Question

我有一种感觉，当我得到回答时，我会感到愚蠢。我有一张客户表和一张订单表。我想要一个查询，它给我一个所有客户的列表，以及它们的最后订单信息（如果有的话），按客户名称排序。

    SELECT c.id, c.name, o.order_time, o.item_name
    FROM clients AS c LEFT JOIN(
        SELECT client_id, max(order_time) AS order_time
        FROM orders GROUP BY client_id
    ) AS o
    ON(c.id = o.client_id)
    ORDER BY UPPER(c.name)"

我的问题是，如果删除o.item_name，我会得到我想要的行，但是所写的查询是无效的，因为如果没有将o.item_name放在GROUP BY中，就无法获取o.item_name。当然，这会导致它为每个客户端返回多行。希望我的意图很明确。

Answer 1

您可以使用窗口功能执行此操作：

SELECT c.id, c.name, o.order_time, o.item_name
FROM clients AS c 
  LEFT JOIN (
    SELECT client_id, 
           item_name,
           order_time,
           row_number() over (partition by client_id order by order_time desc) as rn
    FROM orders 
  ) AS o ON c.id = o.client_id and o.rn = 1
ORDER BY UPPER(c.name);

另一个选择是使用Postgres'distinct on()运算符，它通常比使用窗口函数的解决方案更快：

SELECT c.id, c.name, o.order_time, o.item_name
FROM clients AS c 
  LEFT JOIN (
    SELECT distinct on (client_id) client_id, 
           item_name,
           order_time
    FROM orders 
    order by client_id, order_time desc
  ) AS o ON c.id = o.client_id 
ORDER BY UPPER(c.name);

Answer 2

在Postgres中，您可以使用distinct on：

SELECT DISTINCT ON (c.name) c.id, c.name, o.order_time, o.item_name
FROM clients c LEFT JOIN
     orders o
    ON c.id = o.client_id
ORDER BY UPPER(c.name), o.order_time DESC;

Postgres - 获取所有客户和最新订单

2 个答案: