How to get the difference between two dates?

I have two dates in format(MM/dd/yyyy hh:mm:ss:SS). For the both dates I have converted the two dates to strings by using (stringFromDate) method. But I could not get the difference between them and show them in my console. P开发者_如何学运维lease give me an idea how I should get it? Thank you.

Example

    NSDate *today = [NSDate date];

NSTimeInterval dateTime;


if ([visitDate isEqualToDate:today])   //visitDate is a NSDate

{

NSLog (@"Dates are equal");

}

dateTime = ([visitDate timeIntervalSinceDate:today] / 86400);  

if(dateTime < 0) //Check if visit date is a past date, dateTime returns - val

{

NSLog (@"Past Date");

}

else 

{   
NSLog (@"Future Date");

}

Keep the dates as dates, get the difference between them, then print the difference.

From the docs on NSCalendar and assuming gregorian is an NSCalendar:

NSDate *startDate = ...;

NSDate *endDate = ...;

unsigned int unitFlags = NSMonthCalendarUnit | NSDayCalendarUnit;

NSDateComponents *comps = [gregorian components:unitFlags fromDate:startDate  toDate:endDate  options:0];

int months = [comps month];

int days = [comps day];

Generally I see day delta calculations handled by converting day/year values into flat days (usually days since some starting epoch, like 01/01/1970).

To aid in this, I've found it helpful to create a table of days into the year that each month starts. Here's a class I used for this recently.

namespace {
    // Helper class for figuring out things like day of year
    class month_database {
    public:
        month_database () {

            days_into_year[0] = 0;
            for (int i=0; i<11; i++) {
                days_into_year[i+1] = days_into_year[i] + days_in_month[i];
            }
        };

        // Return the start day of the year for the given month (January = month 1).
        int start_day (int month, int year) const {

            // Account for leap years. Actually, this doesn't get the year 1900 or 2100 right,
            // but should be good enough for a while.
            if ( (year % 4) == 0 && month > 2) {
                return days_into_year[month-1] + 1;
            } else {
                return days_into_year[month-1];
            }
        }
    private:
        static int const days_in_month[12];

        // # of days into the year the previous month ends
        int days_into_year[12];
    };
    // 30 days has September, April, June, and November...
    int const month_database::days_in_month[12] = {31, 28, 31, 30,   31, 30, 31, 31,   30, 31, 30, 31};

    month_database month;
}

As you can see from the start_day method, the main issue you are going to be wrestling with is how many leap days are contained in your range. Within our epoch the calculation I used there is good enough. The actual rule for which years contain leap days is discussed here.

February 29 in the Gregorian calendar, the most widely used today, is a date that occurs only once every four years, in years evenly divisible by 4, such as 1976, 1996, 2000, 2004, 2008, 2012 or 2016 (with the exception of century years not divisible by 400, such as 1900).

If you just want the difference in days, you can do this. (Based on mihir mehta's answer.)

const NSTimeInterval kSecondsPerDay = 60 * 60 * 24;
- (NSInteger)daysUntilDate:(NSDate *)anotherDate {
    NSTimeInterval secondsUntilExpired = [self timeIntervalSinceDate:anotherDate];
    NSTimeInterval days = secondsUntilExpired / kSecondsPerDay;
    return (NSInteger)days;
}