Question

我有一个包含Start_Time和End_Time列的表格。数据存储为123000的{{1}}和Start_Time的{{1}}。

我希望查询将时间值返回为163000和End_Time。这有点容易实现吗？

提前致谢。

Answer 1

Declare @YourTable table (Start_Time varchar(25), End_Time varchar(25))
Insert Into @YourTable values
('123000','163000'),
('090000','170000')


Select Start_Time = convert(varchar(8),convert(time,stuff(left(Start_Time,4),3,0,':')),100)
      ,End_Time   = convert(varchar(8),convert(time,stuff(left(End_Time  ,4),3,0,':')),100)
 From  @YourTable

返回

Start_Time  End_Time
12:30PM     4:30PM
9:00AM      5:00PM

修改 - 在AM / PM之间添加空间

使用两个replace（）可能会减少收益，格式（）实际上可能更快

Declare @YourTable table (Start_Time varchar(25), End_Time varchar(25))
Insert Into @YourTable values
('123000','163000'),
('090000','170000')

Select Start_Time = replace(replace(convert(varchar(8),convert(time,stuff(left(Start_Time,4),3,0,':')),100),'AM',' AM'),'PM',' PM')
      ,End_Time   = replace(replace(convert(varchar(8),convert(time,stuff(left(End_Time  ,4),3,0,':')),100),'AM',' AM'),'PM',' PM')
 From  @YourTable

返回

Start_Time  End_Time
12:30 PM    4:30 PM
9:00 AM     5:00 PM

Answer 2

如果您需要much slower performing option，可以在SQL Server 2012 +中使用format()：

create table t (Start_Time varchar(12), End_Time varchar(12))
insert into t values ('123000','163000'),('090000','170000')

select 
    Start_Time = format(convert(datetime,left(stuff(Start_Time,3,0,':'),5)),'h:mm tt')
  , End_Time   = format(convert(datetime,left(stuff(End_Time  ,3,0,':'),5)),'h:mm tt')
from t

rextester演示：http://rextester.com/ZDSD64086

返回：

+------------+----------+
| Start_Time | End_Time |
+------------+----------+
| 12:30PM    | 4:30 PM  |
| 9:00AM     | 5:00 PM  |
+------------+----------+

format() is nice and all, but… - Aaron Bertrand

Microsoft SQL Server中的时间格式为AM / PM

2 个答案: