Modules & VBA :: Calculate Integer Number Difference From Two Dates

Sep 20, 2014

Basically, I am trying to calculate a integer number difference from two dates (TAT = Due-Date - Result_Date). The number is calculated and excludes weekends and ideally holidays (for that I have a tblHoliday but not sure how to use it). The function below seems to calculate a number but doesn't exclude weekends.

For example, if Due_Date is 9/26/2014 and Result_Date is 9/30/2014, then TAT is calculated to be 5 (should be 2).Since 9/26/2014 is a Friday only Friday and Monday are used in the calculation.

Code:

Option Compare Database
Public Function WorkingDays(StartDate As Date, EndDate As Date) As Long
Dim intCount As Long
intCount = 0

[code]...

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Forms :: Calculate Integer Difference Between Due And Result Date

Sep 20, 2014

Trying to calculate the integer difference between Due_Date and Result_Date excluding weekends and holidays.

I have a table (Holidays) with the dates of the holidays in it. The table looks like:

ID Description Holiday
1 New Years 1/1/2014
2 New Years 1/1/2015

So, if Date_Due: 9/25/2014 and Result_Date: 9/29/2014, then TAT = 2

Since 9/27/2014 and 9/28/2014 are weekends they are excluded from the calculation and only that Thursday and Friday are used in the calculation.

Code:
Private Sub Result_Date_AfterUpdate()
[TAT] = NETWORKDAYS(Due_Date, Result_Date, tblHolidays)
End Sub

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Sep 14, 2014

I have the attached database which calculate the Difference between dates.I have also two problems like i shown in the image

Code:
Option Compare Database
Option Explicit
Sub Recalculate()
Dim rst As DAO.Recordset
Dim dff As Long

[code]...

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For time being, the user have to count manually to get the "TotalPcs". I wonder is there a way to calculate the "TotalPcs" by programming?

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Fields:
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[ReturnedDate]

DESCRIPTION: If ReturnedDate is null, then calculate the datediff "d" between LDW with Now(), if not, then calculate the datediff "d" between LDW with ReturnDate.

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Days Absent: Iif(isnull([ReturnedDate],(DateDiff("d",[LDW],Now())),(DateDiff("d",[LDW],[ReturnedDate])))

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This is a formula I tried to piece together but obviously not correct as it's not returning anything;

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The formula will then be:

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Hello,

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Bij this code you can calculate the difference even over night.

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Code:
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