I have to calculate the difference from StartTime and EndTime. If the EndTime is less then 15 minutes that of StartTime I have to show an error.
我必须计算StartTime和EndTime的差异。如果EndTime小于StartTime的15分钟,我必须显示错误。
CREATE PROCEDURE [Travel].[TravelRequirementValidate]
@Action char(1)
,@CallId int
,@PhaseId int
,@ShipId int
,@CallStartDate datetime
,@CallEndDate DATETIME
,@CallStartTime datetime
,@CallEndTime datetime
,@LanguageId int
,@SessionGroup nvarchar(100)
,@SessionPlace nvarchar(100)
,@ActiveFlg tinyint
,@WarningMessage nvarchar(300)=NULL output
,@Minutes int
as
if @Action in ('I','U')
begin
@Minutes=select DATEDIFF(@CallStartDate,@CallStartTime,@CallEndTime) from [Travel].[TravelRequirement]
if @Minutes<=15
begin
raiserror(3,11,1) --CallEndTime must be equals or higher than 15 minutes
return
end
end
This code doesn't work. I've got an error for the first parameter of DATEDIFF (invalid parameter 1 specified for datediff).
此代码不起作用。我有DATEDIFF的第一个参数错误(为datediff指定的参数1无效)。
How can I fix my code?
我该如何修复我的代码?
EDIT
编辑
I changed @Minutes=select DATEDIFF(@CallStartDate,@CallStartTime,@CallEndTime) from [Travel].[TravelRequirement]
我改变了@minutes =从[Travel]中选择DATEDIFF(@ CallStartDate,@ CallStartTime,@ CallEndTime)。[TravelRequirement]
in
在
declare @Diff int
@Diff=select DATEDIFF(@Minutes,@CallStartTime,@CallEndTime) from [Travel].[TravelRequirement]
but I have the same error
但我有同样的错误
3 个解决方案
#1
2
Function DATEDIFF
wants as first parameter the portion of time.
函数DATEDIFF希望将时间部分作为第一个参数。
The correct use of it is:
正确使用它是:
DATEDIFF(minute, @CallStartTime, @CallEndTime)
#2
1
The correct syntax for that expression would be:
该表达式的正确语法是:
select @Minutes = datediff(minute, @CallStartTime, @CallEndTime)
from [Travel].[TravelRequirement];
However, this does't make sense, because -- presumably -- the table has more than one row. Which row do you want?
但是,这没有意义,因为 - 据推测 - 该表有多行。你想要哪一排?
#3
1
I would suggest using seconds and devide by 60.0 - provides a more accurate result:
我建议使用秒和devide 60.0 - 提供更准确的结果:
select @Minutes = datediff(second, @CallStartTime, @CallEndTime)/60.0
from [Travel].[TravelRequirement];
#1
2
Function DATEDIFF
wants as first parameter the portion of time.
函数DATEDIFF希望将时间部分作为第一个参数。
The correct use of it is:
正确使用它是:
DATEDIFF(minute, @CallStartTime, @CallEndTime)
#2
1
The correct syntax for that expression would be:
该表达式的正确语法是:
select @Minutes = datediff(minute, @CallStartTime, @CallEndTime)
from [Travel].[TravelRequirement];
However, this does't make sense, because -- presumably -- the table has more than one row. Which row do you want?
但是,这没有意义,因为 - 据推测 - 该表有多行。你想要哪一排?
#3
1
I would suggest using seconds and devide by 60.0 - provides a more accurate result:
我建议使用秒和devide 60.0 - 提供更准确的结果:
select @Minutes = datediff(second, @CallStartTime, @CallEndTime)/60.0
from [Travel].[TravelRequirement];