objective-c - 仅获取两个日期之间的周末

Question

我试图只获取两个日期之间的周六和周日，但我不知道为什么要在一周内获取空闲日。

这是我的代码:

- (BOOL)checkForWeekend:(NSDate *)aDate {
    BOOL isWeekendDate = NO;
    NSCalendar *calendar = [NSCalendar currentCalendar];
    NSRange weekdayRange = [calendar maximumRangeOfUnit:NSWeekdayCalendarUnit];
    NSDateComponents *components = [calendar components:NSWeekdayCalendarUnit fromDate:aDate];
    NSUInteger weekdayOfDate = [components weekday];

    if (weekdayOfDate == weekdayRange.location || weekdayOfDate == weekdayRange.length) {
        // The date falls somewhere on the first or last days of the week.
        isWeekendDate = YES;
    }
    return isWeekendDate;
}


- (void)viewWillAppear:(BOOL)animated
{
    [super viewWillAppear:animated];

    NSString *strDateIni = [NSString stringWithString:@"28-01-2012"];
    NSString *strDateEnd = [NSString stringWithString:@"31-01-2012"];

    NSDateFormatter *df = [[NSDateFormatter alloc] init];
    [df setDateFormat:@"dd-MM-yyyy"];
    NSDate *startDate = [df dateFromString:strDateIni];
    NSDate *endDate = [df dateFromString:strDateEnd];

    unsigned int unitFlags = NSMonthCalendarUnit | NSDayCalendarUnit;

    NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
    NSDateComponents *comps = [gregorian components:unitFlags fromDate:startDate  toDate:endDate  options:0];

   // int months = [comps month];
    int days = [comps day];

    for (int i=0; i<days; i++) 
    {

        NSTimeInterval interval = i;
        NSDate * futureDate = [startDate dateByAddingTimeInterval:interval];

        BOOL isWeekend = [self checkForWeekend:futureDate]; // Any date can be passed here.

        if (isWeekend) {
            NSLog(@"Weekend date! Yay!");
        }
        else
        {
            NSLog(@"Not is Weekend");
        }


    }

}

问题:该问题是由 NSTimeInterval Interval = i; 引起的，for 循环的逻辑是按天迭代。将时间间隔设置为 i 以秒为单位迭代。

摘自 NSTimeInterval 文档

NSTimeInterval is always specified in seconds;

答案:

将 NSTimeInterval 行更改为

NSTimeInterval interval = i*24*60*60;

Answer 1

Here is a link to another answer我在SO上发帖(我知道无耻)。它有一些代码可以帮助您确定 future 的日期。这些方法作为 NSDate 的类别实现，这意味着它们成为 NSDate 的方法。

那里有几个功能可以帮助您度过周末。但这两个可能是最有帮助的:

- (NSDate*) theFollowingWeekend;
- (NSDate *) thePreviousWeekend;

它们将周末的日期返回给接收者(自己)。

一般来说，你不应该使用一天是 86400 秒的概念，而应该使用 NSDateComponents 和 NSCalendar。即使日期之间发生夏令时转换，此功能也有效。像这样:

- (NSDate *) dateByAddingDays:(NSInteger) numberOfDays {
    NSDateComponents *dayComponent = [[NSDateComponents alloc] init];
    dayComponent.day = numberOfDays;

    NSCalendar *theCalendar = [NSCalendar currentCalendar];
    return [theCalendar dateByAddingComponents:dayComponent toDate:self options:0];
}