iPhone:将日期字符串转换为相对时间戳

Question

我有一个时间戳作为字符串,如:

星期四,2009年5月21日19:10:09 -0700

而且我想将它转换为像'20分钟前'或'3天前'的相对时间戳.

使用Objective-C为iPhone做到这一点的最佳方法是什么？

Answer 1

-(NSString *)dateDiff:(NSString *)origDate {
    NSDateFormatter *df = [[NSDateFormatter alloc] init];
    [df setFormatterBehavior:NSDateFormatterBehavior10_4];
    [df setDateFormat:@"EEE, dd MMM yy HH:mm:ss VVVV"];
    NSDate *convertedDate = [df dateFromString:origDate];
    [df release];
    NSDate *todayDate = [NSDate date];
    double ti = [convertedDate timeIntervalSinceDate:todayDate];
    ti = ti * -1;
    if(ti < 1) {
        return @"never";
    } else  if (ti < 60) {
        return @"less than a minute ago";
    } else if (ti < 3600) {
        int diff = round(ti / 60);
        return [NSString stringWithFormat:@"%d minutes ago", diff];
    } else if (ti < 86400) {
        int diff = round(ti / 60 / 60);
        return[NSString stringWithFormat:@"%d hours ago", diff];
    } else if (ti < 2629743) {
        int diff = round(ti / 60 / 60 / 24);
        return[NSString stringWithFormat:@"%d days ago", diff];
    } else {
        return @"never";
    }   
}

Answer 2

以下是来自Cocoa的方法,可以帮助您获取相关信息(不确定它们是否都可以通过coca-touch获得).

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

    NSString * str = @"Thu, 21 May 09 19:10:09 -0700";
    NSDate * past = [NSDate dateWithNaturalLanguageString:str
                            locale:[[NSUserDefaults 
                            standardUserDefaults] dictionaryRepresentation]];

    NSLog(@"str: %@", str);
    NSLog(@"past: %@", past);

    NSCalendar *gregorian = [[NSCalendar alloc]
                             initWithCalendarIdentifier:NSGregorianCalendar];
    unsigned int unitFlags = NSYearCalendarUnit | NSMonthCalendarUnit | 
                             NSDayCalendarUnit | 
                             NSHourCalendarUnit | NSMinuteCalendarUnit | 
                             NSSecondCalendarUnit;
    NSDateComponents *components = [gregorian components:unitFlags
                                                fromDate:past
                                                  toDate:today
                                                 options:0];

    NSLog(@"months: %d", [components month]);
    NSLog(@"days: %d", [components day]);
    NSLog(@"hours: %d", [components hour]);
    NSLog(@"seconds: %d", [components second]);

NSDateComponents对象似乎在相关单位中保持差异(如指定的那样).如果指定所有单位,则可以使用此方法:

void dump(NSDateComponents * t)
{
    if ([t year]) NSLog(@"%d years ago", [t year]);
    else if ([t month]) NSLog(@"%d months ago", [t month]);
    else if ([t day]) NSLog(@"%d days ago", [t day]);
    else if ([t minute]) NSLog(@"%d minutes ago", [t minute]);
    else if ([t second]) NSLog(@"%d seconds ago", [t second]);
}

如果你想自己计算,你可以看看:

NSDate timeIntervalSinceDate

然后在算法中使用秒.

免责声明:如果这个界面被弃用(我还没有检查过),苹果公司首选的做法NSDateFormatters,如下面的评论所示,看起来也很漂亮 - 我会保留我的答案,因为历史原因,它可能仍然是有些人可以看一下使用的逻辑.

Answer 3

我还无法编辑,但我采用了Gilean的代码并进行了一些调整,并使其成为NSDateFormatter的一个类别.

它接受一个格式字符串,因此它将使用任意字符串,并且我添加了if子句以使单个事件在语法上正确.

干杯,

卡尔CM

@interface NSDateFormatter (Extras)
+ (NSString *)dateDifferenceStringFromString:(NSString *)dateString
                                  withFormat:(NSString *)dateFormat;

@end

@implementation NSDateFormatter (Extras)

+ (NSString *)dateDifferenceStringFromString:(NSString *)dateString
                                  withFormat:(NSString *)dateFormat
{
  NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
  [dateFormatter setFormatterBehavior:NSDateFormatterBehavior10_4];
  [dateFormatter setDateFormat:dateFormat];
  NSDate *date = [dateFormatter dateFromString:dateString];
  [dateFormatter release];
  NSDate *now = [NSDate date];
  double time = [date timeIntervalSinceDate:now];
  time *= -1;
  if(time < 1) {
    return dateString;
  } else if (time < 60) {
    return @"less than a minute ago";
  } else if (time < 3600) {
    int diff = round(time / 60);
    if (diff == 1) 
      return [NSString stringWithFormat:@"1 minute ago", diff];
    return [NSString stringWithFormat:@"%d minutes ago", diff];
  } else if (time < 86400) {
    int diff = round(time / 60 / 60);
    if (diff == 1)
      return [NSString stringWithFormat:@"1 hour ago", diff];
    return [NSString stringWithFormat:@"%d hours ago", diff];
  } else if (time < 604800) {
    int diff = round(time / 60 / 60 / 24);
    if (diff == 1) 
      return [NSString stringWithFormat:@"yesterday", diff];
    if (diff == 7) 
      return [NSString stringWithFormat:@"last week", diff];
    return[NSString stringWithFormat:@"%d days ago", diff];
  } else {
    int diff = round(time / 60 / 60 / 24 / 7);
    if (diff == 1)
      return [NSString stringWithFormat:@"last week", diff];
    return [NSString stringWithFormat:@"%d weeks ago", diff];
  }   
}

@end

Answer 4

为了完整性,基于@Gilean的答案,这里是NSDate上一个简单类别的完整代码,模仿rails的漂亮日期帮助器.有关类别的更新,这些是您将在NSDate对象上调用的实例方法.所以,如果我有一个代表昨天的NSDate,[myDate distanceOfTimeInWordsToNow] =>"1天".

希望它有用!

@interface NSDate (NSDate_Relativity)

-(NSString *)distanceOfTimeInWordsSinceDate:(NSDate *)aDate;
-(NSString *)distanceOfTimeInWordsToNow;

@end



@implementation NSDate (NSDate_Relativity)


-(NSString *)distanceOfTimeInWordsToNow {
    return [self distanceOfTimeInWordsSinceDate:[NSDate date]];

}

-(NSString *)distanceOfTimeInWordsSinceDate:(NSDate *)aDate {
    double interval = [self timeIntervalSinceDate:aDate];

    NSString *timeUnit;
    int timeValue;

    if (interval < 0) {
        interval = interval * -1;        
    }

    if (interval< 60) {
        return @"seconds";

    } else if (interval< 3600) { // minutes

        timeValue = round(interval / 60);

        if (timeValue == 1) {
            timeUnit = @"minute";

        } else {
            timeUnit = @"minutes";

        }


    } else if (interval< 86400) {
        timeValue = round(interval / 60 / 60);

        if (timeValue == 1) {
            timeUnit = @"hour";

        } else {
            timeUnit = @"hours";
        }


    } else if (interval< 2629743) {
        int days = round(interval / 60 / 60 / 24);

        if (days < 7) {

            timeValue = days;

            if (timeValue == 1) {
                timeUnit = @"day";
            } else {
                timeUnit = @"days";
            }

        } else if (days < 30) {
            int weeks = days / 7;

            timeValue = weeks;

            if (timeValue == 1) {
                timeUnit = @"week";
            } else {
                timeUnit = @"weeks";
            }


        } else if (days < 365) {

            int months = days / 30;
            timeValue = months;

            if (timeValue == 1) {
                timeUnit = @"month";
            } else {
                timeUnit = @"months";
            }

        } else if (days < 30000) { // this is roughly 82 years. After that, we'll say 'forever'
            int years = days / 365;
            timeValue = years;

            if (timeValue == 1) {
                timeUnit = @"year";
            } else {
                timeUnit = @"years";
            }

        } else {
            return @"forever ago";
        }
    }

    return [NSString stringWithFormat:@"%d %@", timeValue, timeUnit];

}

@end

Answer 5

已经有很多答案出现在同一个解决方案中,但选择却不会有什么坏处.这就是我想出来的.

- (NSString *)stringForTimeIntervalSinceCreated:(NSDate *)dateTime
{
    NSDictionary *timeScale = @{@"second":@1,
                                @"minute":@60,
                                @"hour":@3600,
                                @"day":@86400,
                                @"week":@605800,
                                @"month":@2629743,
                                @"year":@31556926};
    NSString *scale;
    int timeAgo = 0-(int)[dateTime timeIntervalSinceNow];
    if (timeAgo < 60) {
        scale = @"second";
    } else if (timeAgo < 3600) {
        scale = @"minute";
    } else if (timeAgo < 86400) {
        scale = @"hour";
    } else if (timeAgo < 605800) {
        scale = @"day";
    } else if (timeAgo < 2629743) {
        scale = @"week";
    } else if (timeAgo < 31556926) {
        scale = @"month";
    } else {
        scale = @"year";
    }

    timeAgo = timeAgo/[[timeScale objectForKey:scale] integerValue];
    NSString *s = @"";
    if (timeAgo > 1) {
        s = @"s";
    } 
    return [NSString stringWithFormat:@"%d %@%@ ago", timeAgo, scale, s];
}