Iphone 查找从现在到目标日期每天将出现多少次

我正在制作一个iPhone倒计时应用程序，需要知道在用户设定日期之前，每个周末（如周六和周日）会发生多少次

例如，我如何确定从现在（8月6日，星期一）到下周有多少个周末我知道答案是1个周末，但我需要能够使用代码解决这个问题

我得到的最接近的结果是使用NSDateComponents和NSCalender，并做一些类似于以下的事情

NSDateComponents *weekendsLeftComponents = [calendar components:NSWeekCalendarUnit
                                                    fromDate:targetDate
                                                      toDate:[NSDate date]
                                                     options:0];

但从那以后，我就撞上了可怕的编码墙——禁止通行

我的问题是不知道该怎么做。这听起来像是一个阵列可以工作，但我担心阵列可能会变得相当大，因为其中一些“东西”可能长达几年。（我正在制作一个应用程序，可以知道一个人毕业前有多少天、周末等。）同时，我的受众主要是使用iPod的儿童到青少年，还有年纪较大的人。我不知道他们的内存（ram）在用完之前会有多满

提前非常感谢你

---

Alex Kafer

您需要指定您的问题。 定义什么是周末？对我来说，周末是周六和周日。因此，每当一个星期天出现时，它就是一个周末（请记住，星期天并不是整个周末——我只想简化要求）

那么，如何在代码中实现这一点呢？ 有很多方法。一种快速方法是创建一个数组，其中包含每一天[周一、周二、…、周日]

现在，您想知道在3周内本周二和周二之间出现了多少个周末。通过减去

targetDateComponent.week currentDateComponent.week

得到数字3。你只需通过这个数组，直到星期二出现3次，并计算所有的星期天

真的很简单。我不确定这是否完全符合你的问题。但是有很多方法

---

只要试着谨慎地想一想，你的

fromDate

和

toDate

看起来是向后的，假设

targetDate

在未来

下面是我解决这个问题的方法

在

NSCalendar

上创建一个类别，以计算特定工作日在两个日期之间出现的次数：

@interface NSCalendar (AlexCategory)

// The number of times weekday number `weekday` (1 = Sunday, 7 = Saturday)
// occurs between `fromDate` and `toDate`.  If `fromDate` falls on the desired
// weekday, it is counted.  If `toDate` falls on the desired weekday, it is NOT counted.
- (NSInteger)countOfWeekday:(NSInteger)weekday fromDate:(NSDate *)fromDate toDate:(NSDate *)toDate;

@end

为了实现这个新方法，我们将从获取fromDate的年、月、日和工作日开始：

@implementation NSCalendar (AlexCategory)

- (NSInteger)countOfWeekday:(NSInteger)weekday fromDate:(NSDate *)fromDate toDate:(NSDate *)toDate {
    NSDateComponents *components = [self components:NSDayCalendarUnit | NSMonthCalendarUnit | NSYearCalendarUnit | NSWeekdayCalendarUnit fromDate:fromDate];

    NSDateComponents *firstDesiredWeekday = [[NSDateComponents alloc] init];
    firstDesiredWeekday.year = components.year;
    firstDesiredWeekday.month = components.month;
    firstDesiredWeekday.day = components.day + daysUntilDesiredWeekday;

    fromDate = [self dateFromComponents:firstDesiredWeekday];
    if ([fromDate compare:toDate] != NSOrderedAscending) {
        return 0;
    }

    NSInteger allDaysCount = [self components:NSDayCalendarUnit
        fromDate:fromDate toDate:toDate options:0].day;

接下来，我们计算从

fromDate

到下一个所需工作日的天数：

    NSInteger daysUntilDesiredWeekday = weekday - components.weekday;
    // If fromDate is a Wednesday and weekday is Monday (for example),
    // daysUntilDesiredWeekday is negative.  Fix that.
    NSRange weekdayRange = [self minimumRangeOfUnit:NSWeekdayCalendarUnit];
    if (daysUntilDesiredWeekday < weekdayRange.location) {
        daysUntilDesiredWeekday += weekdayRange.length;
    }

我们将

fromDate

更新为所需的第一个工作日，如果在

toDate

上或之后，则返回0：

@implementation NSCalendar (AlexCategory)

- (NSInteger)countOfWeekday:(NSInteger)weekday fromDate:(NSDate *)fromDate toDate:(NSDate *)toDate {
    NSDateComponents *components = [self components:NSDayCalendarUnit | NSMonthCalendarUnit | NSYearCalendarUnit | NSWeekdayCalendarUnit fromDate:fromDate];

    NSDateComponents *firstDesiredWeekday = [[NSDateComponents alloc] init];
    firstDesiredWeekday.year = components.year;
    firstDesiredWeekday.month = components.month;
    firstDesiredWeekday.day = components.day + daysUntilDesiredWeekday;

    fromDate = [self dateFromComponents:firstDesiredWeekday];
    if ([fromDate compare:toDate] != NSOrderedAscending) {
        return 0;
    }

    NSInteger allDaysCount = [self components:NSDayCalendarUnit
        fromDate:fromDate toDate:toDate options:0].day;

接下来，我们计算从更新的

fromDate

到

toDate

的所有天数（不仅仅是期望的工作日）：

@implementation NSCalendar (AlexCategory)

- (NSInteger)countOfWeekday:(NSInteger)weekday fromDate:(NSDate *)fromDate toDate:(NSDate *)toDate {
    NSDateComponents *components = [self components:NSDayCalendarUnit | NSMonthCalendarUnit | NSYearCalendarUnit | NSWeekdayCalendarUnit fromDate:fromDate];

    NSDateComponents *firstDesiredWeekday = [[NSDateComponents alloc] init];
    firstDesiredWeekday.year = components.year;
    firstDesiredWeekday.month = components.month;
    firstDesiredWeekday.day = components.day + daysUntilDesiredWeekday;

    fromDate = [self dateFromComponents:firstDesiredWeekday];
    if ([fromDate compare:toDate] != NSOrderedAscending) {
        return 0;
    }

    NSInteger allDaysCount = [self components:NSDayCalendarUnit
        fromDate:fromDate toDate:toDate options:0].day;

我们可以将其除以一周中的天数来计算期望的工作日数。由于我们从一个期望的工作日开始计算，因此任何部分周的剩余部分也将包含期望的工作日，因此我们需要进行四舍五入：

    // Adding weekdayRange.length - 1 makes the integer division round up.
    return (allDaysCount + weekdayRange.length - 1) / weekdayRange.length;
}

@end

我们可以这样测试方法：

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

NSDateComponents *components = [[NSDateComponents alloc] init];
components.year = 2012;
components.month = 1;
components.day = 1;
NSDate *fromDate = [calendar dateFromComponents:components];

for (NSUInteger i = 1; i <= 365; ++i) {
    components.day = i;
    NSDate *toDate = [calendar dateFromComponents:components];
    NSLog(@"%@ to %@: %ld Mondays", fromDate, toDate, [calendar countOfWeekday:2 fromDate:fromDate toDate:toDate]);
}

2012-08-06 16:27:05.751 tuesdays[81152:403] 0 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-01 06:00:00 +0000
2012-08-06 16:27:05.754 tuesdays[81152:403] 0 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-02 06:00:00 +0000
2012-08-06 16:27:05.756 tuesdays[81152:403] 1 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-03 06:00:00 +0000
2012-08-06 16:27:05.758 tuesdays[81152:403] 1 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-04 06:00:00 +0000
2012-08-06 16:27:05.759 tuesdays[81152:403] 1 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-05 06:00:00 +0000
2012-08-06 16:27:05.760 tuesdays[81152:403] 1 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-06 06:00:00 +0000
2012-08-06 16:27:05.762 tuesdays[81152:403] 1 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-07 06:00:00 +0000
2012-08-06 16:27:05.763 tuesdays[81152:403] 1 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-08 06:00:00 +0000
2012-08-06 16:27:05.763 tuesdays[81152:403] 1 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-09 06:00:00 +0000
2012-08-06 16:27:05.764 tuesdays[81152:403] 2 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-10 06:00:00 +0000
2012-08-06 16:27:05.765 tuesdays[81152:403] 2 Mondays from 2012-01-01 06:00:00 +0000 to 2012-01-11 06:00:00 +0000

根据cal 1 2012的输出，这看起来是正确的：

    January 2012
Su Mo Tu We Th Fr Sa
 1  2  3  4  5  6  7
 8  9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31

---

（请记住，

toDate

不计算在内，因此从2012-1-1到2012-1-2有0个星期一，即使2012-1-2是星期一。）

在这种情况下，您是对的，数组可以创建更长的运行时间。但是一个7天的数组在两个日期之间的几周内迭代，它也应该有一个O（n）运行时，这意味着它相当快。。。但我可能错了。内存不会是问题，因为您只在阵列上使用。不确定是否有其他方法，但有几个“日历”问题，请立即解决！为了清楚起见，如果我想将它从周一改为周日，我会将工作日数从2改为1。对的