I'm doing end of term reports for pupils in a school whereby each report
needs to show the age of the pupil on a given date in YEARS and MONTHS (not
a fraction of the year). In other words, 10 yrs and 3 months, (not 10.25).
I've managed to use a function called Diff2Dates Author: © Copyright 2001
Pacific Database Pty Limited Graham R Seach MCP MVP (E-Mail Removed)
to display the pupils age in years and months fine, but I ALSO need to show
the Average Age of the Class in Years and Months.

Simply adding up the results of the Function above and dividing by the
number of children did not work accurately, nor did using AVG as the result
was a fraction of the year and not the exact average of months.

Does anyone have a function for calculating the Average Age in Years and
Months, please?

Thanks
Laura

Why not average and then apply the function called Diff2Dates?

--
Build a little, test a little.


"Laura" wrote:

> I'm doing end of term reports for pupils in a school whereby each report
> needs to show the age of the pupil on a given date in YEARS and MONTHS (not
> a fraction of the year). In other words, 10 yrs and 3 months, (not 10.25).
> I've managed to use a function called Diff2Dates Author: © Copyright 2001
> Pacific Database Pty Limited Graham R Seach MCP MVP (E-Mail Removed)
> to display the pupils age in years and months fine, but I ALSO need to show
> the Average Age of the Class in Years and Months.
>
> Simply adding up the results of the Function above and dividing by the
> number of children did not work accurately, nor did using AVG as the result
> was a fraction of the year and not the exact average of months.
>
> Does anyone have a function for calculating the Average Age in Years and
> Months, please?
>
> Thanks
> Laura
>
>
> .
>

Ken, thanks for replying so quickly, it's so helpful.
Your formula works - thank you so much. I got slightly confused - did you
mean a minus or plus sign before the IIF? I'm not sure of the significance.

_________________________________________________
DateDiff("m",[DoB],Date())+IIf(Day([DoB])>Day(Date()),1,0)
AvgAgeInMonths:
AVG(DateDiff("m",[DoB],Date())-IIf(Day(DoB)>Day(Date()),1,0))
_________________________________________________

Either seems to work. I used it in the Query Builder window and can program
it to "Enter Date" so that the school can use it for each of the 3 terms of
the year to then mailmerge into the School Reports.

Many thanks again.
Laura
Wimbledon
London
UK



"KenSheridan via AccessMonster.com" <u51882@uwe> wrote
Correction. First expression should have been:
DateDiff("m",[DoB],Date())+IIf(Day([DoB])>Day(Date()),1,0)


"KenSheridan via AccessMonster.com" <u51882@uwe> wrote in message
news:a389039a04f4e@uwe...
> Laura:
>
> You can get the total number of months of each pupil's Age with:
>
> DateDiff("m",[DoB],Date())+IIf(Day([DoB])>=Day(Date()),1,0)
>
> So in a query you can average that with:
>
> AvgAgeInMonths:
> AVG(DateDiff("m",[DoB],Date())-IIf(Day(DoB)>Day(Date()),1,0))
>
> You can of course substitute a literal date for the Date() function if you
> want the age on a specific date rather than the current date.
>
> You can convert that to years and month with a combination of integer
> division and the Mod operator. You can do it in the query by repeating
> the
> expression:
>
> AvgAge: AVG(DateDiff("m",[DoB],Date())-IIf(Day(DoB)>Day(Date()),1,0))\12 &
> "
> yrs and " & AVG(DateDiff("m",[DoB],Date())-IIf(Day(DoB)>Day(Date()),1,0))
> Mod
> 12 & " months"
>
> or you do the same in a footer of a report which lists all the pupils, in
> an
> unbound text box, using the same expression as the ControlSource:
>
> =Avg(DateDiff("m",[DoB],Date())-IIf(Day(DoB)>Day(Date()),1,0))\12 & " yrs
> and
> " & Avg(DateDiff("m",[DoB],Date())-IIf(Day(DoB)>Day(Date()),1,0)) Mod 12 &
> "
> months"
>
> Ken Sheridan
> Stafford, England
>
> Laura wrote:
>>I'm doing end of term reports for pupils in a school whereby each report
>>needs to show the age of the pupil on a given date in YEARS and MONTHS
>>(not
>>a fraction of the year). In other words, 10 yrs and 3 months, (not 10.25).
>>I've managed to use a function called Diff2Dates Author: © Copyright
>>2001
>>Pacific Database Pty Limited Graham R Seach MCP MVP
>>(E-Mail Removed)
>>to display the pupils age in years and months fine, but I ALSO need to
>>show
>>the Average Age of the Class in Years and Months.
>>
>>Simply adding up the results of the Function above and dividing by the
>>number of children did not work accurately, nor did using AVG as the
>>result
>>was a fraction of the year and not the exact average of months.
>>
>>Does anyone have a function for calculating the Average Age in Years and
>>Months, please?
>>
>>Thanks
>>Laura
>
> --
> Message posted via AccessMonster.com
> http://www.accessmonster.com/Uwe/For...arted/201002/1
>

Thanks for explaining, Ken - I know the feeling when applying to own
birthdate

I think the same principle applies if you're working out the age of a person
in terms of years.. you have to subtract a year, or something, or it works
out that you're a year older than you are and that's even worse!

Laura
London


"KenSheridan via AccessMonster.com" <u51882@uwe> wrote in message
news:a38a3dca882ac@uwe...
> Laura:
>
> Sorry, it should be a minus sign throughout. Put it down to how
> depressingly
> high the result is when I apply the expression to own date of birth!
>
> The way it works is that the DateDiff function gets the straight
> difference
> in months between the two dates regardless of the day of the month in
> eaither,
> so if the day of the month of the date of birth is after the day of the
> month
> of the current date one month has to be subtracted so that the result is
> whole months only. To see the difference enter the following in the debug
> window:
>
> ?
> DateDiff("m",#2000-10-11#,#2010-02-12#)-IIf(Day(#2000-11-12#)>Day(#2010-02-
> 12#),1,0)
>
> and then enter:
>
> ?
> DateDiff("m",#2000-10-13#,#2010-02-12#)-IIf(Day(#2000-10-13#)>Day(#2010-02-
> 12#),1,0)
>
> Now where did I leave that Zimmer frame?
>
> Ken Sheridan
> Stafford, England
>
> Laura wrote:
>>Ken, thanks for replying so quickly, it's so helpful.
>>Your formula works - thank you so much. I got slightly confused - did you
>>mean a minus or plus sign before the IIF? I'm not sure of the
>>significance.
>>
>>_________________________________________________
>>DateDiff("m",[DoB],Date())+IIf(Day([DoB])>Day(Date()),1,0)
>>AvgAgeInMonths:
>>AVG(DateDiff("m",[DoB],Date())-IIf(Day(DoB)>Day(Date()),1,0))
>>_________________________________________________
>>
>>Either seems to work. I used it in the Query Builder window and can
>>program
>>it to "Enter Date" so that the school can use it for each of the 3 terms
>>of
>>the year to then mailmerge into the School Reports.
>>
>>Many thanks again.
>>Laura
>>Wimbledon
>>London
>>UK
>>
>>"KenSheridan via AccessMonster.com" <u51882@uwe> wrote
>>Correction. First expression should have been:
>>DateDiff("m",[DoB],Date())+IIf(Day([DoB])>Day(Date()),1,0)
>>
>>> Laura:
>>>
>>[quoted text clipped - 56 lines]
>>>>Thanks
>>>>Laura
>
> --
> Message posted via AccessMonster.com
> http://www.accessmonster.com/Uwe/For...arted/201002/1
>

Ken, you're a star - thank you so much.=20

I find that nearly every single time I need to use "date" it's different =
from the previous time and therefore the coding is different too. I =
think one could write a whole book just on "dates". I keep saving =
valuable notes such as yours and appreciate your added input. Thank you =
very much.

Laura
London


"KenSheridan via AccessMonster.com" <u51882@uwe> wrote in message
news:a3a1d70f521c6@uwe...
> Laura:
>
> One other thing worth mentioning is that when you have a parameter in a
> query
> to enter a date time value it's a good idea to declare the parameter in
> the
> query. This then avoids any possibility of a date being entered in short
> date format, e.g. 14/2/2010, being misinterpreted as an arithmetical
> expression. If this does happen it wouldn't raise an error as Access
> implements the date/time data type as a 64 bit floating point number, the
> integer part representing the days and the fractional part the times of
> day,
> so it would be interpreted as the date time value which the number
> represents.
> 14/2/2010 as an arithmetical expression results in a number which
> represents
> a date time value of 30 December 1899 00:05:01. This is because 30
> December
> 1899 is 'day zero' in Access. You can see this by entering the following
> in
> the debug window:
>
> ? Format(CDate(14/2/2010),"dd mmmm yyyy hh:nn:ss")
>
> Parameters can be declared in query design view by selecting Parameters
> from
> the Query menu (or whatever the equivalent is in Access 2007), or in SQL
> view
> by adding a line to the beginning of the query. So for a parameter [Enter
> Date] it would be:
>
> PARAMETERS [Enter Date] DateTime;
> SELECT etc
>
> You can then be assured that however the user enters the date, provided it
> is
> a legitimate date value, it will always be interpreted correctly.
>
> Ken Sheridan
> Stafford, England
>
> Laura wrote:
>>Thanks for explaining, Ken - I know the feeling when applying to own
>>birthdate
>>
>>I think the same principle applies if you're working out the age of a
>>person
>>in terms of years.. you have to subtract a year, or something, or it works
>>out that you're a year older than you are and that's even worse!
>>
>>Laura
>>London
>>
>>> Laura:
>>>
>>[quoted text clipped - 59 lines]
>>>>>>Thanks
>>>>>>Laura
>
> --
> Message posted via AccessMonster.com
> http://www.accessmonster.com/Uwe/For...arted/201002/1
>