Question

I m 写一本用于Pi的GTD仪器。为了完成应有的任务,我要展示“明天”或“昨天”或“18月18日”。显然,即使任务时间不到24小时,我仍需要显示“Tomorrow”(例如星期六11pm的用户检查,并在星期天看到8am的任务)。因此,我写了在两个日期之间获得天数的方法。这部法律......

NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"yyyy-MM-dd-HH-mm"];

NSDate *nowDate = [dateFormatter dateFromString:@"2010-01-01-15-00"];
NSDate *dueDate = [dateFormatter dateFromString:@"2010-01-02-14-00"];

NSLog(@"NSDate *nowDate = %@", nowDate);
NSLog(@"NSDate *dueDate = %@", dueDate);

NSCalendar *calendar = [NSCalendar currentCalendar];

NSDateComponents *differenceComponents = [calendar components:(NSDayCalendarUnit)
                                                     fromDate:nowDate
                                                       toDate:dueDate
                                                      options:0];

NSLog(@"Days between dates: %d", [differenceComponents day]);

......并在此列出产出:

NSDate *nowDate = 2010-01-01 15:00:00 -0700
NSDate *dueDate = 2010-01-02 14:00:00 -0700
Days between dates: 0

如你所知,这种方法的结果不正确。它本应在两天之间恢复1天。我在这里做了什么错误?

EDIT:我写了另一种方法。我没有做过广泛的单位测试,但迄今为止似乎工作:

+ (NSInteger)daysFromDate:(NSDate *)fromDate inTimeZone:(NSTimeZone *)fromTimeZone untilDate:(NSDate *)toDate inTimeZone:(NSTimeZone *)toTimeZone {

    NSCalendar *calendar = [NSCalendar currentCalendar];
    unsigned unitFlags = NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit;

    [calendar setTimeZone:fromTimeZone];
    NSDateComponents *fromDateComponents = [calendar components:unitFlags fromDate:fromDate];

    [calendar setTimeZone:toTimeZone];
    NSDateComponents *toDateComponents = [calendar components:unitFlags fromDate:toDate];

    [calendar setTimeZone:[NSTimeZone defaultTimeZone]];
    NSDate *adjustedFromDate = [calendar dateFromComponents:fromDateComponents];
    NSDate *adjustedToDate = [calendar dateFromComponents:toDateComponents];

    NSTimeInterval timeIntervalBetweenDates = [adjustedToDate timeIntervalSinceDate:adjustedFromDate];
    NSInteger daysBetweenDates = (NSInteger)(timeIntervalBetweenDates / (60.0 * 60.0 * 24.0));

    NSDateComponents *midnightBeforeFromDateComponents = [[NSDateComponents alloc] init];
    [midnightBeforeFromDateComponents setYear:[fromDateComponents year]];
    [midnightBeforeFromDateComponents setMonth:[fromDateComponents month]];
    [midnightBeforeFromDateComponents setDay:[fromDateComponents day]];

    NSDate *midnightBeforeFromDate = [calendar dateFromComponents:midnightBeforeFromDateComponents];
    [midnightBeforeFromDateComponents release];

    NSDate *midnightAfterFromDate = [[NSDate alloc] initWithTimeInterval:(60.0 * 60.0 * 24.0)
                                                               sinceDate:midnightBeforeFromDate];

    NSTimeInterval timeIntervalBetweenToDateAndMidnightBeforeFromDate = [adjustedToDate timeIntervalSinceDate:midnightBeforeFromDate];
    NSTimeInterval timeIntervalBetweenToDateAndMidnightAfterFromDate = [adjustedToDate timeIntervalSinceDate:midnightAfterFromDate];

    if (timeIntervalBetweenToDateAndMidnightBeforeFromDate < 0.0) {

        // toDate is before the midnight before fromDate

        timeIntervalBetweenToDateAndMidnightBeforeFromDate -= daysBetweenDates * 60.0 * 60.0 * 24.0;

        if (timeIntervalBetweenToDateAndMidnightBeforeFromDate < 0.0)
            daysBetweenDates -= 1;
    }
    else if (timeIntervalBetweenToDateAndMidnightAfterFromDate >= 0.0) {

        // toDate is after the midnight after fromDate

        timeIntervalBetweenToDateAndMidnightAfterFromDate -= daysBetweenDates * 60.0 * 60.0 * 24.0;

        if (timeIntervalBetweenToDateAndMidnightAfterFromDate >= 0.0)
            daysBetweenDates += 1;
    }

    [midnightAfterFromDate release];

    return daysBetweenDates;
}

Answer 1

摘自<代码>components: FromDate:toDate:options::

如果没有一个足够小的单位要求完全确定差异,结果就会消失。

由于差异少于整整天,它正确地回归了零天的结果。

Answer 2

如果你们所关心的是明天或昨天的某个具体日期,那么你就可以节省很多工作,并且只是检验日期是否只有一个日历日。

为此,将较早发现的日期和较晚的日期(如果与结果相比是相等的,可以保释)进行比较,然后检验较早日期后1天的日期是否与一年、月份和较晚日期相同。

如果你真的想要知道从一日到另一天,日历日有多少:

Send the calendar a components:fromDate: message to get the year, month, and day-of-the-month of the first date.
Same as #1, but for the second date.
If the two dates are in the same year and month, subtract one day-of-month from the other and pass to abs (see abs(3)) to take the absolute value.
如果他们没有在同一年月,则检验他们是否在相邻的几个月内(例如,2010年12月至2011年1月,或2010年6月至2010年7月)。如果是的话,则在较早的月份增加天数(通过发送日历rangeOfUnit:inUnit:forDate:)。和NSMonthCalendarUnit,分别通过NSDayCalendarUnit和,然后将结果与较早的月份比较。
例如,在比较2010-12-31至2011-01时,你将首先确定这些月是相邻的,然后将31(2010-12年天数)增加到1(2011-01年月的天数),然后从这一数额中减去第31号(2010-12-31个月)。由于差异为1,较早的日期是日后一天。在比较2010-12-30至2011-01-<02时,你将确定在相邻的几个月内,然后将31(2010-12天)增加到2(2011-01-02个月),然后从该数额中减去30(2010-12-30个月)。





无论是哪种方式,我都强烈建议至少为这一守则进行书面单位测试。 我发现,处理日期的代码是最有可能出现仅每年两次出现的微妙的 b。

Answer 3

您可能尝试的一件事是使用<代码>rangeOfUnit:到零小时、分钟和秒钟。

NSCalendar *calendar = [NSCalendar currentCalendar];
NSCalendarUnit range = NSDayCalendarUnit;
NSDateComponents *comps = [[NSDateComponents alloc] init];
NSDate *start = [NSDate date];
NSDate *end;

[comps setDay:1];
[calendar rangeOfUnit:range startDate:&start interval:nil forDate:start];

end = [calendar dateByAddingComponents:comps toDate:start options:0];

举例来说,起首的编号为2010-06-19 00:00 -0400,结尾处为 2010-06-20 00:00:00 -0400<>。我 imagine想,这将更好地采用国家情报和安全局的比较方法,尽管我有过自己的检验。

Answer 4

我正在使用这部法典,它正在很好地发挥作用:

- (NSInteger)daysToDate:(NSDate*)date
{
    if(date == nil) {
        return NSNotFound;
    }
    NSUInteger otherDay = [[NSCalendar currentCalendar] ordinalityOfUnit:NSCalendarUnitDay   inUnit:NSCalendarUnitEra forDate:date];
    NSUInteger currentDay = [[NSCalendar currentCalendar] ordinalityOfUnit:NSCalendarUnitDay inUnit:NSCalendarUnitEra forDate:self];
    return (otherDay-currentDay);
}

Answer 5

Here is the function I ve used in the past its defined in a category on NSDate

- (int) daysToDate:(NSDate*) endDate
{
    //dates needed to be reset to represent only yyyy-mm-dd to get correct number of days between two days.
    NSDateFormatter *temp = [[NSDateFormatter alloc] init];
    [temp setDateFormat:@"yyyy-MM-dd"];
    NSDate *stDt = [temp dateFromString:[temp stringFromDate:self]];
    NSDate *endDt =  [temp dateFromString:[temp stringFromDate:endDate]];
    [temp release]; 
    unsigned int unitFlags = NSMonthCalendarUnit | NSDayCalendarUnit;
    NSCalendar *gregorian = [[NSCalendar alloc]
                             initWithCalendarIdentifier:NSGregorianCalendar];
    NSDateComponents *comps = [gregorian components:unitFlags fromDate:stDt  toDate:endDt  options:0];
    int days = [comps day];
    [gregorian release];
    return days;
}

Answer 6

 -(NSInteger)daysBetweenTwoDates:(NSDate*)fromDateTime andDate:(NSDate*)toDateTime

页: 1

NSDate *fromDate;
NSDate *toDate;

NSCalendar *calendar = [NSCalendar currentCalendar];

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

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

return [difference day];

iii

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