我有3个varchar时间范围,它应该被转换为start&完成时间值,
这是值:
SCH
- 9:00-12:00
- 13-15:00
- 15-17:30
所以我自己尝试分成开始时间和结束并转换为时间值
这是我的分割功能:
CREATE FUNCTION dbo.Split
(
@RowData nvarchar(2000),
@SplitOn nvarchar(5)
)
RETURNS @RtnValue table
(
Id int identity(1,1),
Data nvarchar(100)
)
AS
BEGIN
Declare @Cnt int
Set @Cnt = 1
While (Charindex(@SplitOn,@RowData)>0)
Begin
Insert Into @RtnValue (data)
Select
Data = ltrim(rtrim(Substring(@RowData,1,Charindex(@SplitOn,@RowData)-1)))
Set @RowData = Substring(@RowData,Charindex(@SplitOn,@RowData)+1,len(@RowData))
Set @Cnt = @Cnt + 1
End
Insert Into @RtnValue (data)
Select Data = ltrim(rtrim(@RowData))
Return
END
我得到了开始&通过在我的开发查询中执行此操作来完成值,但是当我尝试转换15-17:30时,我得到了错误转换,因为起始值仅为15:
declare @valueToParse varchar(20) = '15-17:30'
select @schtimestart = data from dbo.split(@valueToParse,'-') where id=1
select @schtimefinish = data from dbo.split(@valueToParse,'-') where id=2
SELECT CAST(@schtimestart AS time)
SELECT CAST(@schtimefinish AS time)
如何只将一个值转换为时间值,还是有任何简单的转换?
答案 0 :(得分:0)
我已经修复了我的分割功能,所以如果开始时间没有分钟值,那么它会自动添加
这是我的代码
USE [Dispatch]
GO
/****** Object: UserDefinedFunction [dbo].[Split] Script Date: 25/10/2013 09:39:51 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
alter FUNCTION dbo.Split
(
@RowData nvarchar(2000),
@SplitOn nvarchar(5)
)
RETURNS @RtnValue table
(
Id int identity(1,1),
Data nvarchar(100)
)
AS
BEGIN
Declare @Cnt int
Set @Cnt = 1
declare @tmp as varchar(20)
While (Charindex(@SplitOn,@RowData)>0)
Begin
SET @tmp = ltrim(rtrim(Substring(@RowData,1,Charindex(@SplitOn,@RowData)-1)))
Insert Into @RtnValue (data)
Select Data = case when ISNUMERIC(SUBSTRING(@tmp,1,2)) = 1 AND ISNUMERIC(SUBSTRING(@tmp,4,2)) = 1 then @tmp else (@tmp + ':00') END
Set @RowData = Substring(@RowData,Charindex(@SplitOn,@RowData)+1,len(@RowData))
Set @Cnt = @Cnt + 1
End
Insert Into @RtnValue (data)
Select Data = ltrim(rtrim(@RowData))
Return
END