設置AccessCount字段可以根據需求在特定的時間范圍內如果是相同IP訪問就在AccessCount上累加。 代碼如下:
Create table Counter
(
CounterID int identity(1,1) not null,
IP varchar(20),
AccessDateTime datetime,
AccessCount int
)
該表在這兒只是演示使用,所以只提供了最基本的字段
現在往表中插入幾條記錄
insert into Counter
select '127.0.0.1',getdate(),1 union all
select '127.0.0.2',getdate(),1 union all
select '127.0.0.3',getdate(),1
1 根據年來查詢,以月為時間單位
通常情況下一個簡單的分組就能搞定
代碼如下:
select
convert(varchar(7),AccessDateTime,120) as Date,
sum(AccessCount) as [Count]
from
Counter
group by
convert(varchar(7),AccessDateTime,120)
像這樣分組後沒有記錄的月份不會顯示,如下:
這當然不是我們想要的,所以得換一種思路來實現,如下:
代碼如下:
declare @Year int
set @Year=2009
select
m as [Date],
sum(
case when datepart(month,AccessDateTime)=m
then AccessCount else 0 end
) as [Count]
from
Counter c,
(
select 1 m
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9
union all select 10
union all select 11
union all select 12
) aa
where
@Year=year(AccessDateTime)
group by
m
查詢結果如下:
2 根據天來查詢,以小時為單位。這個和上面的類似,代碼如下:
代碼如下:
declare @DateTime datetime
set @DateTime=getdate()
select
right(100+a,2)+ ':00 -> '+right(100+b,2)+ ':00 ' as DateSpan,
sum(
case when datepart(hour,AccessDateTime)> =a
and datepart(hour,AccessDateTime) <b
then AccessCount else 0 end
) as [Count]
from Counter c ,
(select 0 a,1 b
union all select 1,2
union all select 2,3
union all select 3,4
union all select 4,5
union all select 5,6
union all select 6,7
union all select 7,8
union all select 8,9
union all select 9,10
union all select 10,11
union all select 11,12
union all select 12,13
union all select 13,14
union all select 14,15
union all select 15,16
union all select 16,17
union all select 17,18
union all select 18,19
union all select 19,20
union all select 20,21
union all select 21,22
union all select 22,23
union all select 23,24
) aa
where datediff(day,@DateTime,AccessDateTime)=0
group by right(100+a,2)+ ':00 -> '+right(100+b,2)+ ':00 '
查詢結果如下圖: