两个NSDates之间的天数[重复]-Java 学习之路

148

这个问题在这里已有答案：

Objective C - calculating the number of days between two dates 6个答案

我怎样才能确定两个 NSDate 值之间的天数（考虑到时间）？

NSDate 值的格式为 [NSDate date] .

具体来说，当用户在我的iPhone应用程序中进入非活动状态时，我存储以下值：

exitDate = [NSDate date];

当他们打开应用程序备份时，我得到当前时间：

NSDate *now = [NSDate date];

现在我想实现以下内容：

-(int)numberOfDaysBetweenStartDate:exitDate andEndDate:now

16 回答

这是我用来确定两个日期之间的日历天数的实现：

+ (NSInteger)daysBetweenDate:(NSDate*)fromDateTime andDate:(NSDate*)toDateTime
{
    NSDate *fromDate;
    NSDate *toDate;

    NSCalendar *calendar = [NSCalendar currentCalendar];

    [calendar rangeOfUnit:NSCalendarUnitDay startDate:&fromDate
        interval:NULL forDate:fromDateTime];
    [calendar rangeOfUnit:NSCalendarUnitDay startDate:&toDate
        interval:NULL forDate:toDateTime];

    NSDateComponents *difference = [calendar components:NSCalendarUnitDay
        fromDate:fromDate toDate:toDate options:0];

    return [difference day];
}

EDIT:

上面的神奇解决方案，这里的Swift版本作为 NSDate 的扩展：

extension NSDate {
  func numberOfDaysUntilDateTime(toDateTime: NSDate, inTimeZone timeZone: NSTimeZone? = nil) -> Int {
    let calendar = NSCalendar.currentCalendar()
    if let timeZone = timeZone {
      calendar.timeZone = timeZone
    }

    var fromDate: NSDate?, toDate: NSDate?

    calendar.rangeOfUnit(.Day, startDate: &fromDate, interval: nil, forDate: self)
    calendar.rangeOfUnit(.Day, startDate: &toDate, interval: nil, forDate: toDateTime)

    let difference = calendar.components(.Day, fromDate: fromDate!, toDate: toDate!, options: [])
    return difference.day
  }
}

根据您的使用情况，您可能想要移除一些力量展开 .

上述解决方案也适用于当前时区以外的时区，非常适合显示世界各地信息的应用程序 .

回复于 2024-05-02T16:52:21+08:00

116
这是我发现的最佳解决方案 . 似乎利用Apple批准的方法来确定NSDates之间的任何数量的单位 .
```
- (int)daysBetween:(NSDate *)dt1 and:(NSDate *)dt2 {
    NSUInteger unitFlags = NSDayCalendarUnit;
    NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar]; 
    NSDateComponents *components = [calendar components:unitFlags fromDate:dt1 toDate:dt2 options:0];
    return [components day]+1;
}
```
例如 . 如果您还想要几个月，那么您可以将'NSMonthCalendarUnit'作为unitFlag包含在内 .

为了归功于原始博主，我在这里找到了这个信息（虽然我上面已经修改了一个小错误）：http://cocoamatic.blogspot.com/2010/09/nsdate-number-of-days-between-two-dates.html?showComment=1306198273659#c6501446329564880344
回复于 2024-05-02T16:52:21+08:00

Swift 3.0 Update

extension Date {

    func differenceInDaysWithDate(date: Date) -> Int {
        let calendar = Calendar.current

        let date1 = calendar.startOfDay(for: self)
        let date2 = calendar.startOfDay(for: date)

        let components = calendar.dateComponents([.day], from: date1, to: date2)
        return components.day ?? 0
    }
}

Swift 2.0 Update

extension NSDate {

    func differenceInDaysWithDate(date: NSDate) -> Int {
        let calendar: NSCalendar = NSCalendar.currentCalendar()

        let date1 = calendar.startOfDayForDate(self)
        let date2 = calendar.startOfDayForDate(date)

        let components = calendar.components(.Day, fromDate: date1, toDate: date2, options: [])
        return components.day
    }

}

Original Solution

Swift的另一个解决方案 .

如果您的目的是获取两个日期之间的确切日期数，您可以解决此问题，如下所示：

// Assuming that firstDate and secondDate are defined
// ...

var calendar: NSCalendar = NSCalendar.currentCalendar()

// Replace the hour (time) of both dates with 00:00
let date1 = calendar.startOfDayForDate(firstDate)
let date2 = calendar.startOfDayForDate(secondDate)

let flags = NSCalendarUnit.DayCalendarUnit
let components = calendar.components(flags, fromDate: date1, toDate: date2, options: nil)

components.day  // This will return the number of day(s) between dates

回复于 2024-05-02T16:52:21+08:00

我将它用作NSDate类的类别方法

// returns number of days (absolute value) from another date (as number of midnights beween these dates)
- (int)daysFromDate:(NSDate *)pDate {
    NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSCalendarIdentifierGregorian];
    NSInteger startDay=[calendar ordinalityOfUnit:NSCalendarUnitDay
                                           inUnit:NSCalendarUnitEra
                                          forDate:[NSDate date]];
    NSInteger endDay=[calendar ordinalityOfUnit:NSCalendarUnitDay
                                         inUnit:NSCalendarUnitEra
                                        forDate:pDate];
    return abs(endDay-startDay);
}

回复于 2024-05-02T16:52:21+08:00

我需要两个日期之间的天数，包括开始日期 . 例如在2012年2月14日至2012年2月2日之间的几天将产生3的结果 .

+ (NSInteger)daysBetween:(NSDate *)dt1 and:(NSDate *)dt2 {
        NSUInteger unitFlags = NSDayCalendarUnit;
        NSCalendar* calendar = [NSCalendar currentCalendar];
        NSDateComponents *components = [calendar components:unitFlags fromDate:dt1 toDate:dt2 options:0];
        NSInteger daysBetween = abs([components day]);
    return daysBetween+1;
}

请注意，提供日期的顺序无关紧要 . 它将始终返回正数 .

回复于 2024-05-02T16:52:21+08:00

NSDate *lastDate = [NSDate date];
NSDate *todaysDate = [NSDate date];
NSTimeInterval lastDiff = [lastDate timeIntervalSinceNow];
NSTimeInterval todaysDiff = [todaysDate timeIntervalSinceNow];
NSTimeInterval dateDiff = lastDiff - todaysDiff;

dateDiff将是两个日期之间的秒数 . 只需除以一天中的秒数即可 .

回复于 2024-05-02T16:52:21+08:00

@布赖恩

布莱恩的答案很好，只计算24小时的天数差异，但不计算日历差异 . 例如12月24日23:59离圣诞节只有1分钟的距离，目的是许多申请被认为仍然是一天 . Brian的daysBetween函数将返回0 .

借用Brian的原始实现和一天的开始/结束，我在我的程序中使用以下内容：（NSDate beginning of day and end of day）

- (NSDate *)beginningOfDay:(NSDate *)date
{
    NSCalendar *cal = [NSCalendar currentCalendar];
    NSDateComponents *components = [cal components:( NSDayCalendarUnit | NSMonthCalendarUnit | NSYearCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit ) fromDate:date];
    [components setHour:0];
    [components setMinute:0];
    [components setSecond:0];
    return [cal dateFromComponents:components];
}

- (NSDate *)endOfDay:(NSDate *)date
{
    NSCalendar *cal = [NSCalendar currentCalendar];
    NSDateComponents *components = [cal components:( NSDayCalendarUnit | NSMonthCalendarUnit | NSYearCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit ) fromDate:date];
    [components setHour:23];
    [components setMinute:59];
    [components setSecond:59];
    return [cal dateFromComponents:components];
}

- (int)daysBetween:(NSDate *)date1 and:(NSDate *)date2 {
    NSDate *beginningOfDate1 = [self beginningOfDay:date1];
    NSDate *endOfDate1 = [self endOfDay:date1];
    NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
    NSDateComponents *beginningDayDiff = [calendar components:NSDayCalendarUnit fromDate:beginningOfDate1 toDate:date2 options:0];
    NSDateComponents *endDayDiff = [calendar components:NSDayCalendarUnit fromDate:endOfDate1 toDate:date2 options:0];
    if (beginningDayDiff.day > 0)
        return beginningDayDiff.day;
    else if (endDayDiff.day < 0)
        return endDayDiff.day;
    else {
        return 0;
    }
}

回复于 2024-05-02T16:52:21+08:00

另一种方法：

NSDateFormatter* dayFmt = [[NSDateFormatter alloc] init];
[dayFmt setTimeZone:<whatever time zone you want>];
[dayFmt setDateFormat:@"g"];
NSInteger firstDay = [[dayFmt stringFromDate:firstDate] integerValue];    
NSInteger secondDay = [[dayFmt stringFromDate:secondDate] integerValue];
NSInteger difference = secondDay - firstDay;

具有优于 timeIntervalSince... 方案的优势，可以考虑时区，并且在一天的短时间或长时间内没有间隔的模糊性 .

并且比NSDateComponents方法更紧凑，更少混乱 .

回复于 2024-05-02T16:52:21+08:00

只需为访问此页面的人添加一个答案，试图在Swift中执行此操作 . 方法几乎相同 .

private class func getDaysBetweenDates(startDate:NSDate, endDate:NSDate) -> NSInteger {

    var gregorian: NSCalendar = NSCalendar.currentCalendar();
    let flags = NSCalendarUnit.DayCalendarUnit
    let components = gregorian.components(flags, fromDate: startDate, toDate: endDate, options: nil)

    return components.day
}

在以下方法的讨论部分中找到了这个答案here：

components(_:fromDate:toDate:options:)

回复于 2024-05-02T16:52:21+08:00

这是在Swift中Brian的函数的实现：

class func daysBetweenThisDate(fromDateTime:NSDate, andThisDate toDateTime:NSDate)->Int?{

    var fromDate:NSDate? = nil
    var toDate:NSDate? = nil

    let calendar = NSCalendar.currentCalendar()

    calendar.rangeOfUnit(NSCalendarUnit.DayCalendarUnit, startDate: &fromDate, interval: nil, forDate: fromDateTime)

    calendar.rangeOfUnit(NSCalendarUnit.DayCalendarUnit, startDate: &toDate, interval: nil, forDate: toDateTime)

    if let from = fromDate {

        if let to = toDate {

            let difference = calendar.components(NSCalendarUnit.DayCalendarUnit, fromDate: from, toDate: to, options: NSCalendarOptions.allZeros)

            return difference.day
        }
    }

    return nil
}

回复于 2024-05-02T16:52:21+08:00

0

你的意思是日历日或24小时？即星期二上午6点，星期三上午6点，或不到一天？

如果你的意思是前者，那么's a bit complicated and you'将不得不诉诸于 NSCalendar 和 NSDateComponent 的操纵，我不记得我的头脑 .

如果你的意思是后者，只需获取自参考日期以来的日期时间间隔，从另一个中减去一个，然后除以24小时（ 24 * 60 * 60 ）得到大致的间隔，不包括闰秒 .

回复于 2024-05-02T16:52:21+08:00
13
有一个，不确定它是你想要的，但它可以帮助你们中的一些人，（帮帮我!!）

我的目标是知道在两个日期之间（差异小于24小时）我是否有“过日”日1：

我做了以下（我承认有点过时）
```
NSDate *startDate = ...
NSDate *endDate = ...
```
NSDate已经被另一个NSDateFormatter格式化了（这个只是为了这个目的:)
```
NSDateFormatter *dayFormater = [[NSDateFormatter alloc]init];
[dayFormater setDateFormat:@"dd"];

int startDateDay = [[dayFormater stringFromDate:startDate]intValue];

int endDateDay = [[dayFormater stringFromDate:dateOn]intValue];

if (endDateDay > startDateDay) {
    NSLog(@"day+1");
} else {
    NSLog(@"same day");
}
```
也许这样的东西已经存在，但没有找到它

蒂姆
回复于 2024-05-02T16:52:21+08:00

我找到的解决方案是：

+(NSInteger)getDaysDifferenceBetween:(NSDate *)dateA and:(NSDate *)dateB {

  if ([dateA isEqualToDate:dateB]) 
    return 0;

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



  NSDate * dateToRound = [dateA earlierDate:dateB];
  int flags = (NSYearCalendarUnit | NSMonthCalendarUnit |  NSDayCalendarUnit);
  NSDateComponents * dateComponents = 
         [gregorian components:flags fromDate:dateToRound];


  NSDate * roundedDate = [gregorian dateFromComponents:dateComponents];

  NSDate * otherDate = (dateToRound == dateA) ? dateB : dateA ;

  NSInteger diff = abs([roundedDate timeIntervalSinceDate:otherDate]);

  NSInteger daysDifference = floor(diff/(24 * 60 * 60));

  return daysDifference;
}

在这里，我有效地将第一个日期从一天开始，然后计算出Jonathan建议的差异......

回复于 2024-05-02T16:52:21+08:00

我已经发布了一个开源类/库来做这件事 .

看看RelativeDateDescriptor，可以用来获得如下的时差......

RelativeDateDescriptor *descriptor = [[RelativeDateDescriptor alloc] initWithPriorDateDescriptionFormat:@"%@ ago" postDateDescriptionFormat:@"in %@"];

// date1: 1st January 2000, 00:00:00
// date2: 6th January 2000, 00:00:00
[descriptor describeDate:date2 relativeTo:date1]; // Returns '5 days ago'
[descriptor describeDate:date1 relativeTo:date2]; // Returns 'in 5 days'

回复于 2024-05-02T16:52:21+08:00

402

为什么不呢：

int days = [date1 timeIntervalSinceDate:date2]/24/60/60;

回复于 2024-05-02T16:52:21+08:00

-1
为什么要注意使用以下NSDate方法：
```
- (NSTimeInterval)timeIntervalSinceDate:(NSDate *)anotherDate
```
这将返回两个日期之间的秒数，你可以除以86,400来获得天数!!
回复于 2024-05-02T16:52:21+08:00

两个NSDates之间的天数[重复]

16 回答

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