I need to compare two dates using best accuracy, for me day will be ok and take into consideration leap years.
It is possible ? I for now only create function to compare month accurancy :/
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4 Answers
You can subtract Date object from another Date object and return microseconds
var d= new Date('2012/11/29');
var a= new Date('2012/11/30');
alert( (a-d) /(1000*24*60*60)) ); /* returns 1 */
Comments
If you have date strings, you can parse them by using Date.parse("datestring"). It will return long(time in milliseconds). And you can compare any two longs.
var date1 = new Date("10/25/2011");
var date2 = new Date("09/03/2010");
var date3 = new Date(Date.parse(date1) - Date.parse(date2));
var dayDiff = date3.getDate() - 1;
var monthDiff = date3.getMonth();
var yearDiff = date3.getFullYear() - 1970;
Here is jsfiddle to test it.
2 Comments
netmajor
Will this miliseconds consider leap years ? How can I convert miliseconds to days ? I have birth date and drive license date and I want to calculate that person have 17 years when pass license...
Mustafa Genç
I've edited my answer with some piece of code. You can test it.
The best JavaScript Time and Date manipulation library I've come across is Moment.js
Getting the # of days between two dates:
d1 = moment('2012-10-31')
d2 = moment('2012-11-02')
Math.abs(moment.duration(d1-d2, 'ms').days())
// => 2
The default precision is milliseconds.
2 Comments
charlietfl
@KKyle Burton have always used date.js which has been around a long time. Really like moment.js docs!
Roel
dayjs is much more lightweight than momentjs
This should work for difference in days with fairly good precision:
function daysSince( past ) {
return 0|( new Date().getTime() - past.getTime() ) * 1.16e-8;
}
console.log( daysSince( new Date('10/03/2012') ) ); //=> 31
Edit: Actually, if you only want to know the difference between two dates you can always return a positive number.
function daysBetweenDates( d1,d2 ) {
return Math.abs( 0|( d1.getTime() - d2.getTime() ) * 1.16e-8 );
}
