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Subtracting dates to gets the number of Months(continued)

>Comment from GwynforWeb
 
continued from > http://www.experts-exchange.com/Web/Web_Languages/JavaScript/Q_20933048.html

>mi_cky, The diffrence between dates in terms of months is not defined properly beacuse of the different number off days in >each month. How would you define the answer to subtract  26/03/2004   from 25/12/2005. The difference between dates in >terms of milli sewcondsseconds  or days etc is easy, eg

so how do i accurately work out the amout of months between dates?
diff=d2.getTime()-d1.getTime()

based on this calculation
/calculate and output anything you want
//alert(diff+' milli secs')  
//alert(diff/1000+' secs')
//alert(diff/(1000*60)+' mins')  
//alert(diff/(1000*60*60)+' hours')
alert(diff/(1000*60*60*24)+' days')

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mi_cky
Asked:
mi_cky
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1 Solution
 
GwynforWebCommented:
What is a month that is the question, this is probably as good as you are going to get

<script>
dateStr2='26/03/2004'
dateStr1='25/12/2005'
dArr1=dateStr1.split('/')
dArr2=dateStr2.split('/')  

months=(dArr1[2]-dArr2[2])*12 + 1*(dArr1[1]-dArr2[1]) + (dArr1[0]-dArr2[0])/28
months=months.toFixed(2)
alert(months)
</script>      
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searlasCommented:
mi_cky, as mentioned in the referenced question, you need to specify your definition of a month.
today - tomorrow: 0 months, or 1 month?
31/03/2004 - 01/04/2004: 0 months or 1 month?
03/03/2004 - 03/04/2004: clearly 1 month
02/03/2004 - 03/04/2004: 1 month or 2 months?
31/01/2004 - 28/02/2004: 0 months or 1 month?
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GwynforWebCommented:
this is better

months=(dArr1[2]-dArr2[2])*12 + 1*(dArr1[1]-dArr2[1]) + 365*(dArr1[0]-dArr2[0])/12
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ZvonkoSystems architectCommented:
365.25
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mi_ckyAuthor Commented:
sorry, wanted to actally award the points to GwynforWeb, :)
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ZvonkoSystems architectCommented:
Make a request here and the question can be again unlocked:
http://www.experts-exchange.com/Community_Support/

Also be aware that grading B brings for nobody an advantage.

Cheers,
Zvonko
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mi_ckyAuthor Commented:
thanks Zvonko
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