Find Non Consecutive date in SQL Server - sql-server

Question

I want to find the missing NON-consecutive dates between two consecutive date.
I am posting my SQL query and temp tables to find out the results.
But I am not getting the proper results
Here is my SQL Query
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
But I want results of non-usage date.
I want to get those AN which has not been used continuously since 10 days.
So here output should be like this:-
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
So I just want to find out only consecutive not used dates count and their min and max dates .

Answer 1

try this :
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
or like this
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

Answer 2

It looks like you're trying to count the number of days not used between the mindate and the maxdate for each an. If that's the case, then this should do the trick:
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

Answer 3

As I understood you need this:
;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
Output:
an MinDate MaxDate ConsecutiveNotUseddays
7472381321 2016-10-14 2016-10-24 10
For 9099457123 I got two rows with 9 in ConsecutiveNotUseddays. You can check results removing WHERE statement.

Answer 4

On any newer version of SQL Server this should be easy:
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