mysql - 跨多行的 WHERE 条件

Question

我有这张 table ...

--------------------------------------
| user_id |  status   |  status_date |
--------------------------------------
|    1    |  Current  |  2012-08-01  |
|    1    |  Referral |  2012-03-14  |
|    2    |  Referral |  2012-04-23  |
|         |           |              |
--------------------------------------

我如何查询才能找到推荐日期在 2012-06-30 之前且当前日期在 2012-06-30 之后或根本没有当前状态记录的不同 user_id？

数据库是MySQL。

Answer 1

您可以使用 LEFT JOIN 执行此操作:

SELECT  DISTINCT T.User_ID
FROM    T
        LEFT JOIN T T2
            ON t.User_ID = T2.User_ID
            AND t2.Status = 'Current'
WHERE   T.Status_Date < '20120630'
AND     T.Status = 'Referral'
AND     (t2.Status_Date > '20120630' OR t2.Status_date IS NULL)

或者，将 GROUP BY 与 HAVING 和 COUNT(CASE ...) 一起使用

SELECT  t.User_ID
FROM    T
GROUP BY t.user_ID
HAVING  COUNT(CASE WHEN t.Status = 'Referral' AND t.Status_Date < '20120630' THEN 1 END) > 0
AND (   COUNT(CASE WHEN t.Status = 'Current' AND t.Status_Date > '20120630' THEN 1 END) > 0
    OR  COUNT(CASE WHEN t.Status = 'Current' THEN 1 ELSE 0 END) = 0
    )

这取决于你的索引和数据量，哪个表现更好，我想在大多数情况下它会是前者