Query Date Help W/Avg work days

```please note that in order to get the date 6/28/1979 I am using average work
days each month of 21.75

I have tried to Add just the total days to the date but it's not correct
because I need it to be based off the 21.75 days per month not actually days.

Total Years of Prior Service	Days	Hire Date	Adjusted Hire Date

7 years 4 months 6 days	2685	6/15/1987	       2/7/1980 (s/b 2/9/80)
1 year 8 months 13 days	620	2/10/2005	       6/1/2003 (s/b 5/27/03)

This is what is being calculated so far once I run the query.
"aldunford" wrote:

> This worked great thanks so much!! I have one more question.
>
> Now that i have the format I need as 7 years 4 months & 19 days for my prior
> service. How can I substract that from new hire date of 11/17/1986 to get
> 6/28/1979?
>
> I appreciate all your time and help!!
>
>
>
> "Ofer Cohen" wrote:
>
> > Look at this link on "A More Complete DateDiff Function"
> >
> > http://www.accessmvp.com/djsteele/Diff2Dates.html
> >
> > --
> > Good Luck
> > BS"D
> >
> >
> > "aldunford" wrote:
> >
> > > I need to create a query that will do the following:
> > >
> > > these are my fields: resign date, hire date
> > >
> > > I need to subtract the resign date from hire date but I want it to return
> > > the answer as years, months & days
> > >
> > > ie. 11/7/1986 - 6/18/1979 = 7 years 4 months & 19 days
> > >
> > > is there a way to use the DateDiff function to return years, monts & days?
```
 0
Utf
3/21/2008 5:05:01 PM
access.queries 6343 articles. 1 followers.

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```in terms it would be a total of 261 work days in a year ALWAYS even in leap
years

"aldunford" wrote:

> please note that in order to get the date 6/28/1979 I am using average work
> days each month of 21.75
>
> I have tried to Add just the total days to the date but it's not correct
> because I need it to be based off the 21.75 days per month not actually days.
>
> Total Years of Prior Service	Days	Hire Date	Adjusted Hire Date
>
> 7 years 4 months 6 days	2685	6/15/1987	       2/7/1980 (s/b 2/9/80)
> 1 year 8 months 13 days	620	2/10/2005	       6/1/2003 (s/b 5/27/03)
>
> This is what is being calculated so far once I run the query.
> "aldunford" wrote:
>
> > This worked great thanks so much!! I have one more question.
> >
> > Now that i have the format I need as 7 years 4 months & 19 days for my prior
> > service. How can I substract that from new hire date of 11/17/1986 to get
> > 6/28/1979?
> >
> > I appreciate all your time and help!!
> >
> >
> >
> > "Ofer Cohen" wrote:
> >
> > > Look at this link on "A More Complete DateDiff Function"
> > >
> > > http://www.accessmvp.com/djsteele/Diff2Dates.html
> > >
> > > --
> > > Good Luck
> > > BS"D
> > >
> > >
> > > "aldunford" wrote:
> > >
> > > > I need to create a query that will do the following:
> > > >
> > > > these are my fields: resign date, hire date
> > > >
> > > > I need to subtract the resign date from hire date but I want it to return
> > > > the answer as years, months & days
> > > >
> > > > ie. 11/7/1986 - 6/18/1979 = 7 years 4 months & 19 days
> > > >
> > > > is there a way to use the DateDiff function to return years, monts & days?
```
 0
Utf
3/21/2008 5:19:00 PM
```On Fri, 21 Mar 2008 10:05:01 -0700, aldunford
<aldunford@discussions.microsoft.com> wrote:

>please note that in order to get the date 6/28/1979 I am using average work
>days each month of 21.75

Since no month contains 21.75 workdays, and since that is an APPROXIMATION,
you will inevitably be off by at least a day or two.

Perhaps you could post your existing query and show how you would do the same
calculation manually.
--

John W. Vinson [MVP]
```
 0
John
3/21/2008 7:41:16 PM
```So a month (on average) has 365.25/12 days.
A month has (on average per your statement) 21.75 days.

So every work day represents 1.3994253 month days.

Every month day represents 0.71457905 work days.

Beyond giving you the ratios I have no idea to calculate what you want.

'====================================================
John Spencer
Access MVP 2002-2005, 2007-2008
Center for Health Program Development and Management
University of Maryland Baltimore County
'====================================================

aldunford wrote:
> please note that in order to get the date 6/28/1979 I am using average work
> days each month of 21.75
>
> I have tried to Add just the total days to the date but it's not correct
> because I need it to be based off the 21.75 days per month not actually days.
>
> Total Years of Prior Service	Days	Hire Date	Adjusted Hire Date
>
> 7 years 4 months 6 days	2685	6/15/1987	       2/7/1980 (s/b 2/9/80)
> 1 year 8 months 13 days	620	2/10/2005	       6/1/2003 (s/b 5/27/03)
>
> This is what is being calculated so far once I run the query.
> "aldunford" wrote:
>
>> This worked great thanks so much!! I have one more question.
>>
>> Now that i have the format I need as 7 years 4 months & 19 days for my prior
>> service. How can I substract that from new hire date of 11/17/1986 to get
>> 6/28/1979?
>>
>> I appreciate all your time and help!!
>>
>>
>>
>> "Ofer Cohen" wrote:
>>
>>> Look at this link on "A More Complete DateDiff Function"
>>>
>>> http://www.accessmvp.com/djsteele/Diff2Dates.html
>>>
>>> --
>>> Good Luck
>>> BS"D
>>>
>>>
>>> "aldunford" wrote:
>>>
>>>> I need to create a query that will do the following:
>>>>
>>>> these are my fields: resign date, hire date
>>>>
>>>> I need to subtract the resign date from hire date but I want it to return
>>>> the answer as years, months & days
>>>>
>>>> ie. 11/7/1986 - 6/18/1979 = 7 years 4 months & 19 days
>>>>
>>>> is there a way to use the DateDiff function to return years, monts & days?
```
 0
John
3/22/2008 12:17:01 AM

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