declare @baseDate datetime
set @baseDate = '1 May 2005'
SELECT
datediff(year, @baseDate, [date]) AS YearBucket
,COUNT(*) AS cnt
FROM logins
GROUP BY datediff(year, @baseDate, [date])
ORDER BY datediff(year, @baseDate, [date])
EDIT - Sorry, you're right. Here is the fixed version (I had to use a test pattern to start ...)
create table logins (date datetime, foo int)
insert logins values ('1 may 2005', 1)
insert logins values ('1 apr 2006', 2)
insert logins values ('1 may 2006', 3)
declare @baseDate datetime
set @baseDate = '1 May 2005'
SELECT
datediff(day, @baseDate, [date]) / 365 AS YearBucket
,COUNT(*) AS cnt
FROM logins
GROUP BY datediff(day, @baseDate, [date]) / 365
ORDER BY datediff(day, @baseDate, [date]) / 365
Change obsolete units if you want to increase the level of detail than days.
EDIT # 2 - , , :)
EDIT # 3 - , . dateadd (, 1, @baseDate) .
declare @baseDate datetime, @interval datetime
select @baseDate = '1 May 2005', @interval = '1901'
declare @timeRanges table (beginIntervalInclusive datetime, endIntervalExclusive datetime)
declare @i int
set @i = 1
while @i <= 10
begin
insert @timeRanges values(@baseDate, @baseDate + @interval)
set @baseDate = @baseDate + @interval
set @i = @i + 1
end
SELECT
tr.beginIntervalInclusive,
tr.endIntervalExclusive,
COUNT(*) AS cnt
FROM logins join @timeRanges as tr
on logins.date >= tr.beginIntervalInclusive
and logins.date < tr.endIntervalExclusive
GROUP BY tr.beginIntervalInclusive, tr.endIntervalExclusive
ORDER BY tr.beginIntervalInclusive