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Counting days for the month

Given the following rows

ID       date             active
1        8/1/2011      1
1        8/10/2011    0
1        8/20/2011    1

i want to write a query that will count the number of days id=1 is active

ie

1-10 = 10
20-31 = 12

Total: 22

Allan
0
acadenilla
Asked:
acadenilla
1 Solution
 
BusyMamaCommented:
SELECT COUNT(*)
FROM TABLENAME
WHERE ID = 1 AND ACTIVE = 1;
0
 
acadenillaAuthor Commented:
BusyMama Thanks for the reply, but that wont work your query would only give me 2.

there are only three rows here there isnt a row for every day.

the query need to count that from 8/1 - 8/10 that is was active and then from 8/21 - 8/31 it was also active for a total of 22

Allan
0
 
ralmadaCommented:
try something like this


;with cte as (
	select a.id, a.[date], isnull(b.nextdate, dateadd(m, dateadiff(m, 0, a.[date])+1, 0)-1) as enddate
	from (select * from yourtable where active = 1) a
	cross outer apply (select id, min([date]) as nextdate from yourtable where active = 0 and id = a.id and [date] > a.[date]) b
), cte2 as (
	select id, [date], enddate, datediff(d, [date], enddate) as ndays
	from cte
)
select id, sum(ndays)
from cte2
where id = 1

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ralmadaCommented:
and actually to limit it to a specific month
;with cte as (
	select a.id, a.[date], isnull(b.nextdate, dateadd(m, dateadiff(m, 0, a.[date])+1, 0)-1) as enddate
	from (select * from yourtable where active = 1 where [date] between '2011-08-01' and '2011-08-31') a
	cross outer apply (select id, min([date]) as nextdate from yourtable where active = 0 and id = a.id and [date] > a.[date]) b
), cte2 as (
	select id, [date], enddate, datediff(d, [date], enddate) as ndays
	from cte
)
select id, sum(ndays)
from cte2
where id = 1

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0
 
LowfatspreadCommented:
show some proper data...

cross a month boundary... can you have


8/20/2011   1
9/04/2011   0    

?
0
 
acadenillaAuthor Commented:
ralmada thx but this only gives me 9, from 8/1-8/10

I would need to add another row

select 1, '9/1/2011' 0

to my dataset to get the correct value which is not a big deal.
; WITH data AS
(
	SELECT 1 AS id, '8/1/2011' AS [date], 1 AS active UNION ALL
	SELECT 1, '8/10/2011', 0 UNION ALL
	SELECT 1, '8/20/2011', 1
), cte as (
	select a.id, a.[date], isnull(b.nextdate, dateadd(m, datediff(m, 0, a.[date])+1, 0)-1) as enddate
	from (select * from data where active = 1) a
	cross apply (select id, min([date]) as nextdate from data where active = 0 and id = a.id and [date] > a.[date] GROUP BY id) b
), cte2 as (
	select id, [date], enddate, datediff(d, [date], enddate) as ndays
	from cte
)

select id, sum(ndays)
from cte2
where id = 1
GROUP BY id

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0
 
acadenillaAuthor Commented:
@Lowfatspread

this is really only specific to one month.
0
 
ralmadaCommented:
Here's the corrected version to give you the extra day
declare @t table (
ID       int,
[date] date,             
active int
)

insert @t values(1,        '8/1/2011',      1),
(1,        '8/10/2011',    0),
(1,        '8/20/2011',    1)

select * from @t

;with cte as (
	select a.id, a.[date], isnull(b.nextdate, dateadd(m, datediff(m, 0, a.[date])+1, 0)-1) as enddate
	from (select * from @t where active = 1 and [date] between '2011-08-01' and '2011-08-31') a
	outer apply (select id, min([date]) as nextdate from @t where active = 0 and id = a.id and [date] > a.[date] group by id) b
), cte2 as (
	select id, [date], enddate, datediff(d, [date], enddate)+1 as ndays
	from cte
)
select id, sum(ndays)
from cte2
where id = 1
group by id

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