NSTimeInterval to HH:mm:ss?

Question

If I have an NSTimeInterval that is set to say 200.0, is there a way to convert that into 00:03:20, I was thinking I could initialise an NSDate with it and then use NSDateFo

Answer 1

To "extend" Matthias Bauch's suggestion, in Swift I would make this a computed property of NSTimeInterval:

extension NSTimeInterval {
  var stringValue: String {
    let interval = Int(self)
    let seconds = interval % 60
    let minutes = (interval / 60) % 60
    let hours = (interval / 3600)
    return String(format: "%02d:%02d:%02d", hours, minutes, seconds)
  }
}

The advantage of this is it's attached to the NSTimeInterval type, not your view controller or wherever else you put that function. To use you'd go something like:

let timeInterval = NSDate().timeIntervalSinceDate(start)        
self.elapsedTimeLabel.text = timeInterval.stringValue

Answer 2

Objective C version of onmyway133's answer

- (NSString*) formatTimeInterval:(NSTimeInterval) timeInterval {

    NSDateComponentsFormatter *formatter = [[NSDateComponentsFormatter alloc] init];

    formatter.zeroFormattingBehavior = NSDateComponentsFormatterZeroFormattingBehaviorPad;

    if (timeInterval > 3600) {
        formatter.allowedUnits = NSCalendarUnitHour | NSCalendarUnitMinute | NSCalendarUnitSecond;
    } else {
        formatter.allowedUnits = NSCalendarUnitMinute | NSCalendarUnitSecond;
    }

    return [formatter stringFromTimeInterval:timeInterval];
}

Answer 3

I guess, the timer fraction should be ceiled out. As Matthias' code was creating that issue in seconds, I use the following slightly modified from that of Matthias

    - (NSString *)stringFromTimeInterval:(NSTimeInterval)interval {
        int ti = (int) ceil(interval);
        int seconds = ti % 60;
        int minutes = (ti / 60) % 60;
        int hours = (ti / 3600);
        return [NSString stringWithFormat:@"%02d:%02d:%02d",hours, minutes, seconds];
    }

Answer 4

Swift 4

DateComponentsFormatter().string(from: <# TimeInterval #>)

ex:

DateComponentsFormatter().string(from: 59.0)

Answer 5

Some extra lines of code, but I feel using NSDateComponents will give a more precise value.

- (NSString *)getTimeRepresentationFromDate:(NSDate *)iDate withTimeInterval:(NSTimeInterval)iTimeInterval {
    NSString *aReturnValue = nil;
    NSDate *aNewDate = [iDate dateByAddingTimeInterval:iTimeInterval]; 

    unsigned int theUnits = NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit;
    NSCalendar *aCalender = [NSCalendar currentCalendar];
    NSDateComponents *aDateComponents = [aCalender components:theUnits fromDate:iDate toDate:aNewDate options:0];

    aReturnValue = [NSString stringWithFormat:@"%d:%d:%d", [aDateComponents hour], [aDateComponents minute], [aDateComponents second]];

    return aReturnValue;
}

Answer 6

NSTimeInterval ti = 3667;
double hours = floor(ti / 60 / 60);
double minutes = floor((ti - (hours * 60 * 60)) / 60);
double seconds = floor(ti - (hours * 60 * 60) - (minutes * 60));