Question

我有一个包含4个cols的安全日志表：

UserID, LOGINDate, LOGINTime, ClickEvents

现在我想尝试过去7天的每小时流量表，就像：

   DAY1 |  DAY2 | DAY3 |  DAY4 |.... |DAY7
1   0   |    1  |   12 |  4567 |     | 43
2
3
4
5
:
:
24

你能告诉我或者让我知道如何使用SQL创建这个表吗？

marc_s感谢您的快速回复。我现在拥有的是：

select  LOGINDate, SUBSTRING(LOGINTime, 1, 2) as 'HoTime', COUNT( *)
From SECLOG
where (CONVERT(varchar( 8) , GETDATE()-7, 112) <= LOGINDate)

group by LOGINDate, SUBSTRING(LOGINTime, 1, 2)
order by LOGINDate, HoTime

这会产生一个像

这样的表格

DATE | HoTime | No of
0926 | 1      | 2
0926 | 2      | 4
0926 | 14     | 6

它也会在没有数据的情况下跳过小时。

Answer 1

首先按日期和小时对最近7天内每小时的点击事件进行分组然后，您需要转动该结果，以便将日期作为列。这可以通过一小时按小时完成，并在字段列表测试中使用case语句来表示日期编号。

要在一小时内没有事件时将零值作为值，您可以使用返回1-24的数字表并离开加入结果集。

对于SQL Server 2008，您可以使用日期数据类型。

with C as
(
  select datediff(day, LoginDate, getdate()) as DD, 
         LoginTime,
         sum(ClickEvents) as ClickEvents
  from YourTable
  where LoginDate >= dateadd(day, -7, cast(getdate() as date))
  group by LoginDate, LoginTime
), H as
(
  select 1 as LoginTime
  union all
  select H.LoginTime+1
  from H
  where H.LoginTime < 24
)  
select H.LoginTime, 
       sum(case DD when 1 then C.ClickEvents else 0 end) as Day1,
       sum(case DD when 2 then C.ClickEvents else 0 end) as Day2,
       sum(case DD when 3 then C.ClickEvents else 0 end) as Day3,
       sum(case DD when 4 then C.ClickEvents else 0 end) as Day4,
       sum(case DD when 5 then C.ClickEvents else 0 end) as Day5,
       sum(case DD when 6 then C.ClickEvents else 0 end) as Day6,
       sum(case DD when 7 then C.ClickEvents else 0 end) as Day7
from H
  left outer join C
    on C.LoginTime = H.LoginTime
group by H.LoginTime
order by H.LoginTime;

在SE Data

上试用

Answer 2

如果您正在使用MS SQL，则可以使用以下内容：

SELECT COUNT(*) FROM SecurityLog WHERE DATEPART(hh, LOGINDate) = 1 AND LOGINDate = [somedate]
SELECT COUNT(*) FROM SecurityLog WHERE DATEPART(hh, LOGINDate) = 2 AND LOGINDate = [somedate]
....
....
SELECT COUNT(*) FROM SecurityLog WHERE DATEPART(hh, LOGINDate) = 24 AND LOGINDate = [somedate]

这将告诉您在任何给定日期，安全日志中有多少记录与任何给定小时匹配。只需将[somedate]替换为相关日期。

参考： http://msdn.microsoft.com/en-us/library/ms174420.aspx http://syntaxhelp.com/SQLServer/Breaking-a-date-by-hour-into-24-parts

Answer 3

这不是一个完整的解决方案。我刚刚提出了一个想法，希望在此基础上构建您的解决方案。如何创建一个表来存储小时信息呢？比如，1,2 ...... 24。那么你可以做这样的事情：

SELECT
    h.[Hour],
    SUM(CASE WHEN  t.LOGINDate BETWEEN GetDate() - 1 AND GetDate()      THEN  1 ELSE 0 END) as DAY1,
    SUM(CASE WHEN  t.LOGINDate BETWEEN GetDate() - 2 AND GetDate() - 1  THEN  1 ELSE 0 END) as DAY2,
    SUM(CASE WHEN  t.LOGINDate BETWEEN GetDate() - 3 AND GetDate() - 2  THEN  1 ELSE 0 END) as DAY3, 
    SUM(CASE WHEN  t.LOGINDate BETWEEN GetDate() - 4 AND GetDate() - 3  THEN  1 ELSE 0 END) as DAY4, 
    SUM(CASE WHEN  t.LOGINDate BETWEEN GetDate() - 5 AND GetDate() - 4  THEN  1 ELSE 0 END) as DAY5, 
    SUM(CASE WHEN  t.LOGINDate BETWEEN GetDate() - 6 AND GetDate() - 5  THEN  1 ELSE 0 END) as DAY6, 
    SUM(CASE WHEN  t.LOGINDate BETWEEN GetDate() - 7 AND GetDate() - 6  THEN  1 ELSE 0 END) as DAY7 
FROM 
    tblHour h LEFT JOIN SecurityLog t 
    ON h.[Hour] = DATEPART(hour, t.LOGINDate)
WHERE 
    t.LOGINDate >= CONVERT(datetime, CONVERT(varchar, getdate() -7, 102)) AND
    t.LOGINDate < DATEADD(day, 1, CONVERT(datetime, CONVERT(varchar, getdate(), 102)))
GROUP BY 
    t.LOGINDate, h.[Hour]

sql表按小时选择

3 个答案: