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

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

其他回答

我的方法要简单得多。您可以根据需要调整返回字符串

    public static string TimeLeft(DateTime utcDate)
    {
        TimeSpan timeLeft = DateTime.UtcNow - utcDate;
        string timeLeftString = "";
        if (timeLeft.Days > 0)
        {
            timeLeftString += timeLeft.Days == 1 ? timeLeft.Days + " day" : timeLeft.Days + " days";
        }
        else if (timeLeft.Hours > 0)
        {
            timeLeftString += timeLeft.Hours == 1 ? timeLeft.Hours + " hour" : timeLeft.Hours + " hours";
        }
        else
        {
            timeLeftString += timeLeft.Minutes == 1 ? timeLeft.Minutes+" minute" : timeLeft.Minutes + " minutes";
        }
        return timeLeftString;
    }

@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?

以某种方式,您可以使用DateTime函数以秒到年计算相对时间,请尝试以下操作:

using System;

public class Program {
    public static string getRelativeTime(DateTime past) {
        DateTime now = DateTime.Today;
        string rt = "";
        int time;
        string statement = "";
        if (past.Second >= now.Second) {
            if (past.Second - now.Second == 1) {
                rt = "second ago";
            }
            rt = "seconds ago";
            time = past.Second - now.Second;
            statement = "" + time;
            return (statement + rt);
        }
        if (past.Minute >= now.Minute) {
            if (past.Second - now.Second == 1) {
                rt = "second ago";
            } else {
                rt = "minutes ago";
            }
            time = past.Minute - now.Minute;
            statement = "" + time;
            return (statement + rt);
        }
        // This process will go on until years
    }
    public static void Main() {
        DateTime before = new DateTime(1995, 8, 24);
        string date = getRelativeTime(before);
        Console.WriteLine("Windows 95 was {0}.", date);
    }
}

不完全有效,但如果您对其进行一点修改和调试,它很可能会完成任务。

/** 
 * {@code date1} has to be earlier than {@code date2}.
 */
public static String relativize(Date date1, Date date2) {
    assert date2.getTime() >= date1.getTime();

    long duration = date2.getTime() - date1.getTime();
    long converted;

    if ((converted = TimeUnit.MILLISECONDS.toDays(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "day" : "days");
    } else if ((converted = TimeUnit.MILLISECONDS.toHours(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "hour" : "hours");
    } else if ((converted = TimeUnit.MILLISECONDS.toMinutes(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "minute" : "minutes");
    } else if ((converted = TimeUnit.MILLISECONDS.toSeconds(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "second" : "seconds");
    } else {
        return "just now";
    }
}
public string getRelativeDateTime(DateTime date)
{
    TimeSpan ts = DateTime.Now - date;
    if (ts.TotalMinutes < 1)//seconds ago
        return "just now";
    if (ts.TotalHours < 1)//min ago
        return (int)ts.TotalMinutes == 1 ? "1 Minute ago" : (int)ts.TotalMinutes + " Minutes ago";
    if (ts.TotalDays < 1)//hours ago
        return (int)ts.TotalHours == 1 ? "1 Hour ago" : (int)ts.TotalHours + " Hours ago";
    if (ts.TotalDays < 7)//days ago
        return (int)ts.TotalDays == 1 ? "1 Day ago" : (int)ts.TotalDays + " Days ago";
    if (ts.TotalDays < 30.4368)//weeks ago
        return (int)(ts.TotalDays / 7) == 1 ? "1 Week ago" : (int)(ts.TotalDays / 7) + " Weeks ago";
    if (ts.TotalDays < 365.242)//months ago
        return (int)(ts.TotalDays / 30.4368) == 1 ? "1 Month ago" : (int)(ts.TotalDays / 30.4368) + " Months ago";
    //years ago
    return (int)(ts.TotalDays / 365.242) == 1 ? "1 Year ago" : (int)(ts.TotalDays / 365.242) + " Years ago";
}

一个月和一年中的天数的转换值取自谷歌。