如何在iPhone中获取两天之间的差异

Question

我想计算iPhone（目标C）中两个不同日期之间的天数，小时数和分钟数。请提供任何代码示例来执行相同的操作。

感谢 桑迪普

Answer 1

如果您的日期位于NSDate对象中，则可以使用timeIntervalSinceDate方法将差异查找为NSTimeInterval，它将以秒为单位给出差异。

然后，您可以通过一个简单的算法转换秒到天，小时和分钟（这是我的头顶部，未测试）：

minutes = seconds/60; 
seconds %= 60; 
hours = minutes/60; 
minutes %= 60; 
days = hours/24; 
hours %= 24; 

Answer 2

NSCalendar* gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar]; 
    unsigned int uintFlags = NSYearCalendarUnit | NSDayCalendarUnit | NSMonthCalendarUnit | NSWeekCalendarUnit | NSWeekdayCalendarUnit | NSHourCalendarUnit; 
    NSDateComponents* differenceComponents = [gregorian components:uintFlags fromDate:firstDate toDate:secondDate options:0]; 

然后就检查differenceComponents的性能。您可以在随XCode提供的文档中找到所有这些属性

Answer 3

我已经写了一个全面的方案，以给差在C中的两个日期之间（不知道这是你的目标C的意思）。请在这里找到原作。

http://tech.bragboy.com/2010/02/different-between-two-days-implemented.html

Answer 4

建立关费鲁乔的回答，这样的事情将让你寻找到这个时间差为一个连接字符串，你可以告诉你有多少天，小时和分钟之日起比较（YOURNSDATE）至今

NSTimeInterval timeDifference = [[NSDate date] timeIntervalSinceDate:YOURNSDATE]; 
     NSInteger seconds = (NSInteger)timeDifference; 
     NSInteger minutes; 
     NSInteger hours; 
     NSInteger days; 

     minutes = seconds/60; 
     seconds %= 60; 
     hours = minutes/60; 
     minutes %= 60; 
     days = hours/24; 
     hours %= 24; 

     NSString* daysString = [NSString stringWithFormat:@"%i days ", days]; 
     if(days == 0){ 
      daysString = @""; 
     }else if(days == 1){ 
      daysString = [NSString stringWithFormat:@"%i day ", days]; 
     } 
     NSString* hoursString = [NSString stringWithFormat:@"%i hrs ", hours]; 
     if(hours == 0){ 
      hoursString = @""; 
     }else if(hours == 1){ 
      hoursString = [NSString stringWithFormat:@"%i hr ", hours]; 
     } 
     NSString* minutesString = [NSString stringWithFormat:@"%i min", minutes]; 
     if(minutes == 0){ 
      minutesString = @""; 
     } 

     NSString *AllDatesString = [NSString stringWithFormat:@"%@%@%@", daysString, hoursString, minutesString];