Question

假设我有一个表MATCHES（OPPONENT，DATE，GOALS_FOR，GOALS_AGAINST）

如果我想要一个返回GOALS_FOR大于2的最近匹配的SQL查询，那么我可以使用

SELECT *
FROM MATCHES
WHERE GOALS_FOR > 2
AND DATE = ( 
SELECT MAX(DATE)
FROM MATCHES
WHERE GOALS_FOR > 2)

如何在不重写/重新计算

的情况下完成此操作

MATCHES
WHERE GOALS_FOR > 2

两次？

Answer 1

只需按日期desc选择前1个订单，例如tsql

SELECT top 1 *
FROM MATCHES
WHERE GOALS_FOR > 2
ORDER BY DATE Desc

（使用rownum代表oracle，限制mysql，首先获取db2等）

Answer 2

您可以简单地将其加入到每个DATE获取最新GOALS_FOR的子查询中。

SELECT  a.*
FROM    `matches` a
        INNER JOIN
        (
            SELECT  GOALS_FOR, MAX(DATE) Date
            FROM    `matches`
            WHERE   GOALS_FOR > 2
            GROUP   BY GOALS_FOR
        ) b ON  a.GOALS_FOR = b.GOALS_FOR
                AND a.Date = b.Date

Answer 3

如果你的数据库支持WITH子句，你可以使用它：

WITH
matches_over_two_goals_for as (
SELECT *
FROM matches
WHERE goals_for > 2
)
SELECT MAX(date)
FROM matches_over_two_goals_for

Answer 4

你有比赛的ID或钥匙吗？你需要GOALS_FOR＆gt;的原因2不止一次是因为在给定日期可能有多个匹配。因此，DATE比较可以提供与GOALS_FOR匹配的＆lt; 3。

如果您可以重写查询，请执行以下操作：

SELECT * FROM MATCHES WHERE MATCH_ID = ( SELECT MATCH_ID FROM MATCHES WHERE GOALS_FOR > 2 GROUP BY MATCH_ID HAVING MAX(DATE) = DATE )

你得到了照片

SQL MAX（）函数

4 个答案: