使用布尔条件计算具有条件 sql 的月份

using a boolean condition count the months that have the condition sql

所以我有一个 table 这样的:

---id---datetime---month---active
    1   2016-12-01  Dec-16  0
    1   2016-12-02  Dec-16  1
    1   2017-01-28  Jan-17  1
    1   2017-02-03  Feb-17  0
    1   2017-02-24  Feb-17  0
    1   2017-03-05  Mar-17  0
    1   2017-03-24  Mar-17  1
    1   2017-04-02  Apr-17  1
    1   2017-04-25  Apr-17  1
    1   2017-05-02  May-17  1
    1   2017-05-28  May-17  0

我想要这样的结果:

---id---monthCount---Active
    1       1          0
    1       2          1
    1       2          0
    1       3          1
    1       1          0

table确实有超过1个id

现在我只是使用分区获取最小日期和最大日期之间的差异,并按 idactive 排序并按 datetime 排序,但这给了我从它第一次变为 1 到它最终变为 0 之间的月份。我希望它通过 active.

的每次更改详细分开

我怎样才能做到这一点?

您可以使用行号的差异来识别组(这是一个间隙和孤岛问题)。那么最终的解决方案需要count(distinct):

select id, count(distinct month) as monthcount, active
from (select t.*,
             row_number() over (partition by id order by month, active) as seqnum_i,
             row_number() over (partition by id, active order by month) as seqnum_ia
      from t
     ) t
group by id, active, (seqnum_i - seqnum_ia);

这是假设数据按 id、月份、活动排序 -- 这是问题中数据的排序。最好有一个单独的列来明确定义行的排序顺序。

在 sql 服务器 2012+ 中,您可以像这样使用 LAG

DECLARE @SampleData AS TABLE
(
    id int,
    [datetime] datetime,
    active bit
)


;WITH temp AS
(
    SELECT   sd.id,
          sd.active, 
          sd.[datetime],
          lag(sd.active) over(PARTITION BY sd.id ORDER BY sd.datetime) AS previousActive,
          lag(sd.id) over( ORDER BY sd.id, sd.datetime) AS previousId
    FROM @SampleData sd
)
,temp1 AS
(
    SELECT *, 
          sum(CASE WHEN t.previousActive IS NULL OR t.previousActive != t.active OR t.id != t.previousId THEN 1 
             ELSE 0 END) 
          OVER(PARTITION BY t.id ORDER BY t.[datetime]) AS groupid
    FROM temp t
)
SELECT t.id, count(DISTINCT month(t.datetime)) AS monthCount, t.active  
FROM temp1 t
GROUP BY id, t.groupid, t.active

演示 link:http://rextester.com/SJZ16279