How to convert an NSTimeInterval (seconds) into minutes

Question

I\'ve got an amount of seconds that passed from a certain event. It\'s stored in a NSTimeInterval data type.

I want to convert it into

Answer 1

Brief Description

1. The answer from Brian Ramsay is more convenient if you only want to convert to minutes.
2. If you want Cocoa API do it for you and convert your NSTimeInterval not only to minutes but also to days, months, week, etc,... I think this is a more generic approach

3. Use NSCalendar method:

   • (NSDateComponents *)components:(NSUInteger)unitFlags fromDate:(NSDate *)startingDate toDate:(NSDate *)resultDate options:(NSUInteger)opts

   • "Returns, as an NSDateComponents object using specified components, the difference between two supplied dates". From the API documentation.

4. Create 2 NSDate whose difference is the NSTimeInterval you want to convert. (If your NSTimeInterval comes from comparing 2 NSDate you don't need to do this step, and you don't even need the NSTimeInterval).

5. Get your quotes from NSDateComponents

Sample Code

// The time interval 
NSTimeInterval theTimeInterval = 326.4;

// Get the system calendar
NSCalendar *sysCalendar = [NSCalendar currentCalendar];

// Create the NSDates
NSDate *date1 = [[NSDate alloc] init];
NSDate *date2 = [[NSDate alloc] initWithTimeInterval:theTimeInterval sinceDate:date1]; 

// Get conversion to months, days, hours, minutes
unsigned int unitFlags = NSHourCalendarUnit | NSMinuteCalendarUnit | NSDayCalendarUnit | NSMonthCalendarUnit;

NSDateComponents *conversionInfo = [sysCalendar components:unitFlags fromDate:date1  toDate:date2  options:0];

NSLog(@"Conversion: %dmin %dhours %ddays %dmoths",[conversionInfo minute], [conversionInfo hour], [conversionInfo day], [conversionInfo month]);

[date1 release];
[date2 release];

Known issues

• Too much for just a conversion, you are right, but that's how the API works.
• My suggestion: if you get used to manage your time data using NSDate and NSCalendar, the API will do the hard work for you.

Answer 2

Brian Ramsay’s code, de-pseudofied:

- (NSString*)formattedStringForDuration:(NSTimeInterval)duration
{
    NSInteger minutes = floor(duration/60);
    NSInteger seconds = round(duration - minutes * 60);
    return [NSString stringWithFormat:@"%d:%02d", minutes, seconds];
}

Answer 3

    NSDate *timeLater = [NSDate dateWithTimeIntervalSinceNow:60*90];

    NSTimeInterval duration = [timeLater  timeIntervalSinceNow];

    NSInteger hours = floor(duration/(60*60));
    NSInteger minutes = floor((duration/60) - hours * 60);
    NSInteger seconds = floor(duration - (minutes * 60) - (hours * 60 * 60));

    NSLog(@"timeLater: %@", [dateFormatter stringFromDate:timeLater]);

    NSLog(@"time left: %d hours %d minutes  %d seconds", hours,minutes,seconds);

Outputs:

timeLater: 22:27
timeLeft: 1 hours 29 minutes  59 seconds

Answer 4

pseudo-code:

minutes = floor(326.4/60)
seconds = round(326.4 - minutes * 60)

Answer 5

Swift 2 version

extension NSTimeInterval {
            func toMM_SS() -> String {
                let interval = self
                let componentFormatter = NSDateComponentsFormatter()

                componentFormatter.unitsStyle = .Positional
                componentFormatter.zeroFormattingBehavior = .Pad
                componentFormatter.allowedUnits = [.Minute, .Second]
                return componentFormatter.stringFromTimeInterval(interval) ?? ""
            }
        }
    let duration = 326.4.toMM_SS()
    print(duration)    //"5:26"

Answer 6

Forgive me for being a Stack virgin... I'm not sure how to reply to Brian Ramsay's answer...

Using round will not work for second values between 59.5 and 59.99999. The second value will be 60 during this period. Use trunc instead...

 double progress;

 int minutes = floor(progress/60);
 int seconds = trunc(progress - minutes * 60);