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Compute number of months between two dates

Posted on 2007-11-22
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Last Modified: 2013-12-29
Hello,

I need to compute the number of months between two dates.
I have 4 variables :
int month1, year1, month2, year2;

for example for respectively 12, 1991, 10, 1991
It must give me 3

Thank you for any help.
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Question by:matthew016
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5 Comments
 
LVL 92

Accepted Solution

by:
objects earned 1000 total points
ID: 20333349
be something like:

int months(int month1, year1, int month2, int year2)
{
    if (year1>year2)
    {
        return months(month2, year2, month1, year1);
    }
    else if (year1==year2)
    {
        return Math.max(month1 - month2) + 1;
    }
    return (12 - month1) + months(1, year1+1, month2, year2);
}

0
 
LVL 4

Assisted Solution

by:jcoombes
jcoombes earned 1000 total points
ID: 20333357
Can't you just:

1. Compute the number of years difference, by subtracting the larger year from the smaller year.
2. Multiply this value by 12 and store it.
3.  Compute the number of months difference between the months fields (in the example this would be 12 - 10 = 2)
4.  Add the stored month value to the value computed in 3.

I would have thought that the correct answer for the two dates you've posted is 2, BTW not 3?

Cheers


JC


0
 
LVL 13

Expert Comment

by:Bart Cremers
ID: 20333366
monthDiff(getDate(1992, 10), getDate(1991, 12)));

private static Date getDate(int year, int month) {
    Calendar cal = Calendar.getInstance();
    cal.set(Calendar.YEAR, year);
    cal.set(Calendar.MONTH, month);
    return cal.getTime();
}
 
private static int monthDiff(Date one, Date two) {
    Calendar calFirst = Calendar.getInstance();
    Calendar calLast = Calendar.getInstance();
 
    if (!one.after(two)) {
        calFirst.setTime(one);
        calLast.setTime(two);
    } else {
        calFirst.setTime(two);
        calLast.setTime(one);
    }
 
    int diff = 0;
    while (!equalMonthAndYear(calFirst, calLast)) {
        diff++;
        calFirst.add(Calendar.MONTH, 1);
    }
    return diff;
}
 
private static boolean equalMonthAndYear(Calendar first, Calendar last) {
    return first.get(Calendar.YEAR) == last.get(Calendar.YEAR) 
           && first.get(Calendar.MONTH) == last.get(Calendar.MONTH);
}

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0
 
LVL 2

Expert Comment

by:Ronald112197
ID: 21076568
well, I guess all of the above works, but why make life so difficult?
Just compute the "total number of months" for each date and subtract them from each other... then add one to include both the first and the last month.

Like this:
public static int calDiff(int year1, int month1, int year2, int month2) {
	int monthsDiff = (year1 * 12 + month1) - (year2 * 12 + month2);
	return Math.abs(monthsDiff) + 1;
}
 
// or even:
int diff = Math.abs((year1 * 12 + month1) - (year2 * 12 + month2)) + 1;
 
A bit more compact than the solutions above, but at least as readable IMHO...
 
Cheers...

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0
 
LVL 2

Expert Comment

by:Ronald112197
ID: 21076586
btw... I just now realized what the accepted solution above does....
It performs really badly and uses recursion for no obvious reason...(it's not like some years have more months than others, right?)  You probably won't do that, but if you compute the months between October 1500 and December 2500, it will take at least 5000 times (!) longer (!) than my suggestion and use a LOT of memory in the process... (several megabytes)

Oh yes... AND it returns incorrect results if the two dates are not in the same year... (one month missing) which just shows that it's much too complex for the problem at hand...

P.S.: Just tried this just for fun: For a timespan of more than ~5500 years, you even get an OutOfMemory problem... (or rather a StackOverflowError because this does 5500 nested method calls!)...
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