I
igbert
Is there a fuction to return the days in a given year?
Entry Return
2008 366
2010 365
Igbert
Entry Return
2008 366
2010 365
Igbert
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B
Brad
assume 2008 is in a2
in b2
=DATE(A2+1,1,0)-DATE(A2,1,0)
Success - click yes.
in b2
=DATE(A2+1,1,0)-DATE(A2,1,0)
Success - click yes.
D
Dave Peterson
=date(2010,1,1)-date(2009,1,1)
would give the number of days in 2009.
If you put the interested year in A1, you could use:
=date(a1+1,1,1)-date(a1,1,1)
(well, except for 1900.)
would give the number of days in 2009.
If you put the interested year in A1, you could use:
=date(a1+1,1,1)-date(a1,1,1)
(well, except for 1900.)
M
Ms-Exl-Learner
Assume that you are having the Year in A1. Copy and paste the below formula
in B1 cell.
=IF(A1="","",DATEDIF(DATE(A1,1,1),DATE(A1+1,1,1),"D"))
Remember to Click Yes, if this post helps!
in B1 cell.
=IF(A1="","",DATEDIF(DATE(A1,1,1),DATE(A1+1,1,1),"D"))
Remember to Click Yes, if this post helps!
I
igbert
Thanks, formula works preferct for any year.
Brad said:assume 2008 is in a2
in b2
=DATE(A2+1,1,0)-DATE(A2,1,0)
Success - click yes.
B
Brad
You're welcome
Another approach that would also work
=365+(MONTH(DATE(A2,2,29))=2)
Success click yes
Another approach that would also work
=365+(MONTH(DATE(A2,2,29))=2)
Success click yes
M
Mike H
Micky,
I like the second formula, very neat, but I'm afraid the first gives errors,
you would need to do this to get the correct result using MOD
=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A1,100)<>0)),366, 365)
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
I like the second formula, very neat, but I'm afraid the first gives errors,
you would need to do this to get the correct result using MOD
=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A1,100)<>0)),366, 365)
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
B
Bernd P
You're welcome
Another approach that would also work
=365+(MONTH(DATE(A2,2,29))=2)
Success click yes
=337+DAY(DATE(A1,3,0))
Regards,
Bernd
M
Mike H
Mickey,
Of course
1908 = leap year
2008 = leap year
2108 - leap year
2200 - Not a leap year
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
Would you be so kind and check your suggested formula for the years:
1908, 2008, 2108 and 2200(!)
Of course
1908 = leap year
2008 = leap year
2108 - leap year
2200 - Not a leap year
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
M
Mike H
Mickey,
For clarification:-
A leap year is every 4 years, but not every 100 years, then again every 400
years
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
For clarification:-
A leap year is every 4 years, but not every 100 years, then again every 400
years
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
M
Mike H
Micky,
No I don't recognize that formula! I posted :-
=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A1,100)<>0)),366, 365)
You posted:-
=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A9,100)<>0)),366, 365)
Note the error in your formula referring to A1 & A9 and in the one in the
image you posted (A7 & A14) and hence the erronious results you are getting.
My formula returns 366 for 1908, 2008 & 2108 & 365 for year 2200 which isn't
a leap year by the following definition:-
A year will be a leap year if it is divisible by 4 but not by 100. If a year
is divisible by 4 and by 100, it is not a leap year unless it is also
divisible by 400.
Regards,
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
If you recognize the following formula as the one suggested by you - check it
No I don't recognize that formula! I posted :-
=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A1,100)<>0)),366, 365)
You posted:-
=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A9,100)<>0)),366, 365)
Note the error in your formula referring to A1 & A9 and in the one in the
image you posted (A7 & A14) and hence the erronious results you are getting.
My formula returns 366 for 1908, 2008 & 2108 & 365 for year 2200 which isn't
a leap year by the following definition:-
A year will be a leap year if it is divisible by 4 but not by 100. If a year
is divisible by 4 and by 100, it is not a leap year unless it is also
divisible by 400.
Regards,
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
M
Mike H
Micky.
I'm glad we resolved that.
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.
I'm glad we resolved that.
--
Mike
When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.