如何比较两个日期并返回相差的天数

Question

如何比较两个日期并返回天数。 例如：错过了杯赛的X天。 看看我的代码。

  NSDateFormatter *df = [[NSDateFormatter alloc]init];  
  [df setDateFormat:@"d MMMM,yyyy"];  
  NSDate *date1 = [df dateFromString:@"11-05-2010"];  
  NSDate *date2 = [df dateFromString:@"11-06-2010"];  
  NSTimeInterval interval = [date2 timeIntervalSinceDate:date1];  
  //int days = (int)interval / 30;  
  //int months = (interval - (months/30)) / 30;  
  NSString *timeDiff = [NSString stringWithFormat:@"%dMissing%d days of the Cup",date1,date2, fabs(interval)];  

  label.text = timeDiff; // output (Missing X days of the Cup)  

- Dans

---

你可以查看这个答案：https://dev59.com/-mrWa4cB1Zd3GeqP_Xl9，它帮助我解决了我的问题。 - Sergey Pekar

Answer 1

基本上使用NSCalendar，参考Apple的示例

NSDate * date1 = <however you initialize this>;
NSDate * date2 = <...>;

NSCalendar *gregorian = [[NSCalendar alloc]
                 initWithCalendarIdentifier:NSGregorianCalendar];

NSUInteger unitFlags = NSMonthCalendarUnit | NSDayCalendarUnit;

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

NSInteger months = [components month];
NSInteger days = [components day];

- darelf

---

我第一次错过了这个。 这也许就是他没有得到想要的原因。 - darelf

Answer 2

您可以使用下一个类别：

@interface NSDate (Additions)

-(NSInteger)numberOfDaysUntilDay:(NSDate *)aDate; 
-(NSInteger)numberOfHoursUntilDay:(NSDate *)aDate;

@end

@implementation NSDate (Additions)
const NSInteger secondPerMunite = 60;       
const NSInteger munitePerHour = 60;
const NSInteger hourPerDay = 24;

-(NSInteger)numberOfDaysUntilDay:(NSDate *)aDate 
{
    NSInteger selfTimeInterval = [aDate timeIntervalSinceDate:self];   
    return abs(selfTimeInterval / (secondPerMunite * munitePerHour * hourPerDay));    
}

-(NSInteger)numberOfHoursUntilDay:(NSDate *)aDate 
{
    NSInteger selfTimeInterval = [aDate timeIntervalSinceDate:self];
    return abs(selfTimeInterval / (secondPerMunite * munitePerHour));
}
@end

- Igor

---

这是一种非常错误的方法。改用NSCalendar方法。 - Andrew

Answer 3

尝试这个分类：

@interface NSDate (DateUtils)

-(NSInteger)numberOfDaysUntilDay:(NSDate *)aDate; 

@end

@implementation NSDate (DateUtils)

-(NSInteger)numberOfDaysUntilDay:(NSDate *)aDate
{

    NSCalendar *calendar = [[NSCalendar alloc]
                             initWithCalendarIdentifier:NSCalendarIdentifierGregorian];

    NSDateComponents *components = [calendar components:NSCalendarUnitDay
                                                fromDate:self
                                                  toDate:aDate options:kNilOptions];

    return [components day];

}

@end

您可以通过添加导入来使用此类别：

#import "NSDate+DateUtils.h"

然后从您的代码中调用它：

NSInteger days = [myDate numberOfDaysUntilDay:someOtherDate];

- reshat2

Answer 4

需要调用此方法并仅设置我们想要以不同方式计算的2个日期。

-(void) calculateSleepHours:(NSDate *)sleepDate :(NSDate *)wakeStr 
{

   if (sleepDate !=nil && wakeStr!=nil) {`enter code here`

    NSDateFormatter *dateFormatter1 = [[NSDateFormatter alloc] init];
    [dateFormatter1 setDateFormat:@"yyyy-MM-dd HH:mm:ssZ"];
    NSDate *date1 = [ApplicationManager getInstance].sleepTime;;

    NSDate *date2 = [ApplicationManager getInstance].wakeupTime;

    NSCalendar *gregorian = [[NSCalendar alloc]
                             initWithCalendarIdentifier:NSGregorianCalendar];

    NSUInteger unitFlags = NSMonthCalendarUnit | NSDayCalendarUnit|NSHourCalendarUnit|NSMinuteCalendarUnit|NSSecondCalendarUnit;

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

    NSInteger months = [components month];
    NSInteger days = [components day];
    NSInteger hours = [components hour];
    NSInteger minute=[components minute];
    NSInteger second=[components second];
    DLog(@"Month %ld day %ld hour is %ld min %ld sec %ld  ",(long)months,(long)days,(long)hours,(long)minute,(long)second);

    sleepHours.text=[NSString stringWithFormat:@"Hour %ld Min %ld Sec %ld",(long)hours,(long)minute,(long)second];
    }
}

- Chandni - Systematix