Average Age

[Laura]

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 address 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

 

[KARL DEWEY]

Why not average and then apply the function called Diff2Dates?

 

[KenSheridan via AccessMonster.com]

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

 

[KenSheridan via AccessMonster.com]

Correction. First expression should have been:

DateDiff("m",[DoB],Date())+IIf(Day([DoB])>Day(Date()),1,0)

KenSheridan
Stafford, England

 

[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



Correction. First expression should have been:
DateDiff("m",[DoB],Date())+IIf(Day([DoB])>Day(Date()),1,0)

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

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 address 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

 

[KenSheridan via AccessMonster.com]

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

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

Correction. First expression should have been:
DateDiff("m",[DoB],Date())+IIf(Day([DoB])>Day(Date()),1,0)

Laura:

[quoted text clipped - 56 lines]

 

[Laura]

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:

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

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

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

 

[KenSheridan via AccessMonster.com]

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

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]

 

[Laura]

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

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

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