给定特定的DateTime值,如何显示相对时间,例如:

2小时前3天前一个月前


当前回答

如果您想获得类似“2天4小时12分钟前”的输出,则需要一个时间跨度:

TimeSpan timeDiff = DateTime.Now-CreatedDate;

然后您可以访问您喜欢的值:

timeDiff.Days
timeDiff.Hours

其他回答

@杰夫

var ts=新时间跨度(DateTime.UtcNow.Ticks-dt.Ticks);

对DateTime执行减法仍会返回TimeSpan。

所以你可以这样做

(DateTime.UtcNow - dt).TotalSeconds

我也很惊讶地看到常数用手相乘,然后注释加上乘法。这是错误的优化吗?

jquery.timeago插件

Jeff,因为Stack Overflow广泛使用jQuery,所以我建议使用jQuery.timeago插件。

优点:

即使页面是在10分钟前打开的,也不要使用“1分钟前”的时间戳;timeago自动刷新。您可以充分利用web应用程序中的页面和/或片段缓存,因为时间戳不是在服务器上计算的。你可以像酷孩子一样使用微格式。

只需将其附加到DOM就绪的时间戳:

jQuery(document).ready(function() {
    jQuery('abbr.timeago').timeago();
});

这将在标题中使用timeago类和ISO 8601时间戳转换所有缩写元素:

<abbr class="timeago" title="2008-07-17T09:24:17Z">July 17, 2008</abbr>

变成这样:

<abbr class="timeago" title="July 17, 2008">4 months ago</abbr>

结果:4个月前。随着时间的推移,时间戳将自动更新。

免责声明:我写了这个插件,所以我有偏见。

文森特接受的答案做出了许多武断的决定。为什么45分钟舍入为一小时,而45秒不舍入为一分钟?在年和月的计算中,它的圈复杂度增加了,这使得遵循逻辑变得更加复杂。它假设TimeSpan是相对于过去(2天前)的,而它很可能是在未来(2天后)。它定义了不必要的常量,而不是使用TimeSpan.TicksPerSecond等。

此实现解决了上述问题,并更新了语法以使用开关表达式和关系模式

/// <summary>
/// Convert a <see cref="TimeSpan"/> to a natural language representation.
/// </summary>
/// <example>
/// <code>
/// TimeSpan.FromSeconds(10).ToNaturalLanguage();
/// // 10 seconds
/// </code>
/// </example>
public static string ToNaturalLanguage(this TimeSpan @this)
{
    const int daysInWeek = 7;
    const int daysInMonth = 30;
    const int daysInYear = 365;
    const long threshold = 100 * TimeSpan.TicksPerMillisecond;
    @this = @this.TotalSeconds < 0
        ? TimeSpan.FromSeconds(@this.TotalSeconds * -1)
        : @this;
    return (@this.Ticks + threshold) switch
    {
        < 2 * TimeSpan.TicksPerSecond => "a second",
        < 1 * TimeSpan.TicksPerMinute => @this.Seconds + " seconds",
        < 2 * TimeSpan.TicksPerMinute => "a minute",
        < 1 * TimeSpan.TicksPerHour => @this.Minutes + " minutes",
        < 2 * TimeSpan.TicksPerHour => "an hour",
        < 1 * TimeSpan.TicksPerDay => @this.Hours + " hours",
        < 2 * TimeSpan.TicksPerDay => "a day",
        < 1 * daysInWeek * TimeSpan.TicksPerDay => @this.Days + " days",
        < 2 * daysInWeek * TimeSpan.TicksPerDay => "a week",
        < 1 * daysInMonth * TimeSpan.TicksPerDay => (@this.Days / daysInWeek).ToString("F0") + " weeks",
        < 2 * daysInMonth * TimeSpan.TicksPerDay => "a month",
        < 1 * daysInYear * TimeSpan.TicksPerDay => (@this.Days / daysInMonth).ToString("F0") + " months",
        < 2 * daysInYear * TimeSpan.TicksPerDay => "a year",
        _ => (@this.Days / daysInYear).ToString("F0") + " years"
    };
}

/// <summary>
/// Convert a <see cref="DateTime"/> to a natural language representation.
/// </summary>
/// <example>
/// <code>
/// (DateTime.Now - TimeSpan.FromSeconds(10)).ToNaturalLanguage()
/// // 10 seconds ago
/// </code>
/// </example>
public static string ToNaturalLanguage(this DateTime @this)
{
    TimeSpan timeSpan = @this - DateTime.Now;
    return timeSpan.TotalSeconds switch
    {
        >= 1 => timeSpan.ToNaturalLanguage() + " until",
        <= -1 => timeSpan.ToNaturalLanguage() + " ago",
        _ => "now",
    };
}

可以使用NUnit对其进行如下测试:

[TestCase("a second", 0)]
[TestCase("a second", 1)]
[TestCase("2 seconds", 2)]
[TestCase("a minute", 0, 1)]
[TestCase("5 minutes", 0, 5)]
[TestCase("an hour", 0, 0, 1)]
[TestCase("2 hours", 0, 0, 2)]
[TestCase("a day", 0, 0, 24)]
[TestCase("a day", 0, 0, 0, 1)]
[TestCase("6 days", 0, 0, 0, 6)]
[TestCase("a week", 0, 0, 0, 7)]
[TestCase("4 weeks", 0, 0, 0, 29)]
[TestCase("a month", 0, 0, 0, 30)]
[TestCase("6 months", 0, 0, 0, 6 * 30)]
[TestCase("a year", 0, 0, 0, 365)]
[TestCase("68 years", int.MaxValue)]
public void NaturalLanguageHelpers_TimeSpan(
    string expected,
    int seconds,
    int minutes = 0,
    int hours = 0,
    int days = 0
)
{
    // Arrange
    TimeSpan timeSpan = new(days, hours, minutes, seconds);

    // Act
    string result = timeSpan.ToNaturalLanguage();

    // Assert
    Assert.That(result, Is.EqualTo(expected));
}

[TestCase("now", 0)]
[TestCase("10 minutes ago", 0, -10)]
[TestCase("10 minutes until", 10, 10)]
[TestCase("68 years until", int.MaxValue)]
[TestCase("68 years ago", int.MinValue)]
public void NaturalLanguageHelpers_DateTime(
    string expected,
    int seconds,
    int minutes = 0,
    int hours = 0,
    int days = 0
)
{
    // Arrange
    TimeSpan timeSpan = new(days, hours, minutes, seconds);
    DateTime now = DateTime.Now;
    DateTime dateTime = now + timeSpan;

    // Act
    string result = dateTime.ToNaturalLanguage();

    // Assert
    Assert.That(result, Is.EqualTo(expected));
}

或者作为要点:https://gist.github.com/StudioLE/2dd394e3f792e79adc927ede274df56e

在Java中有没有一种简单的方法可以做到这一点?java.util.Date类似乎相当有限。

下面是我的快速而肮脏的Java解决方案:

import java.util.Date;
import javax.management.timer.Timer;

String getRelativeDate(Date date) {     
  long delta = new Date().getTime() - date.getTime();
  if (delta < 1L * Timer.ONE_MINUTE) {
    return toSeconds(delta) == 1 ? "one second ago" : toSeconds(delta) + " seconds ago";
  }
  if (delta < 2L * Timer.ONE_MINUTE) {
    return "a minute ago";
  }
  if (delta < 45L * Timer.ONE_MINUTE) {
    return toMinutes(delta) + " minutes ago";
  }
  if (delta < 90L * Timer.ONE_MINUTE) {
    return "an hour ago";
  }
  if (delta < 24L * Timer.ONE_HOUR) {
    return toHours(delta) + " hours ago";
  }
  if (delta < 48L * Timer.ONE_HOUR) {
    return "yesterday";
  }
  if (delta < 30L * Timer.ONE_DAY) {
    return toDays(delta) + " days ago";
  }
  if (delta < 12L * 4L * Timer.ONE_WEEK) { // a month
    long months = toMonths(delta); 
    return months <= 1 ? "one month ago" : months + " months ago";
  }
  else {
    long years = toYears(delta);
    return years <= 1 ? "one year ago" : years + " years ago";
  }
}

private long toSeconds(long date) {
  return date / 1000L;
}

private long toMinutes(long date) {
  return toSeconds(date) / 60L;
}

private long toHours(long date) {
  return toMinutes(date) / 60L;
}

private long toDays(long date) {
  return toHours(date) / 24L;
}

private long toMonths(long date) {
  return toDays(date) / 30L;
}

private long toYears(long date) {
  return toMonths(date) / 365L;
}

如果您想获得类似“2天4小时12分钟前”的输出,则需要一个时间跨度:

TimeSpan timeDiff = DateTime.Now-CreatedDate;

然后您可以访问您喜欢的值:

timeDiff.Days
timeDiff.Hours