Question

我已经阅读了所有问题和答案以及有关此主题的所有教程，但由于某种原因，它不适合我。总是告诉我两个日期是相同的日期！

请有人帮我弄明白，我只是想检查一个人是否比另一个人大（包括日期和时间 - 没有秒数）或是否相等。

这是我的代码：

- (BOOL)isEndDateIsBiggerThanCurrectDate:(NSDate *)checkEndDate
{
    NSString *endd = [NSDateFormatter localizedStringFromDate:checkEndDate
                                                      dateStyle:NSDateFormatterShortStyle
                                                      timeStyle:NSDateFormatterShortStyle];

    NSString *curreeeent = [NSDateFormatter localizedStringFromDate:[NSDate date]
                                                      dateStyle:NSDateFormatterShortStyle
                                                      timeStyle:NSDateFormatterShortStyle];

    NSDateFormatter * df = [[NSDateFormatter alloc]init];;

    NSDate * newCurrent = [df dateFromString:endd];
    NSDate * newEnd = [df dateFromString:curreeeent];

    switch ([newCurrent compare:newEnd])
    {
        case NSOrderedAscending:
            return YES;
            break;
        case NSOrderedSame:
            return NO;
            break;
        case NSOrderedDescending:
            return NO;
            break;
    }
}

非常感谢！

Answer 1

为此，您必须使用NSCalender。

NSCalendar *calendar = [NSCalendar currentCalendar];
NSInteger desiredComponents = (NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit);


NSDateComponents *firstComponents = [calendar components:desiredComponents fromDate:[NSDate date]];
NSDateComponents *secondComponents = [calendar components:desiredComponents fromDate: checkEndDate];

NSDate *first = [calendar dateFromComponents:firstComponents];
NSDate *second = [calendar dateFromComponents:secondComponents];

NSComparisonResult result = [first compare:second];
if (result == NSOrderedAscending) {
    //checkEndDate is before now
} else if (result == NSOrderedDescending) {
    //checkEndDate is after now
}  else {
    //both are same
}

Answer 2

您应该使用时间间隔，而不是在日期和字符串之间进行转换。

以下内容应符合您的需求：

//current time
NSDate *now = [NSDate date];

//time in the future
NSDate *distantFuture = [NSDate distantFuture];

//gather time interval
if([now timeIntervalSinceDate:distantFuture] > 0)
{
    //huzzah!
}

Answer 3

我得到了答案，只是检查两个日期之间的确切时间并进行比较。

- (BOOL)isEndDateIsSmallerThanCurrent:(NSDate *)checkEndDate
{
    NSDate* enddate = checkEndDate;
    NSDate* currentdate = [NSDate date];
    NSTimeInterval distanceBetweenDates = [enddate timeIntervalSinceDate:currentdate];
    double secondsInMinute = 60;
    NSInteger secondsBetweenDates = distanceBetweenDates / secondsInMinute;

    if (secondsBetweenDates == 0)
        return YES;
    else if (secondsBetweenDates < 0)
        return YES;
    else
        return NO;
}

Answer 4

为什么不将日期更改为自1970年以来的时间间隔并按此排序。非常简单的数字比较，比字符串比较快得多，并且它们总是排序正确，而不是像1,10,11,2,21,22,3 ...... ....

NSDate *now = [NSDate date];
NSTimeInterval ti = [now timeIntervalSince1970];

多数民众赞成。没有新的对象创建，更快，更少的CPU负担。

请看这里你如何摆脱秒，但这很简单，因为你有数字，几秒钟。见How to set seconds to zero for NSDate

NSDate比较效果不佳

4 个答案: