e.reason IS NULL)) AS resend_count ,
SUM((SELECT COUNT(*) FROM exchanges AS e2 WHERE e2.order_id = e.order_id AND e.type = :countExchange AND NOT e.reason IS NULL)) AS exchange_count
FROM orders AS o
JOIN sales_channel_config AS s ON o.sales_channel = s.sales_channel AND o.sub_sales_channel = s.sub_sales_channel
JOIN courier_service AS cs ON o.courier_service = cs.code
LEFT JOIN refunds AS r ON o.orders_id = r.order_id
JOIN orders_items AS oi ON o.orders_id = oi.orders_id
JOIN third_party_config AS tc ON SUBSTRING(oi.product_id_new, 3, 2) = tc.code
LEFT JOIN exchanges AS e ON o.orders_id = e.order_id
WHERE 1 = 1
AND o.tracking_num NOT IN (:cancelStatus)
AND (o.order_date >= :startDate AND o.order_date <= :endDate)
AND o.courier_service = :courier
AND SUBSTRING(oi.product_id_new, 3, 2) = :supplier
AND (NOT r.reason IS NULL OR NOT e.reason IS NULL)

我遇到的问题是各种SUM((query))条款正在计算重复订单，这很难解决。例如：

SUM((SELECT COUNT(DISTINCT r1.order_id) FROM refunds AS r1 WHERE r1.order_id = r.order_id AND NOT r.reason IS NULL)) AS refund_count ,

和

SUM((SELECT COUNT(*) FROM refunds AS r1 WHERE r1.order_id = r.order_id AND NOT r.reason IS NULL GROUP BY r1.order_id)) AS refund_count ,

根本不要降低结果SUM。我已经确认返回的数据将包含通过另一个结构相同的查询重复，该查询返回父查询中的行。如果在没有重复过滤的情况下运行其他查询，则计数会正确匹配，因此除了包含重复的订单ID之外，我确信我的问题查询是准确的。

那么有人可以建议我尝试另一种方法吗？

Answer 1

对于任何可能受益的人：

我删除了大部分选择逻辑并在orders_id上分组，这为我提供了完整准确的相关订单列表：

SELECT o.orders_id AS order_id, r.id AS refund_id, e.id AS exchange_id, e.type AS exchange_type
FROM orders AS o
JOIN sales_channel_config AS s ON o.sales_channel = s.sales_channel AND o.sub_sales_channel = s.sub_sales_channel
JOIN courier_service AS cs ON o.courier_service = cs.code
LEFT JOIN refunds AS r ON o.orders_id = r.order_id
JOIN orders_items AS oi ON o.orders_id = oi.orders_id
JOIN third_party_config AS tc ON SUBSTRING(oi.product_id_new, 3, 2) = tc.code
LEFT JOIN exchanges AS e ON o.orders_id = e.order_id
WHERE 1 = 1
AND o.tracking_num NOT IN (:cancelStatus)
AND (o.order_date >= :startDate
AND o.order_date <= :endDate)
AND o.courier_service = :courier
AND SUBSTRING(oi.product_id_new, 3, 2) = :supplier
AND (NOT r.reason IS NULL OR NOT e.reason IS NULL)
GROUP BY (o.orders_id)

我在这里咬了一口子弹。我要做一些后期处理来自己获取计数，这对我来说至少是可能的。

仍然不明白为什么在子选择中获取不同的值失败了。

聚合子查询中重复的ID的问题

1 个答案: