Question

我有一张大约有800万行的表格。行包含ID，日期和事件代码。我想选择事件代码等于1的所有ID和日期，并且在过去的某个时间有一个等于2的事件代码。

例如，我的表格如下：

ID    Date       Code
----------------------
1     4/16/2016   6
1     4/10/2016   1
1     3/1/2016    13
1     1/26/2016   2
2     5/2/2016    8
2     3/14/2016   1
2     1/13/2016   14

我希望ID = 1并返回Date = 4/10/2016，但我不希望返回ID = 2的任何内容，因为ID = 2从未有过等于2的事件代码。

我应该如何编写SELECT语句来获得这些结果？

Answer 1

您可以使用exists。

select *
from t 
where code = 1 and
      exists (select 1 from t t1 where t.id = t1.id and t.dt > t1.dt and t1.code=2)

Answer 2

如果您只想为每个ID选择最长日期：

WITH Cte AS(
    SELECT *,
        rn = ROW_NUMBER() OVER(PARTITION BY ID ORDER BY Date DESC)
    FROM tbl
    WHERE Code = 1
)
SELECT
    ID, Date, Code
FROM Cte c
WHERE
    rn = 1
    AND EXISTS(
        SELECT 1
        FROM tbl t
        WHERE
            t.ID = c.ID
            AND t.Date < c.Date
            AND t.Code = 2
    )
;

ONLINE DEMO

使用MAX，GROUP BY和HAVING：

SELECT
    ID, Date = MAX(Date)
FROM tbl  t1
WHERE Code = 1
GROUP BY t1.ID
HAVING MAX(Date) > (SELECT Date FROM tbl t2 WHERE t2.ID = t1.ID AND t2.Code = 2)

ONLINE DEMO

Answer 3

select id
, max(date)
from table to
where to.code = 1
and exists (select 1 from table ti where ti.id = to.id AND ti.code = 2)
group by id

Answer 4

@vkp有最好的答案。但是，这是一种使用窗口函数的方法：

select t.id, t.code, t.dt
from (select t.*, min(case when code = 2 then dt end) over (partition by id) as dt_2
      from t
      where code in (1, 2)
     ) t
where t.code = 1 and dt_2 < dt;

SQL Server 2012根据多个条件选择最大日期

4 个答案: