Comparing dates only by year, month and date?

Question

Say we have 2 Date objects in Javascript;

var d1 = new Date('...');
var d2 = new Date('...');

We do a comparison:

d1 < d2;

This comparison will always take into account hours, minutes, seconds.

I want it to only take into account the year, month and date for comparison.

What's the easiest way to do this?

jQuery is allowed aswell.

Answer 1

Reset the hours, minutes, seconds, and milliseconds:

var d1 = new Date();
d1.setHours(0);
d1.setMinutes(0);
d1.setSeconds(0);
d1.setMilliseconds(0);

Or using setHours, which is less verbose:

var d1= new Date();
d1.setHours(0, 0, 0, 0);

And finally, to compare if the resulting Dates are the same use getTime():

d1.getTime() == d2.getTime()

Answer 2

As an algebraic solution, you could just run a bit of math:

function sameDay(d1, d2) {
    return d1 - d1 % 86400000 == d2 - d2 % 86400000
}

The equation actually breaks down as:

function sameDay(d1, d2) {
    var d1HMS, //hours, minutes, seconds & milliseconds
        d2HMS,
        d1Day,
        d2Day,
        result;
    //d1 and d2 will be implicitly cast to Number objects
    //this is to be explicit
    d1 = +d1;
    d2 = +d2;
    //1000 milliseconds in a second
    //60 seconds in a minute
    //60 minutes in an hour
    //24 hours in a day
    //modulus used to find remainder of hours, minutes, seconds, and milliseconds
    //after being divided into days
    d1HMS = d1 % (1000 * 60 * 60 * 24);
    d2HMS = d2 % (1000 * 60 * 60 * 24);
    //remove the remainder to find the timestamp for midnight of that day
    d1Day = d1 - d1HMS;
    d2Day = d2 - d2HMS;
    //compare the results
    result = d1Day == d2Day;
    return result;
}

This has the advantage of not losing data on the original Date objects, as setHours and the like will modify the referenced object.

Alternatively, a safe sameDay function using setHours could be written as:

function sameDay(d1, d2) {
    var a,
        b;
    a = new Date(+d1);
    b = new Date(+d2);
    a.setHours(0, 0, 0, 0);
    b.setHours(0, 0, 0, 0);
    return +a == +b;
}