在 SQL 服务器中查找非连续日期
Find Non Consecutive date in SQL Server
我想找到两个连续日期之间的missing NON-consecutive dates
。
我正在发布我的 SQL 查询和临时表以找出结果。
但是我没有得到正确的结果
这是我的SQL查询
drop table #temp
create table #temp(an varchar(20),dt date)
insert into #temp
select '2133783715' , '2016-10-16' union all
select '5107537880' , '2016-10-15' union all
select '6619324250' , '2016-10-15' union all
select '7146586717' , '2016-10-15' union all
select '7472381321' , '2016-10-12' union all
select '7472381321' , '2016-10-13' union all
select '7472381321' , '2016-10-14' union all
select '7472381321' , '2016-10-24' union all
select '8186056340' , '2016-10-15' union all
select '9099457123' , '2016-10-12' union all
select '9099457123' , '2016-10-13' union all
select '9099457123' , '2016-10-14' union all
select '9099457123' , '2016-10-23' union all
select '9099457123' , '2016-11-01' union all
select '9099457123' , '2016-11-02' union all
select '9099457123' , '2016-11-03' union all
select '9165074784' , '2016-10-16'
drop table #final
SELECT an,MIN(dt) AS MinDate,MAX(dt) AS MaxDate, COUNT(*) AS ConsecutiveUsage
--DateDiff(Day,LAG(MAX(dt)) OVER (partition by an ORDER BY an),MAX(dt)) nonusageDate
into #final
FROM(
SELECT an,dt,
DATEDIFF(D, ROW_NUMBER() OVER(partition by an ORDER BY dt),dt) AS Diff
FROM #temp c
)P
GROUP BY an,diff
select * from #final order by 1
an MinDate MaxDate ConsecutiveUsage
2133783715 2016-10-16 2016-10-16 1
5107537880 2016-10-15 2016-10-15 1
6619324250 2016-10-15 2016-10-15 1
7146586717 2016-10-15 2016-10-15 1
7472381321 2016-10-12 2016-10-14 3
7472381321 2016-10-24 2016-10-24 1
7472381321 2016-10-27 2016-10-28 1
8186056340 2016-10-15 2016-10-15 1
9099457123 2016-10-12 2016-10-14 3
9099457123 2016-10-23 2016-10-23 1
9165074784 2016-10-16 2016-10-16 1
但我想要未使用日期的结果。
我想获取那些自 10 天以来未连续使用的 AN。
所以这里输出应该是这样的:-
an minusagesdate maxusagedate ConsecutiveNotUseddays
7472381321 2016-10-15 2016-10-23 9
7472381321 2016-10-25 2016-10-26 2
9099457123 2016-10-15 2016-10-22 8
所以我只想找出连续未使用的日期计数及其最小和最大日期。
试试这个:
with ranked as (
select f1.*,
ROW_NUMBER() over(partition by an order by dt) rang
from #temp f1
where exists
(select * from #temp f2
where f1.an=f2.an and datediff( day, f2.dt, f1.dt) >1
)
)
select an, minusagesdate, maxusagesdate, ConsecutiveNotUseddays
from (
select f1.*,
DATEADD(DAY,1, (select f2.dt from ranked f2 where f1.an=f2.an and f2.rang+1=f1.rang)) minusagesdate ,
DATEADD(DAY,-1, f1.dt) maxusagesdate ,
datediff( day, (select f2.dt from ranked f2 where f1.an=f2.an and f2.rang+1=f1.rang), f1.dt) - 1 ConsecutiveNotUseddays
from ranked f1
) tmp
where tmp.ConsecutiveNotUseddays>0
或者像这样
with ranked as (
select f1.*,
ROW_NUMBER() over(partition by an order by dt) rang
from #temp f1
where exists
(select * from #temp f2
where f1.an=f2.an and datediff( day, f2.dt, f1.dt) >1
)
)
select f1.an,
DATEADD(DAY,1, f3.dtbefore) minusagesdate ,
DATEADD(DAY,-1, f1.dt) maxusagesdate ,
datediff( day, f3.dtbefore, f1.dt) - 1 ConsecutiveNotUseddays
from ranked f1
outer apply
(
select top 1 f2.dt as dtbefore from ranked f2
where f1.an=f2.an and f2.rang+1=f1.rang
) f3
where datediff( day, f3.dtbefore, f1.dt) - 1>0
您似乎在计算每个 an
的 mindate 和 maxdate 之间未使用的天数。如果是这样,那么这应该可以解决问题:
select an, min(dt) as min_dt, max(dt) as max_dt
, count(distinct dt) as daysused --this counts each day used, but only once
, datediff(day,min(dt),max(dt)) as totaldays --this is the total number of days between min and max date
, datediff(day,min(dt),max(dt)) - count(distinct dt) as daysnotused
--This takes total days - used days to give non-used days
from #temp c
group by an
having datediff(day,min(dt),max(dt)) - count(distinct dt) >= 10
据我所知,您需要这个:
;WITH cte AS (
SELECT an,
dt,
ROW_NUMBER() OVER (PARTITION BY an ORDER BY dt) as rn
FROM #temp
)
SELECT c1.an,
c1.dt MinDate,
c2.dt MaxDate,
DATEDIFF(day,c1.dt,c2.dt) as ConsecutiveNotUseddays
FROM cte c1
INNER JOIN cte c2
ON c1.an = c2.an AND c1.rn = c2.rn-1
WHERE DATEDIFF(day,c1.dt,c2.dt) >= 10
输出:
an MinDate MaxDate ConsecutiveNotUseddays
7472381321 2016-10-14 2016-10-24 10
对于 9099457123
,我在 ConsecutiveNotUseddays
中得到了两行 9
。您可以检查删除 WHERE 语句的结果。
在任何较新版本的 SQL 服务器上,这应该很容易:
with x as (
select *, lag(dt) over(partition by an order by dt) dt_lag
from #temp
)
select *, datediff(day, dt_lag, dt)
from x
where datediff(day, dt_lag, dt) >= 10
我想找到两个连续日期之间的missing NON-consecutive dates
。
我正在发布我的 SQL 查询和临时表以找出结果。
但是我没有得到正确的结果
这是我的SQL查询
drop table #temp
create table #temp(an varchar(20),dt date)
insert into #temp
select '2133783715' , '2016-10-16' union all
select '5107537880' , '2016-10-15' union all
select '6619324250' , '2016-10-15' union all
select '7146586717' , '2016-10-15' union all
select '7472381321' , '2016-10-12' union all
select '7472381321' , '2016-10-13' union all
select '7472381321' , '2016-10-14' union all
select '7472381321' , '2016-10-24' union all
select '8186056340' , '2016-10-15' union all
select '9099457123' , '2016-10-12' union all
select '9099457123' , '2016-10-13' union all
select '9099457123' , '2016-10-14' union all
select '9099457123' , '2016-10-23' union all
select '9099457123' , '2016-11-01' union all
select '9099457123' , '2016-11-02' union all
select '9099457123' , '2016-11-03' union all
select '9165074784' , '2016-10-16'
drop table #final
SELECT an,MIN(dt) AS MinDate,MAX(dt) AS MaxDate, COUNT(*) AS ConsecutiveUsage
--DateDiff(Day,LAG(MAX(dt)) OVER (partition by an ORDER BY an),MAX(dt)) nonusageDate
into #final
FROM(
SELECT an,dt,
DATEDIFF(D, ROW_NUMBER() OVER(partition by an ORDER BY dt),dt) AS Diff
FROM #temp c
)P
GROUP BY an,diff
select * from #final order by 1
an MinDate MaxDate ConsecutiveUsage
2133783715 2016-10-16 2016-10-16 1
5107537880 2016-10-15 2016-10-15 1
6619324250 2016-10-15 2016-10-15 1
7146586717 2016-10-15 2016-10-15 1
7472381321 2016-10-12 2016-10-14 3
7472381321 2016-10-24 2016-10-24 1
7472381321 2016-10-27 2016-10-28 1
8186056340 2016-10-15 2016-10-15 1
9099457123 2016-10-12 2016-10-14 3
9099457123 2016-10-23 2016-10-23 1
9165074784 2016-10-16 2016-10-16 1
但我想要未使用日期的结果。
我想获取那些自 10 天以来未连续使用的 AN。
所以这里输出应该是这样的:-
an minusagesdate maxusagedate ConsecutiveNotUseddays
7472381321 2016-10-15 2016-10-23 9
7472381321 2016-10-25 2016-10-26 2
9099457123 2016-10-15 2016-10-22 8
所以我只想找出连续未使用的日期计数及其最小和最大日期。
试试这个:
with ranked as (
select f1.*,
ROW_NUMBER() over(partition by an order by dt) rang
from #temp f1
where exists
(select * from #temp f2
where f1.an=f2.an and datediff( day, f2.dt, f1.dt) >1
)
)
select an, minusagesdate, maxusagesdate, ConsecutiveNotUseddays
from (
select f1.*,
DATEADD(DAY,1, (select f2.dt from ranked f2 where f1.an=f2.an and f2.rang+1=f1.rang)) minusagesdate ,
DATEADD(DAY,-1, f1.dt) maxusagesdate ,
datediff( day, (select f2.dt from ranked f2 where f1.an=f2.an and f2.rang+1=f1.rang), f1.dt) - 1 ConsecutiveNotUseddays
from ranked f1
) tmp
where tmp.ConsecutiveNotUseddays>0
或者像这样
with ranked as (
select f1.*,
ROW_NUMBER() over(partition by an order by dt) rang
from #temp f1
where exists
(select * from #temp f2
where f1.an=f2.an and datediff( day, f2.dt, f1.dt) >1
)
)
select f1.an,
DATEADD(DAY,1, f3.dtbefore) minusagesdate ,
DATEADD(DAY,-1, f1.dt) maxusagesdate ,
datediff( day, f3.dtbefore, f1.dt) - 1 ConsecutiveNotUseddays
from ranked f1
outer apply
(
select top 1 f2.dt as dtbefore from ranked f2
where f1.an=f2.an and f2.rang+1=f1.rang
) f3
where datediff( day, f3.dtbefore, f1.dt) - 1>0
您似乎在计算每个 an
的 mindate 和 maxdate 之间未使用的天数。如果是这样,那么这应该可以解决问题:
select an, min(dt) as min_dt, max(dt) as max_dt
, count(distinct dt) as daysused --this counts each day used, but only once
, datediff(day,min(dt),max(dt)) as totaldays --this is the total number of days between min and max date
, datediff(day,min(dt),max(dt)) - count(distinct dt) as daysnotused
--This takes total days - used days to give non-used days
from #temp c
group by an
having datediff(day,min(dt),max(dt)) - count(distinct dt) >= 10
据我所知,您需要这个:
;WITH cte AS (
SELECT an,
dt,
ROW_NUMBER() OVER (PARTITION BY an ORDER BY dt) as rn
FROM #temp
)
SELECT c1.an,
c1.dt MinDate,
c2.dt MaxDate,
DATEDIFF(day,c1.dt,c2.dt) as ConsecutiveNotUseddays
FROM cte c1
INNER JOIN cte c2
ON c1.an = c2.an AND c1.rn = c2.rn-1
WHERE DATEDIFF(day,c1.dt,c2.dt) >= 10
输出:
an MinDate MaxDate ConsecutiveNotUseddays
7472381321 2016-10-14 2016-10-24 10
对于 9099457123
,我在 ConsecutiveNotUseddays
中得到了两行 9
。您可以检查删除 WHERE 语句的结果。
在任何较新版本的 SQL 服务器上,这应该很容易:
with x as (
select *, lag(dt) over(partition by an order by dt) dt_lag
from #temp
)
select *, datediff(day, dt_lag, dt)
from x
where datediff(day, dt_lag, dt) >= 10