iPhone：将日期字符串转换为相对时间戳

Question

我有一个像这样的字符串的时间戳：

09 年 5 月 21 日星期四 19:10:09 -0700

我想将其转换为相对时间戳，例如“20 分钟前”或“3 天前”。

在 iPhone 上使用 Objective-C 的最佳方法是什么？

Answer 1

在 Swift 中

用法：

let time = NSDate(timeIntervalSince1970: timestamp).timeIntervalSinceNow
let relativeTimeString = NSDate.relativeTimeInString(time)
println(relativeTimeString)

扩展名：

extension NSDate {
    class func relativeTimeInString(value: NSTimeInterval) -> String {
        func getTimeData(value: NSTimeInterval) -> (count: Int, suffix: String) {
            let count = Int(floor(value))
            let suffix = count != 1 ? "s" : ""
            return (count: count, suffix: suffix)
        }

        let value = -value
        switch value {
            case 0...15: return "just now"

            case 0..<60:
                let timeData = getTimeData(value)
                return "\(timeData.count) second\(timeData.suffix) ago"

            case 0..<3600:
                let timeData = getTimeData(value/60)
                return "\(timeData.count) minute\(timeData.suffix) ago"

            case 0..<86400:
                let timeData = getTimeData(value/3600)
                return "\(timeData.count) hour\(timeData.suffix) ago"

            case 0..<604800:
                let timeData = getTimeData(value/86400)
                return "\(timeData.count) day\(timeData.suffix) ago"

            default:
                let timeData = getTimeData(value/604800)
                return "\(timeData.count) week\(timeData.suffix) ago"
        }
    }
}

Answer 2

已经有很多答案可以得出相同的解决方案，但有选择也无妨。这是我想出的。

- (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];
}

Answer 3

我在 Stack Overflow 上看到好几次之前的代码 sn-ps 中的函数，我想要一个真正给出最清晰的时间感的函数（因为发生了一些动作）。对我来说，这意味着短时间间隔（5 分钟前，2 小时前）的“时间前”样式和较长时间段的特定日期（2011 年 4 月 15 日而不是 2 年前）。基本上，我认为 Facebook 在这方面做得非常好，我想以他们的例子为例（因为我确信他们对此进行了很多思考，从消费者的角度来看，这很容易理解）。

在谷歌上搜索了很长时间后，我很惊讶地发现，据我所知，没有人实施过这个。决定我希望它足够糟糕以花时间写作并认为我会分享。

希望你喜欢:)

在此处获取代码：https://github.com/nikilster/NSDate-Time-Ago

Answer 4

我采用了 Carl Coryell-Martin 的代码并制作了一个更简单的 NSDate 类别，它没有关于单数字符串格式的警告，并且还整理了一周前的单数：

@interface NSDate (Extras)
- (NSString *)differenceString;
@end

@implementation NSDate (Extras)

- (NSString *)differenceString{
    NSDate* date = self;
    NSDate *now = [NSDate date];
    double time = [date timeIntervalSinceDate:now];
    time *= -1;
    if (time < 60) {
        int diff = round(time);
        if (diff == 1)
            return @"1 second ago";
        return [NSString stringWithFormat:@"%d seconds ago", diff];
    } else if (time < 3600) {
        int diff = round(time / 60);
        if (diff == 1)
            return @"1 minute ago";
        return [NSString stringWithFormat:@"%d minutes ago", diff];
    } else if (time < 86400) {
        int diff = round(time / 60 / 60);
        if (diff == 1)
            return @"1 hour ago";
        return [NSString stringWithFormat:@"%d hours ago", diff];
    } else if (time < 604800) {
        int diff = round(time / 60 / 60 / 24);
        if (diff == 1)
            return @"yesterday";
        if (diff == 7)
            return @"a week ago";
        return[NSString stringWithFormat:@"%d days ago", diff];
    } else {
        int diff = round(time / 60 / 60 / 24 / 7);
        if (diff == 1)
            return @"a week ago";
        return [NSString stringWithFormat:@"%d weeks ago", diff];
    }   
}

@end

Answer 5

我的解决方案：

- (NSString *) dateToName:(NSDate*)dt withSec:(BOOL)sec {

    NSLocale *locale = [NSLocale currentLocale];
    NSTimeInterval tI = [[NSDate date] timeIntervalSinceDate:dt];
    if (tI < 60) {
      if (sec == NO) {
           return NSLocalizedString(@"Just Now", @"");
       }
       return [NSString stringWithFormat:
                 NSLocalizedString(@"%d seconds ago", @""),(int)tI];
     }
     if (tI < 3600) {
       return [NSString stringWithFormat:
                 NSLocalizedString(@"%d minutes ago", @""),(int)(tI/60)];
     }
     if (tI < 86400) {
      return [NSString stringWithFormat:
                 NSLocalizedString(@"%d hours ago", @""),(int)tI/3600];
     }

     NSDateFormatter *relativeDateFormatter = [[NSDateFormatter alloc] init];
     [relativeDateFormatter setTimeStyle:NSDateFormatterNoStyle];
     [relativeDateFormatter setDateStyle:NSDateFormatterMediumStyle];
     [relativeDateFormatter setDoesRelativeDateFormatting:YES];
     [relativeDateFormatter setLocale:locale];

     NSString * relativeFormattedString = 
            [relativeDateFormatter stringForObjectValue:dt];
     return relativeFormattedString;
}

Answer 6

为了完整起见，根据@Gilean 的回答，这里是 NSDate 上一个简单类别的完整代码，它模仿了 rails 的漂亮日期助手。对于类别的复习，这些是您可以在 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 day"。

希望有用！

@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 7

以下是来自 Cocoa 的方法，可帮助您获取相关信息（不确定它们是否都在 coca-touch 中可用）。

    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 对象似乎包含相关单位的差异（如指定的那样）。 如果您指定所有单位，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：如果此接口被弃用（我尚未检查），Apple 通过NSDateFormatters 执行此操作的首选方式（如下面的 cmets 建议）看起来也很简洁 - 我出于历史原因，我会保留我的答案，对于某些人来说看看使用的逻辑可能仍然有用。

Answer 8

我还不能编辑，但我采用了 Gilean 的代码并进行了一些调整，并将其设为 NSDateFormatter 的类别。

它接受格式字符串，因此可以使用任意字符串，我添加了 if 子句以使单数事件在语法上是正确的。

干杯，

卡尔 C-M

@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 9

-(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 10

不知道为什么这不是可可触摸，我很好的标准方法会很棒。

设置一些类型来保存数据，如果您需要更多地本地化它会更容易。 （如果您需要更多时间段，显然可以扩展）

typedef struct DayHours {
    int Days;
    double Hours;
} DayHours;


+ (DayHours) getHourBasedTimeInterval:(double) hourBased withHoursPerDay:(double) hpd
{
    int NumberOfDays = (int)(fabs(hourBased) / hpd);
    float hoursegment = fabs(hourBased) - (NumberOfDays * hpd);
    DayHours dh;
    dh.Days = NumberOfDays;
    dh.Hours = hoursegment;
    return dh;
}

注意：我使用的是基于小时的计算，因为这是我的数据所在。NSTimeInterval 是基于秒的。我还必须在两者之间进行转换。

Answer 11

使用 NSDate 类：

timeIntervalSinceDate

以秒为单位返回间隔。

在objective-c中实现这个的快速练习：

1. “现在”获取时间 NSDate
2. 获取您希望与之比较的 NSDate
3. 使用 timeIntervalSinceDate 获取间隔（以秒为单位）

然后实现这个伪代码：

if (x < 60) // x seconds ago

else if( x/60 < 60) // floor(x/60) minutes ago

else if (x/(60*60) < 24) // floor(x/(60*60) hours ago

else if (x/(24*60*60) < 7) // floor(x(24*60*60) days ago

等等……

那么您需要决定一个月是 30,31 还是 28 天。保持简单 - 选择 30。

可能有更好的方法，但它是凌晨 2 点，这是我想到的第一件事......