Question

我有一个网络服务,可以收回一名JSON物体。答复包括UNIX(milliseconds)格式的复读时间。我将UNIX时间邮票改为NSDate,其代码如下:

NSDate *date = [NSDate dateWithTimeIntervalSince1970:[[timeInUNIX objectAtIndex:i]doubleValue]/1000];

这使我有时间以形式发言。

2009-11-13 19:47:49 +0000

我只想到这个日期。这一点很重要,因为这一天用于国家安全局的描述人员,也作为表层的主人,因此,我把日期和时间与以下法典分开:

+ (NSDate *)dateWithOutTime:(NSDate *)datDate {
    if( datDate == nil ) {
        datDate = [NSDate date];
    }
    NSDateComponents* comps = [[NSCalendar currentCalendar] components:NSYearCalendarUnit|NSMonthCalendarUnit|NSDayCalendarUnit fromDate:datDate];
    return [[NSCalendar currentCalendar] dateFromComponents:comps];
}

The problem here is the returned NSDate object still contains a junk time value. Is there any way to store only the date in NSDate format?

Can someone help me with this?

Regards, iSee

Answer 1

NSDate必须有一个时间——这是自参考日期以来的几秒钟,它代表着一个具体的时间点。

If you are only interested in the day/month/year sections, the correct way to deal with this is to use a date formatter or date components when presenting the date, rather then when you store it. Alternatively, store the date as a string, but this way you lose any date functionality.

Answer 2

两种可能的解决办法,两者都不是非常粗略的:

Take the NSTimeInterval from the date, then truncate it down to the nearest 60*60*24 (seconds*minutes*hours). Something like interval = floor(interval - interval % (60*60*24)). Create a new date from that.
Convert the date into a string showing only the date, then convert that string back to NSDate (might still have junk time components, though, but you can pad the string with 0 values).

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