Question

我想写一个模糊日期方法来计算iPhone的Objective-C中的日期。这里有一个流行的解释：

但它包含缺少的参数。怎么能在Objective-C中使用它？感谢。

const int SECOND = 1;
const int MINUTE = 60 * SECOND;
const int HOUR = 60 * MINUTE;
const int DAY = 24 * HOUR;
const int MONTH = 30 * DAY;

if (delta < 1 * MINUTE)
{
  return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2 * MINUTE)
{
  return "a minute ago";
}
if (delta < 45 * MINUTE)
{
  return ts.Minutes + " minutes ago";
}
if (delta < 90 * MINUTE)
{
  return "an hour ago";
}
if (delta < 24 * HOUR)
{
  return ts.Hours + " hours ago";
}
if (delta < 48 * HOUR)
{
  return "yesterday";
}
if (delta < 30 * DAY)
{
  return ts.Days + " days ago";
}
if (delta < 12 * MONTH)
{
  int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
  return months <= 1 ? "one month ago" : months + " months ago";
}
else
{
  int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
  return years <= 1 ? "one year ago" : years + " years ago";
}

Answer 1

使用NSDate类在Cocoa中表示日期。在NSDate中实现了一种方便的方法，可以在两个日期实例timeIntervalSinceDate:之间获得以秒为单位的增量。这是在NSDate实例上调用的，将另一个NSDate对象作为参数。它返回NSTimeInterval（这是double的typedef），它代表两个日期之间的秒数。

鉴于此，将上面给出的代码调整为Objective-C / Cocoa上下文会相当简单。由于NSDate计算的增量以秒为单位给出，给定两个日期，您可以轻松调整上述代码：

//Constants
#define SECOND 1
#define MINUTE (60 * SECOND)
#define HOUR (60 * MINUTE)
#define DAY (24 * HOUR)
#define MONTH (30 * DAY)

- (NSString*)timeIntervalWithStartDate:(NSDate*)d1 withEndDate:(NSDate*)d2
{
    //Calculate the delta in seconds between the two dates
    NSTimeInterval delta = [d2 timeIntervalSinceDate:d1];

    if (delta < 1 * MINUTE)
    {
        return delta == 1 ? @"one second ago" : [NSString stringWithFormat:@"%d seconds ago", (int)delta];
    }
    if (delta < 2 * MINUTE)
    {
        return @"a minute ago";
    }
    if (delta < 45 * MINUTE)
    {
        int minutes = floor((double)delta/MINUTE);
        return [NSString stringWithFormat:@"%d minutes ago", minutes];
    }
    if (delta < 90 * MINUTE)
    {
        return @"an hour ago";
    }
    if (delta < 24 * HOUR)
    {
        int hours = floor((double)delta/HOUR);
        return [NSString stringWithFormat:@"%d hours ago", hours];
    }
    if (delta < 48 * HOUR)
    {
        return @"yesterday";
    }
    if (delta < 30 * DAY)
    {
        int days = floor((double)delta/DAY);
        return [NSString stringWithFormat:@"%d days ago", days];
    }
    if (delta < 12 * MONTH)
    {
        int months = floor((double)delta/MONTH);
        return months <= 1 ? @"one month ago" : [NSString stringWithFormat:@"%d months ago", months];
    }
    else
    {
        int years = floor((double)delta/MONTH/12.0);
        return years <= 1 ? @"one year ago" : [NSString stringWithFormat:@"%d years ago", years];
    }
}

然后调用它，将起始和结束NSDate对象作为参数传递，并返回带有时间间隔的NSString。

Answer 2

您可以使用NSDate方法获取两个timeIntervalSinceDate:对象之间的增量。那将在几秒钟内给你增量。

从那里你可以通过除以适当的数量来计算出分钟/小时/天/飞蛾/年。

Answer 3

作为替代方案，您可以依靠可以从两个日期之间的差异中提取的日历组件来避免容易出错的日历算法：

NSDate *nowDate =    [[NSDate alloc] init];
NSDate *targetDate = nil; // some other date here of your choosing, obviously nil isn't going to get you very far

NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSUInteger unitFlags = NSMonthCalendarUnit | NSWeekCalendarUnit | NSDayCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit;
NSDateComponents *components = [gregorian components:unitFlags
                                            fromDate:dateTime
                                              toDate:nowDate options:0];
NSInteger months = [components month];
NSInteger weeks = [components week];
NSInteger days = [components day];
NSInteger hours = [components hour];
NSInteger minutes = [components minute];

关键是单位标志的设置 - 这允许您设置您希望将日期/时间细分为哪个时间单位。如果您只想要几个小时就可以设置NSHourCalendarUnit，并且随着日期的进一步分离，该值将继续增加，因为没有更大的单位可以开始递增。

一旦有了组件，就可以通过修改@ alex的条件流来继续你选择的逻辑。

这就是我一起扔的东西：

if (months > 1) {
    // Simple date/time
    if (weeks >3) {
        // Almost another month - fuzzy
        months++;
    }
    return [NSString stringWithFormat:@"%ld months ago", (long)months];
}
else if (months == 1) {
    if (weeks > 3) {
        months++;
        // Almost 2 months
        return [NSString stringWithFormat:@"%ld months ago", (long)months];
    }
    // approx 1 month
    return [NSString stringWithFormat:@"1 month ago"];
}
// Weeks
else if (weeks > 1) {
    if (days > 6) {
        // Almost another month - fuzzy
        weeks++;
    }
    return [NSString stringWithFormat:@"%ld weeks ago", (long)weeks];
}
else if (weeks == 1 ||
         days > 6) {
    if (days > 6) {
        weeks++;
        // Almost 2 weeks
        return [NSString stringWithFormat:@"%ld weeks ago", (long)weeks];
    }
    return [NSString stringWithFormat:@"1 week ago"];
}
// Days
else if (days > 1) {
    if (hours > 20) {
        days++;
    }
    return [NSString stringWithFormat:@"%ld days ago", (long)days];
}
else if (days == 1) {
    if (hours > 20) {
        days++;
        return [NSString stringWithFormat:@"%ld days ago", (long)days];
    }
    return [NSString stringWithFormat:@"1 day ago"];
}
// Hours
else if (hours > 1) {
    if (minutes > 50) {
        hours++;
    }
    return [NSString stringWithFormat:@"%ld hours ago", (long)hours];
}
else if (hours == 1) {
    if (minutes > 50) {
        hours++;
        return [NSString stringWithFormat:@"%ld hours ago", (long)hours];
    }
    return [NSString stringWithFormat:@"1 hour ago"];
}
// Minutes
else if (minutes > 1) {
    return [NSString stringWithFormat:@"%ld minutes ago", (long)minutes];
}
else if (minutes == 1) {
    return [NSString stringWithFormat:@"1 minute ago"];
}
else if (minutes < 1) {
    return [NSString stringWithFormat:@"Just now"];
}

Answer 4

您可以参考Apple的此开发者指南：

http://developer.apple.com/mac/library/documentation/cocoa/Conceptual/DatesAndTimes/Articles/dtCalendricalCalculations.html#//apple_ref/doc/uid/TP40007836-SW1

Objective-C中的模糊数据算法

4 个答案: