本文介绍了DateDiff问题(VB .net 4.0)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

嗨!!,

对于我的一个项目,我在计时器中使用DateDiff功能,如下所述: -



NoOfValues = DateDiff(DateInterval.Second,StartTime,EndTime)



StartTime,EndTime&执行周期的NoOfValues如下所述



Hi!!,
For one of my project i'm using DateDiff function in a timer as mentioned below:-

NoOfValues = DateDiff(DateInterval.Second, StartTime, EndTime)

the values of StartTime, EndTime & NoOfValues for a cycle of execution are as mentioned below

Start Time= 11/18/2013 4:28:46 PM   End Time= 11/18/2013 4:29:37 PM
NoOfValues= 51

Start Time= 11/18/2013 4:29:38 PM   End Time= 11/18/2013 4:29:49 PM
NoOfValues= 10

Start Time= 11/18/2013 4:29:50 PM   End Time= 11/18/2013 4:30:00 PM
NoOfValues= 10





我无法理解



I'm not able to understand how the difference of

[11/18/2013 4:29:38 PM & 11/18/2013 4:29:49 PM] =10 





Plz指导我是否我已正确使用DateDiff函数。



Plz guide me whether i've used DateDiff function correctly.

推荐答案

Dim myDate1 As DateTime = new DateTime(2013, 11, 18, 4, 29, 38);
Dim myDate2 As DateTime = new DateTime(2013, 11, 18, 4, 29, 49);
TimeSpan ts = myDate2 - myDate1;
long elapsedInMilliseconds = ts.Elapsed;
float valueInSeconds = (float)elapsedInMilliseconds / 1000f;
int value = Math.Floor(valueInSeconds);


这篇关于DateDiff问题(VB .net 4.0)的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

10-23 18:52