Question

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Finding date by subtracting X number of days from a particular date in Javascript

I have a jQuery datepicker which allows users to pick a Friday from any given week. I want to populate fields based off that Friday and input the dates from Mon-Thurs as well.

I.e. the user picks this coming Friday (10-26-2012), I want to subtract from the day by 1 to get Thursday date, and then Wed, etc etc.

How would I approach this?

    $(document).ready(function(){
        $("#friDate").Zebra_DatePicker({
            format: 'm-d-Y',
            disabled_dates: ['* * * 0-4,6'],
            first_day_of_week: 0,
            onSelect: function(){
                var date = $("#friDate").val();
                $("#mon").val('INSERT MON DATE');
                $("#tue").val('INSERT TUE DATE');
                $("#wed").val('INSERT WED DATE');
                $("#thu").val('INSERT THU DATE');
                $("#fri").val(date);                    
            }
        });
    });

Any help is greatly appreciated! Thanks :).

Answer 1

To subtract a day from an existing date, just subtract the number of milliseconds in a day.

var today = new Date();
var yesterday = new Date(today.getTime() - 1000*60*60*24);

// As pointed out by Waxen, we need to account for daylight savings changes
function addDays(date, days) {
    var newDate = new Date(date.getTime() + (1000*60*60*24*days) );
    // To account for daylight savings differences
    // The hours won't be the same when crossing the daylight savings changes
    var hourDifference =  newDate.getHours() - date.getHours();
    var msInHour = 60*60*1000;
    if (hourDifference > 0) {
        newDate = new Date(newDate.getTime() + msInHour) ;
    } else if (hourDifference < 0){
        newDate = new Date(newDate.getTime() - msInHour) ;    
    }    
    return newDate;    
}

So for your code, you could do the following

$("#mon").val(addDays(date, -4));
$("#tue").val(addDays(date, -3));
$("#wed").val(addDays(date, -2);
$("#thu").val(addDays(date, -1));
$("#fri").val(date);             

UPDATE Though my code works, Waxen’s solution is simpler and I would use that. I’ll leave this answer here for reference