55

我有一个跨越数年的时间间隔,我想要从一年到秒的所有时间组件。

我的第一个想法是将时间间隔除以一年中的秒数,从运行总秒数中减去,再除以一个月中的秒数,从运行总数中减去,等等。

这看起来很复杂,而且我读到,每当你做一些看起来很复杂的事情时,可能有一个内置的方法。

在那儿?

我将 Alex 的第二种方法集成到我的代码中。

它在我的界面中由 UIDatePicker 调用的方法中。

NSDate *now = [NSDate date];
NSDate *then = self.datePicker.date;
NSTimeInterval howLong = [now timeIntervalSinceDate:then];

NSDate *date = [NSDate dateWithTimeIntervalSince1970:howLong];
NSString *dateStr = [date description];
const char *dateStrPtr = [dateStr UTF8String];
int year, month, day, hour, minute, sec;

sscanf(dateStrPtr, "%d-%d-%d %d:%d:%d", &year, &month, &day, &hour, &minute, &sec);
year -= 1970;

NSLog(@"%d years\n%d months\n%d days\n%d hours\n%d minutes\n%d seconds", year, month, day, hour, minute, sec);

当我将日期选择器设置为过去 1 年零 1 天的日期时,我得到:

1年1个月1天16小时0分20秒

这是1个月零16小时的休息时间。如果我将日期选择器设置为过去 1 天,我会离开相同的数量。

更新:我有一个应用程序可以计算你的年龄,给定你的生日(从 UIDatePicker 设置),但它经常关闭。这证明有一个不准确的地方,但我无法弄清楚它来自哪里,你能吗?

4

9 回答 9

102

Brief Description

  1. Just another approach to complete the answer of JBRWilkinson but adding some code. It can also offers a solution to Alex Reynolds's comment.

  2. Use NSCalendar method:

    • (NSDateComponents *)components:(NSUInteger)unitFlags fromDate:(NSDate *)startingDate toDate:(NSDate *)resultDate options:(NSUInteger)opts

    • "Returns, as an NSDateComponents object using specified components, the difference between two supplied dates". (From the API documentation).

  3. Create 2 NSDate whose difference is the NSTimeInterval you want to break down. (If your NSTimeInterval comes from comparing 2 NSDate you don't need to do this step, and you don't even need the NSTimeInterval, just apply the dates to the NSCalendar method).

  4. Get your quotes from NSDateComponents

Sample Code

// The time interval 
NSTimeInterval theTimeInterval = ...;

// Get the system calendar
NSCalendar *sysCalendar = [NSCalendar currentCalendar];

// Create the NSDates
NSDate *date1 = [[NSDate alloc] init];
NSDate *date2 = [[NSDate alloc] initWithTimeInterval:theTimeInterval sinceDate:date1]; 

// Get conversion to months, days, hours, minutes
NSCalendarUnit unitFlags = NSHourCalendarUnit | NSMinuteCalendarUnit | NSDayCalendarUnit | NSMonthCalendarUnit;

NSDateComponents *breakdownInfo = [sysCalendar components:unitFlags fromDate:date1  toDate:date2  options:0];
NSLog(@"Break down: %i min : %i hours : %i days : %i months", [breakdownInfo minute], [breakdownInfo hour], [breakdownInfo day], [breakdownInfo month]);
于 2009-09-06T21:18:50.870 回答
17

这段代码知道夏令时和其他可能的讨厌的事情。

NSCalendar *gregorianCalendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *components = [gregorianCalendar components: (NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit | NSDayCalendarUnit | NSMonthCalendarUnit | NSYearCalendarUnit )
                                                    fromDate:startDate
                                                      toDate:[NSDate date]
                                                     options:0];


NSLog(@"%ld", [components year]);
NSLog(@"%ld", [components month]);
NSLog(@"%ld", [components day]);
NSLog(@"%ld", [components hour]);
NSLog(@"%ld", [components minute]);
NSLog(@"%ld", [components second]);
于 2013-02-17T14:11:18.223 回答
5

从iOS8及以上你可以使用NSDateComponentsFormatter

它具有在用户友好的格式化字符串中转换时差的方法。

NSDateComponentsFormatter *formatter = [[NSDateComponentsFormatter alloc] init];
formatter.unitsStyle = NSDateComponentsFormatterUnitsStyleFull;

NSLog(@"%@", [formatter stringFromTimeInterval:1623452]);

这给出了输出 - 2 周、4 天、18 小时、57 分钟、32 秒

于 2016-10-19T12:52:37.213 回答
4

将您的时间间隔转换为NSDateusing +dateWithIntervalSince1970,从中获取日期组件 usingNSCalendar-componentsFromDate方法。

开发工具包参考

于 2009-08-06T09:20:15.630 回答
3

或者有我的课堂方法。它不能处理年,但可以很容易地添加它,尽管它更适合像天、小时和分钟这样的小时间间隔。它考虑了复数,只显示需要的内容:

+(NSString *)TimeRemainingUntilDate:(NSDate *)date {

    NSTimeInterval interval = [date timeIntervalSinceNow];
    NSString * timeRemaining = nil;

    if (interval > 0) {

        div_t d = div(interval, 86400);
        int day = d.quot;
        div_t h = div(d.rem, 3600);
        int hour = h.quot;
        div_t m = div(h.rem, 60);
        int min = m.quot;

        NSString * nbday = nil;
        if(day > 1)
            nbday = @"days";
        else if(day == 1)
            nbday = @"day";
        else
            nbday = @"";
        NSString * nbhour = nil;
        if(hour > 1)
            nbhour = @"hours";
        else if (hour == 1)
            nbhour = @"hour";
        else
            nbhour = @"";
        NSString * nbmin = nil;
        if(min > 1)
            nbmin = @"mins";
        else
            nbmin = @"min";

        timeRemaining = [NSString stringWithFormat:@"%@%@ %@%@ %@%@",day ? [NSNumber numberWithInt:day] : @"",nbday,hour ? [NSNumber numberWithInt:hour] : @"",nbhour,min ? [NSNumber numberWithInt:min] : @"00",nbmin];
    }
    else
        timeRemaining = @"Over";

    return timeRemaining;
}
于 2011-11-08T19:49:44.503 回答
3

这对我有用:

    float *lenghInSeconds = 2345.234513;
    NSDate *date = [NSDate dateWithTimeIntervalSinceReferenceDate:lenghInSeconds];
    NSDateFormatter *formatter = [[NSDateFormatter alloc] init];


    [formatter setTimeZone:[NSTimeZone timeZoneForSecondsFromGMT:0.0]];

    [formatter setDateFormat:@"HH:mm:ss"];
    NSLog(@"%@", [formatter stringFromDate:date]); 
    [formatter release];

这里的主要区别是您需要调整时区。

于 2010-04-06T02:00:40.520 回答
2
NSDate *date = [NSDate dateWithTimeIntervalSince1970:timeInterval];

// format: YYYY-MM-DD HH:MM:SS ±HHMM
NSString *dateStr = [date description];
NSRange range;

// year
range.location = 0;
range.length = 4;
NSString *yearStr = [dateStr substringWithRange:range];
int year = [yearStr intValue] - 1970;

// month
range.location = 5;
range.length = 2;
NSString *monthStr = [dateStr substringWithRange:range];
int month = [monthStr intValue];

// day, etc.
...
于 2009-08-06T09:26:02.187 回答
0
- (NSString *)convertTimeFromSeconds:(NSString *)seconds {

    // Return variable.
    NSString *result = @"";

    // Int variables for calculation.
    int secs = [seconds intValue];
    int tempHour    = 0;
    int tempMinute  = 0;
    int tempSecond  = 0;

    NSString *hour      = @"";
    NSString *minute    = @"";
    NSString *second    = @"";

    // Convert the seconds to hours, minutes and seconds.
    tempHour    = secs / 3600;
    tempMinute  = secs / 60 - tempHour * 60;
    tempSecond  = secs - (tempHour * 3600 + tempMinute * 60);

    hour    = [[NSNumber numberWithInt:tempHour] stringValue];
    minute  = [[NSNumber numberWithInt:tempMinute] stringValue];
    second  = [[NSNumber numberWithInt:tempSecond] stringValue];

    // Make time look like 00:00:00 and not 0:0:0
    if (tempHour < 10) {
        hour = [@"0" stringByAppendingString:hour];
    } 

    if (tempMinute < 10) {
        minute = [@"0" stringByAppendingString:minute];
    }

    if (tempSecond < 10) {
        second = [@"0" stringByAppendingString:second];
    }

    if (tempHour == 0) {

        NSLog(@"Result of Time Conversion: %@:%@", minute, second);
        result = [NSString stringWithFormat:@"%@:%@", minute, second];

    } else {

        NSLog(@"Result of Time Conversion: %@:%@:%@", hour, minute, second); 
        result = [NSString stringWithFormat:@"%@:%@:%@",hour, minute, second];

    }

    return result;

}
于 2012-04-18T19:36:29.377 回答
-2

这是另一种可能性,更简洁:

NSDate *date = [NSDate dateWithTimeIntervalSince1970:timeInterval];
NSString *dateStr = [date description];
const char *dateStrPtr = [dateStr UTF8String];

// format: YYYY-MM-DD HH:MM:SS ±HHMM
int year, month, day, hour, minutes, seconds;
sscanf(dateStrPtr, "%d-%d-%d %d:%d:%d", &year, &month, &day, &hour, &minutes, &seconds);
year -= 1970;
于 2009-08-06T09:38:55.860 回答