*编辑（希望更清楚）
下面的表格，我想计算ID并计算重复的ID，其中createddate的ID为3个月或更长的差距。
查询我到目前为止......
if object_id('tempdb..#temp') is not null
begin drop table #temp end
select
top 100
a.id, a.CreatedDate
into #temp
from tbl a
where 1=1
--and year(CreatedDate) = '2015'
if object_id('tempdb..#temp2') is not null
begin drop table #temp2 end
select t.id, count(t.id) as Total_Cnt
into #temp2
from #temp t
group by id
select distinct #temp2.Total_Cnt, #temp2.id, #temp.CreatedDate, DENSE_RANK() over (partition by #temp.id order by createddate) RK
from #temp2
inner join #temp on #temp2.id = #temp.id
where 1=1
order by Total_Cnt desc
结果：
Total_cnt id createddate rk
3 1 01-01-2015 1
3 1 03-02-2015 2
3 1 01-02-2015 3
2 2 05-01-2015 1
2 2 05-02-2015 2
1 3 06-01-2015 1
1 4 07-01-2015 1
计算ID并且仅当id的createddate大于3个月时才计算重复ID。
像这样......
Total_cnt id Countwith3monthgap
3 1 2
2 2 1
1 3 1
1 4 1
答案 0 :(得分：1)
您可以使用cte和ROW_NUMBER获取订单，并根据订单自行加入cte ..
WITH cte AS
( SELECT
*,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY CreatedDate) Rn
FROM
Test
)
SELECT
c1.ID,
COUNT(CASE WHEN c2.CreatedDate IS NULL THEN 1
WHEN c1.CreatedDate >= DATEADD(month,3,c2.CreatedDate) THEN 1
END)
FROM
cte c1
LEFT JOIN cte c2 ON c1.ID = c2.ID
AND c1.RN = c2.RN + 1
GROUP BY
c1.ID
您还需要使用条件计数，其中Previous CreatedDate为null或者Current CreatedDate为＆gt; = Previous CreatedDate + 3个月
如果你碰巧使用的是SQL 2012+，你也可以在这里使用LAG来获得相同的结果
SELECT
ID,
COUNT(*)
FROM
(SELECT
ID,
CreatedDate CurrentDate,
LAG(CreatedDate) OVER (PARTITION BY ID ORDER BY CreatedDate) PreviousDate
FROM
Test
) T
WHERE
PreviousDate IS NULL
OR CurrentDate >= DATEADD(month, 3, PreviousDate)
GROUP BY
ID
答案 1 :(得分：0)
您可以使用滞后来获取上一个日期，Null为列表中的第一个
SELECT
id,
lag(CreatedDate,1) OVER (PARTITION BY Id ORDER BY CreatedDate) AS PreviousCreateDate,
CreatedDate
FROM @t
您可以将其用作子查询，并使用DATEDIFF
获取月份差异SELECT sub.id,DATEDiff(month, sub.PreviousCreateDate ,sub.CreatedDate)
FROM (SELECT
id,
lag(CreatedDate,1) OVER (PARTITION BY Id ORDER BY CreatedDate) AS PreviousCreateDate,
CreatedDate
FROM @t) sub
WHERE DATEDiff(month, sub.PreviousCreateDate ,sub.CreatedDate) >=3
OR sub.PreviousCreateDate IS NULL
然后你可以拿走你的总数
SELECT sub.id,COUNT(sub.id) as cnt
FROM (SELECT
id,
lag(CreatedDate,1) OVER (PARTITION BY Id ORDER BY CreatedDate) AS PreviousCreateDate,
CreatedDate
FROM @t) sub
WHERE DATEDIFF(month, sub.PreviousCreateDate ,sub.CreatedDate) >=3
OR sub.PreviousCreateDate IS NULL
GROUP BY sub.id
请注意，在1月的最后一天使用datediff是在游行的第一天之前的三个月。这似乎是你追求的逻辑。
您可能希望将三个月的差距标准定义为
WHERE sub.PreviousCreateDate <= DATEADD(month, -3, sub.CreatedDate)
OR sub.PreviousCreateDate IS NULL
或
WHERE sub.CreatedDate >= DATEADD(month, +3, sub.PreviousCreateDate )
OR sub.PreviousCreateDate IS NULL
答案 2 :(得分：0)
我猜你所希望的三个月差距的定义与datediff()不一致。这里的大多数逻辑是回顾前一个日期，并确定差距是否足以达到合格。
当datediff()计算三个月差异时，我们仍需要确保一个月中的某一天晚于第一天（每个示例和ID 5 ）。如果差异超过三个月，那么我们自动就好了。
但我也假设您希望将11月30日至2月28日（或闰年29日）的距离视为整整三个月，因为结束日期是在该月的最后一天。通过将结束日期调整一天，这是一个容易出现的情况，因为它会将日期推迟到下个月，并将月差异也增加一个。如果这不是您想要的，那么只需删除dateadd(day, 1, ...)部分，并仅使用原始CreatedDate值。
您的样本数据是有限的，因此我也假设在连续日期之间测量差距。如果您想要找到整个集合中不超过三个月的运行块，那么这是一个不同的问题，您应该提供更多信息。
由于您已经表明您可能在SQL Server 2008上，因此您将不得不使用lag()功能。虽然可以调整第一个查询，但最后可能更容易使用第二种方法。
with diffs as (
select
ID,
row_number() over (partition by ID order by CreatedDate) as RN,
case when
datediff(
month,
lag(CreatedDate, 1) over (partition by ID order by CreatedDate),
CreatedDate
) = 3
and
datepart(
day,
lag(CreatedDate, 1) over (partition by ID order by CreatedDate)
) <= datepart(day, CreatedDate)
or
datediff(
month,
lag(CreatedDate, 1) over (partition by ID order by CreatedDate),
/* adding one day to handle gaps like Nov30 - Feb28/29 and Jan31 - Apr30 */
dateadd(day, 1, CreatedDate)
) >= 4
then 1
else 0
end as GapFlag
from <T> /* <--- your table name here */
), gaps as (
select
ID, RN,
sum(1 + GapFlag) over (partition by ID order by RN) as Counter
from diffs
)
select ID, count(distinct Counter - RN) as "Count"
from gaps
group by ID
其余逻辑是典型的间隙和岛屿场景，寻找sum(1 + GapCount)序列中的漏洞，偏移量为1，与row_number()非常相似。
http://sqlfiddle.com/#!6/61b12/3
JamieD77的方法也有效。我原先认为你的问题不仅仅是按顺序查看行。这是我如何根据我一直在运行的差距定义进行调整：
with data as (
select ID, CreatedDate, row_number() over (partition by ID order by CreatedDate) as RN
from T
)
select ID, count(*) as "Count"
from data d1 left outer join data d0
on d0.ID = d1.ID and d0.RN = d1.RN - 1 /* connect to the one before */
where
datediff(month, d0.CreatedDate, d1.CreatedDate) = 3
and datepart(day, d0.CreatedDate) <= datepart(day, d0.CreatedDate)
or datediff(month, d0.CreatedDate, dateadd(day, 1, d0.CreatedDate)) >= 4
or d0.ID is null
group by ID
修改：自昨天以来，您已经更改了问题。
在第一个查询中更改此行以包含总计数：
...
select count(*) as TotalCnt, ID, count(distinct Counter - RN) as GapCount
...
第二个看起来像：
with data as (
select ID, CreatedDate, row_number() over (partition by ID order by CreatedDate) as RN
from T
)
select
count(*) as TotalCnt, ID,
count(case when
datediff(month, d0.CreatedDate, d1.CreatedDate) = 3
and datepart(day, d0.CreatedDate) <= datepart(day, d0.CreatedDate)
or datediff(month, d0.CreatedDate, dateadd(day, 1, d0.CreatedDate)) >= 4
or d0.ID is null then 1 end
) as GapCount
from data d1 left outer join data d0
on d0.ID = d1.ID and d0.RN = d1.RN - 1 /* connect to the one before */
where
group by ID