sql - 如何从select语句中的2个日期中获取最后一个日期

Question

我正在尝试在过程中创建选择语句。

这是我的代码：

SELECT u.id, max(a.time), max(b.time)
FROM buy a, sell b, users u
WHERE a.user_id = u.id AND b.user_id = u.id
GROUP BY u.id;

我需要在max(a.time)和之间取最后一个日期max(b.time)。

提前致谢

Answer 1

那这个呢？

SELECT 
  u.id, 
  CASE 
    WHEN max(a.time) > max(b.time) THEN max(a.time)
    ELSE max(b.time)
  END
FROM buy a, sell b, users u
WHERE a.user_id = u.id AND b.user_id = u.id
GROUP BY u.id;

Answer 2

SELECT u.id, max(case when a.time > b.time then a.time else b.time end)
FROM buy a, sell b, users u
WHERE a.user_id = u.id AND b.user_id = u.id
GROUP BY u.id;

如果你的数据库支持别名查询，试试这个会更有效：

SELECT u.id, case when a_max > b_max then a_max else b_max end
FROM (SELECT u.id, max(a.time) as a_max, max(b.time) as b_max
    FROM buy a, sell b, users u
    WHERE a.user_id = u.id AND b.user_id = u.id
    GROUP BY u.id) x;

如果存在索引，则内部查询可以使用索引来快速获取每列的最大值。

---

当我们这样做时，让我们使用现代连接语法重新编码：

SELECT u.id, max(case when a.time > b.time then a.time else b.time end)
FROM buy a
JOIN sell b ON b.user_id = u.id
JOIN users u ON a.user_id = u.id
GROUP BY u.id;

Answer 3

SELECT U.ID, 
CASE WHEN MAX(A.TIME) > MAX(B.TIME) THEN MAX(A.TIME)
     ELSE MAX(B.TIME) 
END
FROM BUY A, SELL B, USERS U
WHERE A.USER_ID = U.ID AND B.USER_ID = U.ID
GROUP BY U.ID;