11

我有一个方法,它通过将小时和秒组件分解为 NSDateComponents 来提取 NSDate。我的代码如下...

unsigned hourAndMinuteFlags = NSHourCalendarUnit | NSMinuteCalendarUnit;
NSCalendar* calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[calendar setTimeZone:[NSTimeZone timeZoneWithName:@"GMT"]];
NSDateComponents* travelDateTimeComponents = [calendar components:hourAndMinuteFlags fromDate:travelDate];
NSString* hours = [NSString stringWithFormat:@"%02i", [travelDateTimeComponents hour]];
NSString* minutes = [NSString stringWithFormat:@"%02i", [travelDateTimeComponents minute]];

我的问题是我在转换中损失了一个小时。我怀疑这是由于时区造成的,但据我所知,传入的日期和使用的日历都是格林威治标准时间。

例如,如果我传入以下 NSDate 对象(这是 [NSDate 描述] 的日志)...

2010-08-02 08:00:00 +0100

我希望我得到 8 小时和 0 分钟,但我实际上得到了 7 小时。

我的系统时间是格林威治标准时间,因此上面的 NSDate 目前是英国夏令时的 +0100。

4

3 回答 3

15

另请记住,NSDate对象始终根据 GMT 显示时间,而您从该日期提取的组件会被timeZoneWithName部件更改。

例子:

NSDate *today = [NSDate date];
NSLog(@"Today's date: %@",today);

unsigned hourAndMinuteFlags = NSHourCalendarUnit | NSMinuteCalendarUnit;
NSCalendar* calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[calendar setTimeZone:[NSTimeZone timeZoneWithName:@"GMT"]];
NSDateComponents* travelDateTimeComponents = [calendar components:hourAndMinuteFlags fromDate:today];
NSString* hours = [NSString stringWithFormat:@"%02i", [travelDateTimeComponents hour]];
NSString* minutes = [NSString stringWithFormat:@"%02i", [travelDateTimeComponents minute]];

NSLog(@"Calendar: %@",calendar);
NSLog(@"Travel Components: %@",travelDateTimeComponents);
NSLog(@"Hours: %@",hours);
NSLog(@"Minutes: %@",minutes);

控制台输出:

[Session started at 2011-01-23 12:59:17 -0700.]
2011-01-23 12:59:19.755 testttt[5697:207] Today's date: 2011-01-23 19:59:19 +0000
2011-01-23 12:59:19.756 testttt[5697:207] Calendar: <__NSCFCalendar: 0x4b2ca70>
2011-01-23 12:59:19.757 testttt[5697:207] Travel Components: <NSDateComponents: 0x4b2de20>
2011-01-23 12:59:19.757 testttt[5697:207] Hours: 19
2011-01-23 12:59:19.758 testttt[5697:207] Minutes: 59

现在,如果我们更改时区...

NSDate *today = [NSDate date];
NSLog(@"Today's date: %@",today);

unsigned hourAndMinuteFlags = NSHourCalendarUnit | NSMinuteCalendarUnit;
NSCalendar* calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[calendar setTimeZone:[NSTimeZone timeZoneWithName:@"MST"]];
NSDateComponents* travelDateTimeComponents = [calendar components:hourAndMinuteFlags fromDate:today];
NSString* hours = [NSString stringWithFormat:@"%02i", [travelDateTimeComponents hour]];
NSString* minutes = [NSString stringWithFormat:@"%02i", [travelDateTimeComponents minute]];

NSLog(@"Calendar: %@",calendar);
NSLog(@"Travel Components: %@",travelDateTimeComponents);
NSLog(@"Hours: %@",hours);
NSLog(@"Minutes: %@",minutes);

输出:

2011-01-23 13:05:29.896 testttt[5723:207] Today's date: 2011-01-23 20:05:29 +0000
2011-01-23 13:05:29.897 testttt[5723:207] Calendar: <__NSCFCalendar: 0x4e10020>
2011-01-23 13:05:29.897 testttt[5723:207] Travel Components: <NSDateComponents: 0x4e0f6d0>
2011-01-23 13:05:29.898 testttt[5723:207] Hours: 13
2011-01-23 13:05:29.898 testttt[5723:207] Minutes: 05

请注意当地小时和分钟如何变化,但“今天的日期:”部分仍反映 GMT。如果你问我,这对程序员有点误导。

于 2011-01-23T20:09:21.743 回答
5

问题是,GMT 不等于英国夏令时。日期2010-08-02 08:00:00 +0100等于2010-08-02 07:00:00 GMT所以 7 小时是您应该期望的结果。

于 2011-01-23T12:47:09.687 回答
4

我同意水。这是非常具有误导性的,并导致我几天都在用头撞屏幕。它的工作原理如下: [NSDate date] 总是返回当前日期。所以在上面的例子中,它将是 2010-08-02 08:00:00。当我们转换为 NSDateComponents 时,实际时间会丢失并转换为一组不再包含时区的整数(小时、年等)。

转换回 NSDate 对象时,它将获取年、月、日等的值,并相对于 GMT 将其提供给您,因此会损失小时。这只是框架中决定的方式。我要做的是:

+(NSDateComponents *)getComponentFromDate:(NSDate *)date {

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

    unsigned unitFlags = NSYearCalendarUnit | NSMonthCalendarUnit | NSWeekdayCalendarUnit | NSDayCalendarUnit;
    NSDateComponents* components = [gregorian components:unitFlags fromDate:date];

    NSTimeZone* timeZone = [NSTimeZone localTimeZone];
    if(timeZone.isDaylightSavingTime) components.hour = ((int)timeZone.daylightSavingTimeOffset/3600);

    [gregorian release];

    return components;
}

这将为我提供四舍五入到当天的组件的“更正”版本,当传递给日历以检索日期时,将始终设置为作为参数传递的日期当天的 00:00:00。

于 2011-09-14T13:15:16.000 回答