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])
編輯 - 抱歉,你是正確的。這裏有一個固定的版本(我應該使用的測試表...開始),如果你想比天更精細
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
更改DATEDIFF單位。
編輯#2 - 好吧,這裏是一個更強大的解決方案,處理閏年:) 編輯#3 - 實際上這不處理閏年,而是允許指定可變的時間間隔。使用dateadd(year,1,@baseDate)進行閏年安全方法。
declare @baseDate datetime, @interval datetime
[email protected] is expressed as time above 0 time (1/1/1900)
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
給出完全相同的結果,只是沒有前一年。 – cdeszaq 2009-05-06 17:50:59