Got Wrong Value From DateDiff function

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I am migrating one application VB6 to VB.Net.

I am getting wrong value in DateDiff function.

I have a piece of code which I need to modify:

in VB6 Code,

Dim n as Integer
Dim CurDate as Date
Dim FindDate AS Date
CurDate = Now()
FindDate = DateAdd("n", -540, CurDate)
n = DateDiff("d", FindDate, CurDate)

==> here returns n value as 1

in VB.Net Code,

Dim CurDate as Date
Dim FindDate AS Date
CurDate = Now
FindDate = CurDate.AddMinutes(-540)
Dim dateDiff AS TimeSpan = FindDate.Subtract(CurDate)
n = dateDiff.Days

==> here returns n value as 0

I can't change VB6 code. But answer should be same in VB.Net code, so what i do for that.
Thanks for Any Help!

Posted 26-Jun-11 22:43pm

Ramanujam Shankar

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2 solutions

Solution 1

Why are you expecting that subtracting nine hours will give you a day's worth of difference? Are you sure that you just didn't test at before 09:00 for the VB6 and after 09:00 for the VB.NET code?

Posted 26-Jun-11 22:55pm

OriginalGriff

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Ramanujam Shankar 27-Jun-11 8:26am

i am getting "-540" value from one variable. without subtract then it will working fine. but in vb code was written like subtract "-540" value. then find the difference

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Ramanujam Shankar · Accepted Answer · 2011-06-27T02:43:00

Actually in VB6, DateDiff function considered date but in VB.Net considered with time.

for Example:

CurDate = "06/27/2011 4:01:15 AM"
FindDate = "06/26/2011 7:01:15 PM"    '==> with subtraction of "-540" minutes

in VB,

n = DateDiff("d", FindDate, CurDate)

'==> it's returned "1" value because i mentioned interval as "d". so it's consider only date.

in VB.Net,

Dim dateDiff AS TimeSpan = FindDate.Subtract(CurDate)
n = dateDiff.Days

'==> it's returned "0" value because it's consider with time. so it's not reached 24 hours.

Solution:
i changed VB.Net Code,

n = DateDiff(DateInterval.Day, FindDate.Date, CurDate.Date)

So now it's consider only date.
Now i am getting "1" value