如何在iPhone中獲取兩天之間的差異

Question

我想計算iPhone（目標C）中兩個不同日期之間的天數，小時數和分鐘數。請提供任何代碼示例來執行相同的操作。

感謝 桑迪普

Answer 1

如果您的日期位於NSDate對象中，則可以使用timeIntervalSinceDate方法將差異查找爲NSTimeInterval，它將以秒爲單位給出差異。

然後，您可以通過一個簡單的算法轉換秒到天，小時和分鐘（這是我的頭頂部，未測試）：

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]; 

然後就檢查differenceComponents的性能。您可以在隨XCode提供的文檔中找到所有這些屬性

Answer 3

我已經寫了一個全面的方案，以給差在C中的兩個日期之間（不知道這是你的目標C的意思）。請在這裏找到原作。

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

Answer 4

建立關費魯喬的回答，這樣的事情將讓你尋找到這個時間差爲一個連接字符串，你可以告訴你有多少天，小時和分鐘之日起比較（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];