Question

请参阅下表了解更多详情。

1   vishal  10  7/4/2017 1:00:00 PM
2   rahul   12  7/4/2017 2:00:00 PM
3   sudha   15  7/4/2017 3:00:00 PM
4   vishal  10  7/4/2017 4:00:00 PM
5   vishal  10  7/4/2017 2:30:00 PM
6   vishal  10  7/4/2017 12:30:00 PM
7   vishal  10  6/4/2017 10:30:00 PM
8   vishal  10  10/4/2017 12:00:00 AM

我想要低于输出

vishal  10  7/4/2017   12:30:00 PM 4:00:00 PM 3.5hrs

注意：必须在SQL Server 2003上运行查询。

Answer 1

如果我理解你的问题，这可以做到这一点

create table #table1 (field1 int, field2 varchar(10), field3 int, field4 datetime)
insert into #table1 values
   (1, 'vishal', 10, '20170704 1:00:00'),
   (2, 'rahul',  12, '20170704 2:00:00'),
   (3, 'sudha',  15, '20170704 3:00:00'),
   (4, 'vishal', 10, '20170704 4:00:00'),
   (5, 'vishal', 10, '20170704 2:30:00'),
   (6, 'vishal', 10, '20170704 00:30:00'),
   (7, 'vishal', 10, '20170604 10:30:00'),
   (8, 'vishal', 10, '20171004 00:00:00')

select field2, 
       field3, 
       convert(date, field4), 
       min(field4), 
       max(field4), 
       Datediff(second,Min(field4),Max(field4))/3600.00 
from   #table1
group by field2, field3, convert(date, field4)

drop table #table1

结果如下：

rahul   12  2017-07-04  4/07/2017 2:00:00   4/07/2017 2:00:00   0   
sudha   15  2017-07-04  4/07/2017 3:00:00   4/07/2017 3:00:00   0   
vishal  10  2017-06-04  4/06/2017 10:30:00  4/06/2017 10:30:00  0   
vishal  10  2017-07-04  4/07/2017 0:30:00   4/07/2017 4:00:00   3.5 
vishal  10  2017-10-04  4/10/2017           4/10/2017           0

Answer 2

希望这会奏效，

SELECT Name,
       Max(Value) AS Value,
       Cast(Date_Time as Date) AS Date,
       Max(Date_Time) AS Max_Date_Time,
       Min(Date_Time) AS Min_Date_Time,
       Datediff(second,Min(Date_Time),Max(Date_Time))/3600.00 AS Time_Diff
 FROM  #Your_Table 
 Group by Name, Cast(Date_Time as Date)

注意：在SQL Server 2008中工作

Answer 3

使用DATEDIFF，LTRIM，RIGHT和Group By

select name as Name,Number as Value,
convert(date,startdate) as [Date],
LTRIM(RIGHT(CONVERT(VARCHAR(20),MIN(convert(datetime,convert(varchar,startdate,109))),22),11)) AS MIN_Date,
LTRIM(RIGHT(CONVERT(VARCHAR(20),MAX(convert(datetime,convert(varchar,startdate,109))),22),11)) AS MAX_Date,
convert(varchar(50),convert(numeric(8,1),(DATEDIFF(second,MIN(convert(datetime,convert(varchar,startdate,109))),max(startdate))/3600.00)))+' hrs' as TimeDifference
from my_table_new group by name,convert(date,startdate),Number

输出：

Name    Value   Date    MIN_Date    MAX_Date    TimeDifference
rahul   12  2017-07-04  2:00:00 PM  2:00:00 PM  0.0 hrs
sudha   15  2017-07-04  3:00:00 PM  3:00:00 PM  0.0 hrs
vishal  10  2017-06-04  10:30:00 PM 10:30:00 PM 0.0 hrs
vishal  10  2017-07-04  12:30:00 PM 4:00:00 PM  3.5 hrs
vishal  10  2017-10-04  12:00:00 AM 12:00:00 AM 0.0 hrs

特定用户的第一个和最后一个条目以及该条目之间的特定日期和小时差异

3 个答案: