Question

以下是我的问题的简化版本。我有两张桌子。每个表都有一个唯一的ID字段，但在这种情况下它无关紧要。

shipments 有3个字段：shipment_id，receive_by_datetime和qty。

deliveries 有4个字段：delivery_id，shipment_id，delivered_on_datetime和qty。

在shipments中，shipment_id和receive_by_datetime字段始终匹配。表中有很多行看起来是基于这两列的重复行（但它们不是......其他字段不同）。

在deliveries中，shipment_id与shipments表匹配。还有许多行似乎是基于delivery_id和delivered_on_datetime字段的重复行（但它们不再存在......其他字段存在，我没有列出）。

我试图为每个聚合delivered_on_datetime和receive_by_datetime拉一行，但由于多对多关系，这很难。这些问题的某个地方的查询是否正确？

SELECT d.delivered_on_datetime, s.receive_by_datetime, SUM(d.qty)
FROM deliveries d
LEFT JOIN (
  SELECT DISTINCT s1.shipment_id, s1.receive_by_datetime
  FROM shipments s1
) s ON (s.shipment_id = d.shipment_id)
GROUP BY d.delivered_on_datetime, s.receive_by_datetime

Answer 1

如果总SUM(d.qty)大于SELECT SUM(qty) FROM deliveries

的值，您将遇到问题

这样的事情可能更适合你：

SELECT d.delivered_on_datetime, s.receive_by_datetime, SUM(d.qty) AS delivered_qty, SUM(d.qty) AS shipped_qty
FROM deliveries d
LEFT JOIN (
  SELECT s1.shipment_id, s1.receive_by_datetime, SUM(s1.qty) AS qty
  FROM shipments s1
  GROUP BY s1.shipment_id, s1.received_by_datetime
) s ON (s.shipment_id = d.shipment_id)
GROUP BY d.delivered_on_datetime, s.receive_by_datetime

如果您以某种方式拥有（或可能拥有）具有shipment_id的多个值的received_by_datetime，并且最佳做法是假设其他内容可能会略微损坏数据，那么就要防止其中的行deliveries表被复制，同时仍然返回您可以使用的有效结果：

SELECT d.delivered_on_datetime, s.receive_by_datetime, SUM(d.qty) AS delivered_qty, SUM(d.qty) AS shipped_qty
FROM deliveries d
LEFT JOIN (
  SELECT s1.shipment_id, MAX(s1.receive_by_datetime) AS receive_by_datetime, SUM(s1.qty) AS qty
  FROM shipments s1
  GROUP BY s1.shipment_id
) s ON (s.shipment_id = d.shipment_id)
GROUP BY d.delivered_on_datetime, s.receive_by_datetime

Answer 2

是的，多对多的问题是你得到行的笛卡尔积，所以你不止一次计算同一行。每隔一行匹配一次。

在货件中，shipment_id和receive_by_datetime字段始终匹配

如果这意味着不能有两个具有相同ID但日期不同的货件，那么您的查询将起作用。但总的来说它并不安全。即如果subselect distinct可以为每个货件ID返回多于一行，则您将受到重复计算问题的影响。一般来说，这是一个非常棘手的问题需要解决 - 实际上我认为这种数据模型无法实现。

如何在多对多关系中加入表？

2 个答案: