Cómo calcular el tiempo en horas entre dos fechas en iOS

Question

¿Cómo puedo calcular el tiempo transcurrido en horas entre dos veces (posiblemente en días diferentes) en iOS?

Answer 1

El NSDate función timeIntervalSinceDate: le dará la diferencia de dos fechas en cuestión de segundos.

NSDate* date1 = someDate; 
NSDate* date2 = someOtherDate; 
NSTimeInterval distanceBetweenDates = [date1 timeIntervalSinceDate:date2]; 
double secondsInAnHour = 3600; 
NSInteger hoursBetweenDates = distanceBetweenDates/secondsInAnHour; 

Sede, la biblioteca de referencia manzanahttp://developer.apple.com/library/mac/navigation/ o si está usando Xcode simplemente seleccione ayuda/documentación en el menú.

Ver: how-to-convert-an-nstimeinterval-seconds-into-minutes

--edit: Ver ÐąrέÐέvil's answer below for correctly handling daylight savings/leap seconds

Answer 2

NSCalendar *c = [NSCalendar currentCalendar]; 
NSDate *d1 = [NSDate date]; 
NSDate *d2 = [NSDate dateWithTimeIntervalSince1970:1340323201];//2012-06-22 
NSDateComponents *components = [c components:NSHourCalendarUnit fromDate:d2 toDate:d1 options:0]; 
NSInteger diff = components.minute; 

NSDayCalendarUnit|NSHourCalendarUnit|NSMinuteCalendarUnit 

cambiar los componentes necesarios a día, hora o minuto, lo que diferencia desea. Si NSDayCalendarUnit se selecciona entonces va a devolver el número de días entre dos fechas de manera similar para NSHourCalendarUnit y NSMinuteCalendarUnit

Answer 3

-(NSMutableString*) timeLeftSinceDate: (NSDate *) dateT { 

    NSMutableString *timeLeft = [[NSMutableString alloc]init]; 

    NSDate *today10am =[NSDate date]; 

    NSInteger seconds = [today10am timeIntervalSinceDate:dateT]; 

    NSInteger days = (int) (floor(seconds/(3600 * 24))); 
    if(days) seconds -= days * 3600 * 24; 

    NSInteger hours = (int) (floor(seconds/3600)); 
    if(hours) seconds -= hours * 3600; 

    NSInteger minutes = (int) (floor(seconds/60)); 
    if(minutes) seconds -= minutes * 60; 

    if(days) { 
     [timeLeft appendString:[NSString stringWithFormat:@"%ld Days", (long)days*-1]]; 
    } 

    if(hours) { 
     [timeLeft appendString:[NSString stringWithFormat: @"%ld H", (long)hours*-1]]; 
    } 

    if(minutes) { 
     [timeLeft appendString: [NSString stringWithFormat: @"%ld M",(long)minutes*-1]]; 
    } 

    if(seconds) { 
     [timeLeft appendString:[NSString stringWithFormat: @"%lds", (long)seconds*-1]]; 
    } 

    return timeLeft; 
} 

Answer 4

Pedido: Se encarga de horario de verano, año bisiesto, ya que utiliza el calendario para calcular el IOS. Puede cambiar la cadena y las condiciones para incluir minutos con horas y días.

+(NSString*)remaningTime:(NSDate*)startDate endDate:(NSDate*)endDate { 

    NSDateComponents *components; 
    NSInteger days; 
    NSInteger hour; 
    NSInteger minutes; 
    NSString *durationString; 

    components = [[NSCalendar currentCalendar] components: NSCalendarUnitDay|NSCalendarUnitHour|NSCalendarUnitMinute 
               fromDate: startDate toDate: endDate options: 0]; 
    days = [components day]; 
    hour = [components hour]; 
    minutes = [components minute]; 

    if (days > 0) { 

     if (days > 1) { 
      durationString = [NSString stringWithFormat:@"%d days", days]; 
     } 
     else { 
      durationString = [NSString stringWithFormat:@"%d day", days]; 
     } 
     return durationString; 
    } 

    if (hour > 0) { 

     if (hour > 1) { 
      durationString = [NSString stringWithFormat:@"%d hours", hour]; 
     } 
     else { 
      durationString = [NSString stringWithFormat:@"%d hour", hour]; 
     } 
     return durationString; 
    } 

    if (minutes > 0) { 

     if (minutes > 1) { 
      durationString = [NSString stringWithFormat:@"%d minutes", minutes]; 
     } 
     else { 
      durationString = [NSString stringWithFormat:@"%d minute", minutes]; 
     } 
     return durationString; 
    } 

    return @""; 
}