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
我希望将其转换为“20分钟前”或“3天前”的相对时间戳。
使用Objective-C为iPhone执行此操作的最佳方法是什么？
答案 0 :(得分：72)
-(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";
}
}
答案 1 :(得分：22)
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对象似乎保持相关单位（如指定的）的差异。 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]);
}
如果您想自己计算一下，可以查看：
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），Apple正在通过NSDateFormatters执行此操作的首选方式，如下面的评论所示，看起来也很漂亮 - 由于历史原因，我会保留我的答案，对某些人来说，查看所使用的逻辑可能仍然有用。
答案 2 :(得分：14)
我无法编辑，但是我接受了Gilean的代码并进行了一些调整，并将其作为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，
Carl 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
答案 3 :(得分：8)
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的漂亮日期助手。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] =＆gt; “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
答案 4 :(得分：6)
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];
}
答案 5 :(得分：4)
我采用了Carl Coryell-Martin的代码并制作了一个更简单的NSDate类别，该类别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
答案 6 :(得分：3)
在Swift中
用法：
let time = NSDate(timeIntervalSince1970: timestamp).timeIntervalSinceNow
let relativeTimeString = NSDate.relativeTimeInString(time)
println(relativeTimeString)
扩展：
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"
}
}
}
答案 7 :(得分：1)
使用NSDate类：
timeIntervalSinceDate
以秒为单位返回间隔。
快速练习在objective-c中实现这一点：
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
依旧......
然后你需要决定一个月是30,31还是28天。保持简单 - 选择30。
可能有更好的方法，但凌晨2点，这是我想到的第一件事......
答案 8 :(得分：1)
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;
}
答案 9 :(得分：0)
不知道为什么这不是可可触摸，我很好的标准方式这样做会很棒。
设置一些类型以保存数据，如果您需要更多地进行本地化，它将更容易。 （如果你需要更多的时间段，显然会扩大）
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：我使用基于小时的计算，因为这是我的数据.NSTimeInterval是第二个。我还必须在两者之间进行转换。
答案 10 :(得分：0)
我看到Stack Overflow上的代码片段中有几段时间的函数，我想要一个真正给出最清晰的时间感（因为某些动作发生）。对我来说，这意味着“时间前”风格的短时间间隔（5分钟前，2小时前）和特定日期的较长时间段（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，我想通过他们的榜样（因为我确信他们对此有很多想法，从消费者的角度来理解这一点非常简单明了）。 / p>
经过长时间的谷歌搜索，我很惊讶地看到，据我所知，没有人实现过这一点。决定我希望它足够糟糕，花时间写作并认为我会分享。
希望你喜欢：）