Given a specific DateTime value, how do I display relative time, like:
- 2 hours ago
- 3 days ago
- a month ago
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187994 次
40 回答
1050
杰夫，您的代码很好，但使用常量可能更清晰（如代码完成中所建议的那样）。
const int SECOND = 1;
const int MINUTE = 60 * SECOND;
const int HOUR = 60 * MINUTE;
const int DAY = 24 * HOUR;
const int MONTH = 30 * DAY;
var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
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";
}
于 2008-08-04T13:57:26.097 回答
365
jquery.timeago 插件
Jeff，因为 Stack Overflow 广泛使用 jQuery，我推荐jquery.timeago 插件。
好处：
- 即使页面是在 10 分钟前打开的，也要避免使用“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 上的时间戳即可：
jQuery(document).ready(function() {
jQuery('abbr.timeago').timeago();
});
这将在标题中转换所有abbr具有 timeago 类和ISO 8601时间戳的元素：
<abbr class="timeago" title="2008-07-17T09:24:17Z">July 17, 2008</abbr>
变成这样的东西：
<abbr class="timeago" title="July 17, 2008">4 months ago</abbr>
产生：4个月前。随着时间的推移，时间戳会自动更新。
免责声明：我写了这个插件，所以我有偏见。
于 2008-09-21T16:04:55.783 回答
340
Here's how I do it
var ts = new TimeSpan(DateTime.UtcNow.Ticks - dt.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
if (delta < 60)
{
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 60 * 2)
{
return "a minute ago";
}
if (delta < 45 * 60)
{
return ts.Minutes + " minutes ago";
}
if (delta < 90 * 60)
{
return "an hour ago";
}
if (delta < 24 * 60 * 60)
{
return ts.Hours + " hours ago";
}
if (delta < 48 * 60 * 60)
{
return "yesterday";
}
if (delta < 30 * 24 * 60 * 60)
{
return ts.Days + " days ago";
}
if (delta < 12 * 30 * 24 * 60 * 60)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months <= 1 ? "one month ago" : months + " months ago";
}
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years <= 1 ? "one year ago" : years + " years ago";
Suggestions? Comments? Ways to improve this algorithm?
于 2008-07-31T23:56:41.303 回答
96
public static string RelativeDate(DateTime theDate)
{
Dictionary<long, string> thresholds = new Dictionary<long, string>();
int minute = 60;
int hour = 60 * minute;
int day = 24 * hour;
thresholds.Add(60, "{0} seconds ago");
thresholds.Add(minute * 2, "a minute ago");
thresholds.Add(45 * minute, "{0} minutes ago");
thresholds.Add(120 * minute, "an hour ago");
thresholds.Add(day, "{0} hours ago");
thresholds.Add(day * 2, "yesterday");
thresholds.Add(day * 30, "{0} days ago");
thresholds.Add(day * 365, "{0} months ago");
thresholds.Add(long.MaxValue, "{0} years ago");
long since = (DateTime.Now.Ticks - theDate.Ticks) / 10000000;
foreach (long threshold in thresholds.Keys)
{
if (since < threshold)
{
TimeSpan t = new TimeSpan((DateTime.Now.Ticks - theDate.Ticks));
return string.Format(thresholds[threshold], (t.Days > 365 ? t.Days / 365 : (t.Days > 0 ? t.Days : (t.Hours > 0 ? t.Hours : (t.Minutes > 0 ? t.Minutes : (t.Seconds > 0 ? t.Seconds : 0))))).ToString());
}
}
return "";
}
我更喜欢这个版本，因为它简洁，并且能够添加新的刻度点。这可以通过Latest()对 Timespan 的扩展而不是那个长的 1 行来封装，但是为了发布的简洁起见，这样做可以。
这修复了一个小时前，1 小时前，提供一个小时直到 2 小时过去
于 2008-08-14T05:43:45.330 回答
71
public static string ToRelativeDate(DateTime input)
{
TimeSpan oSpan = DateTime.Now.Subtract(input);
double TotalMinutes = oSpan.TotalMinutes;
string Suffix = " ago";
if (TotalMinutes < 0.0)
{
TotalMinutes = Math.Abs(TotalMinutes);
Suffix = " from now";
}
var aValue = new SortedList<double, Func<string>>();
aValue.Add(0.75, () => "less than a minute");
aValue.Add(1.5, () => "about a minute");
aValue.Add(45, () => string.Format("{0} minutes", Math.Round(TotalMinutes)));
aValue.Add(90, () => "about an hour");
aValue.Add(1440, () => string.Format("about {0} hours", Math.Round(Math.Abs(oSpan.TotalHours)))); // 60 * 24
aValue.Add(2880, () => "a day"); // 60 * 48
aValue.Add(43200, () => string.Format("{0} days", Math.Floor(Math.Abs(oSpan.TotalDays)))); // 60 * 24 * 30
aValue.Add(86400, () => "about a month"); // 60 * 24 * 60
aValue.Add(525600, () => string.Format("{0} months", Math.Floor(Math.Abs(oSpan.TotalDays / 30)))); // 60 * 24 * 365
aValue.Add(1051200, () => "about a year"); // 60 * 24 * 365 * 2
aValue.Add(double.MaxValue, () => string.Format("{0} years", Math.Floor(Math.Abs(oSpan.TotalDays / 365))));
return aValue.First(n => TotalMinutes < n.Key).Value.Invoke() + Suffix;
}
http://refactormycode.com/codes/493-twitter-esque-relative-dates
C# 6 版本：
static readonly SortedList<double, Func<TimeSpan, string>> offsets =
new SortedList<double, Func<TimeSpan, string>>
{
{ 0.75, _ => "less than a minute"},
{ 1.5, _ => "about a minute"},
{ 45, x => $"{x.TotalMinutes:F0} minutes"},
{ 90, x => "about an hour"},
{ 1440, x => $"about {x.TotalHours:F0} hours"},
{ 2880, x => "a day"},
{ 43200, x => $"{x.TotalDays:F0} days"},
{ 86400, x => "about a month"},
{ 525600, x => $"{x.TotalDays / 30:F0} months"},
{ 1051200, x => "about a year"},
{ double.MaxValue, x => $"{x.TotalDays / 365:F0} years"}
};
public static string ToRelativeDate(this DateTime input)
{
TimeSpan x = DateTime.Now - input;
string Suffix = x.TotalMinutes > 0 ? " ago" : " from now";
x = new TimeSpan(Math.Abs(x.Ticks));
return offsets.First(n => x.TotalMinutes < n.Key).Value(x) + Suffix;
}
于 2008-09-17T03:19:19.487 回答
71
这里是 Jeffs Script for PHP 的重写：
define("SECOND", 1);
define("MINUTE", 60 * SECOND);
define("HOUR", 60 * MINUTE);
define("DAY", 24 * HOUR);
define("MONTH", 30 * DAY);
function relativeTime($time)
{
$delta = time() - $time;
if ($delta < 1 * MINUTE)
{
return $delta == 1 ? "one second ago" : $delta . " seconds ago";
}
if ($delta < 2 * MINUTE)
{
return "a minute ago";
}
if ($delta < 45 * MINUTE)
{
return floor($delta / MINUTE) . " minutes ago";
}
if ($delta < 90 * MINUTE)
{
return "an hour ago";
}
if ($delta < 24 * HOUR)
{
return floor($delta / HOUR) . " hours ago";
}
if ($delta < 48 * HOUR)
{
return "yesterday";
}
if ($delta < 30 * DAY)
{
return floor($delta / DAY) . " days ago";
}
if ($delta < 12 * MONTH)
{
$months = floor($delta / DAY / 30);
return $months <= 1 ? "one month ago" : $months . " months ago";
}
else
{
$years = floor($delta / DAY / 365);
return $years <= 1 ? "one year ago" : $years . " years ago";
}
}
于 2009-02-01T19:22:41.450 回答
51
这是我作为扩展方法添加到 DateTime 类的一个实现，它处理未来和过去的日期，并提供一个近似选项，允许您指定您正在寻找的详细程度（“3 小时前”与“3 小时， 23 分 12 秒前”）：
using System.Text;
/// <summary>
/// Compares a supplied date to the current date and generates a friendly English
/// comparison ("5 days ago", "5 days from now")
/// </summary>
/// <param name="date">The date to convert</param>
/// <param name="approximate">When off, calculate timespan down to the second.
/// When on, approximate to the largest round unit of time.</param>
/// <returns></returns>
public static string ToRelativeDateString(this DateTime value, bool approximate)
{
StringBuilder sb = new StringBuilder();
string suffix = (value > DateTime.Now) ? " from now" : " ago";
TimeSpan timeSpan = new TimeSpan(Math.Abs(DateTime.Now.Subtract(value).Ticks));
if (timeSpan.Days > 0)
{
sb.AppendFormat("{0} {1}", timeSpan.Days,
(timeSpan.Days > 1) ? "days" : "day");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Hours > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Hours, (timeSpan.Hours > 1) ? "hours" : "hour");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Minutes > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Minutes, (timeSpan.Minutes > 1) ? "minutes" : "minute");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Seconds > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Seconds, (timeSpan.Seconds > 1) ? "seconds" : "second");
if (approximate) return sb.ToString() + suffix;
}
if (sb.Length == 0) return "right now";
sb.Append(suffix);
return sb.ToString();
}
于 2009-03-09T22:02:29.940 回答
44
Nuget 上还有一个名为Humanizr的包，它实际上运行良好，并且位于 .NET Foundation 中。
DateTime.UtcNow.AddHours(-30).Humanize() => "yesterday"
DateTime.UtcNow.AddHours(-2).Humanize() => "2 hours ago"
DateTime.UtcNow.AddHours(30).Humanize() => "tomorrow"
DateTime.UtcNow.AddHours(2).Humanize() => "2 hours from now"
TimeSpan.FromMilliseconds(1299630020).Humanize() => "2 weeks"
TimeSpan.FromMilliseconds(1299630020).Humanize(3) => "2 weeks, 1 day, 1 hour"
Scott Hanselman 在他的博客上有一篇文章
于 2014-04-09T11:50:10.300 回答
37
I would recommend computing this on the client side too. Less work for the server.
The following is the version that I use (from Zach Leatherman)
/*
* Javascript Humane Dates
* Copyright (c) 2008 Dean Landolt (deanlandolt.com)
* Re-write by Zach Leatherman (zachleat.com)
*
* Adopted from the John Resig's pretty.js
* at http://ejohn.org/blog/javascript-pretty-date
* and henrah's proposed modification
* at http://ejohn.org/blog/javascript-pretty-date/#comment-297458
*
* Licensed under the MIT license.
*/
function humane_date(date_str){
var time_formats = [
[60, 'just now'],
[90, '1 minute'], // 60*1.5
[3600, 'minutes', 60], // 60*60, 60
[5400, '1 hour'], // 60*60*1.5
[86400, 'hours', 3600], // 60*60*24, 60*60
[129600, '1 day'], // 60*60*24*1.5
[604800, 'days', 86400], // 60*60*24*7, 60*60*24
[907200, '1 week'], // 60*60*24*7*1.5
[2628000, 'weeks', 604800], // 60*60*24*(365/12), 60*60*24*7
[3942000, '1 month'], // 60*60*24*(365/12)*1.5
[31536000, 'months', 2628000], // 60*60*24*365, 60*60*24*(365/12)
[47304000, '1 year'], // 60*60*24*365*1.5
[3153600000, 'years', 31536000], // 60*60*24*365*100, 60*60*24*365
[4730400000, '1 century'] // 60*60*24*365*100*1.5
];
var time = ('' + date_str).replace(/-/g,"/").replace(/[TZ]/g," "),
dt = new Date,
seconds = ((dt - new Date(time) + (dt.getTimezoneOffset() * 60000)) / 1000),
token = ' ago',
i = 0,
format;
if (seconds < 0) {
seconds = Math.abs(seconds);
token = '';
}
while (format = time_formats[i++]) {
if (seconds < format[0]) {
if (format.length == 2) {
return format[1] + (i > 1 ? token : ''); // Conditional so we don't return Just Now Ago
} else {
return Math.round(seconds / format[2]) + ' ' + format[1] + (i > 1 ? token : '');
}
}
}
// overflow for centuries
if(seconds > 4730400000)
return Math.round(seconds / 4730400000) + ' centuries' + token;
return date_str;
};
if(typeof jQuery != 'undefined') {
jQuery.fn.humane_dates = function(){
return this.each(function(){
var date = humane_date(this.title);
if(date && jQuery(this).text() != date) // don't modify the dom if we don't have to
jQuery(this).text(date);
});
};
}
于 2008-10-23T10:44:13.777 回答
33
@jeff
IMHO yours seems a little long. However it does seem a little more robust with support for "yesterday" and "years". But in my experience when this is used, the person is most likely to view the content in the first 30 days. It is only the really hardcore people that come after that. So, I usually elect to keep this short and simple.
This is the method I am currently using in one of my websites. This returns only a relative day, hour and time. And then the user has to slap on "ago" in the output.
public static string ToLongString(this TimeSpan time)
{
string output = String.Empty;
if (time.Days > 0)
output += time.Days + " days ";
if ((time.Days == 0 || time.Days == 1) && time.Hours > 0)
output += time.Hours + " hr ";
if (time.Days == 0 && time.Minutes > 0)
output += time.Minutes + " min ";
if (output.Length == 0)
output += time.Seconds + " sec";
return output.Trim();
}
于 2008-08-01T12:17:19.357 回答
25
晚会晚了几年，但我有要求在过去和未来的日期都这样做，所以我把Jeff和Vincent合并到了这个中。这是一场三重盛宴！:)
public static class DateTimeHelper
{
private const int SECOND = 1;
private const int MINUTE = 60 * SECOND;
private const int HOUR = 60 * MINUTE;
private const int DAY = 24 * HOUR;
private const int MONTH = 30 * DAY;
/// <summary>
/// Returns a friendly version of the provided DateTime, relative to now. E.g.: "2 days ago", or "in 6 months".
/// </summary>
/// <param name="dateTime">The DateTime to compare to Now</param>
/// <returns>A friendly string</returns>
public static string GetFriendlyRelativeTime(DateTime dateTime)
{
if (DateTime.UtcNow.Ticks == dateTime.Ticks)
{
return "Right now!";
}
bool isFuture = (DateTime.UtcNow.Ticks < dateTime.Ticks);
var ts = DateTime.UtcNow.Ticks < dateTime.Ticks ? new TimeSpan(dateTime.Ticks - DateTime.UtcNow.Ticks) : new TimeSpan(DateTime.UtcNow.Ticks - dateTime.Ticks);
double delta = ts.TotalSeconds;
if (delta < 1 * MINUTE)
{
return isFuture ? "in " + (ts.Seconds == 1 ? "one second" : ts.Seconds + " seconds") : ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2 * MINUTE)
{
return isFuture ? "in a minute" : "a minute ago";
}
if (delta < 45 * MINUTE)
{
return isFuture ? "in " + ts.Minutes + " minutes" : ts.Minutes + " minutes ago";
}
if (delta < 90 * MINUTE)
{
return isFuture ? "in an hour" : "an hour ago";
}
if (delta < 24 * HOUR)
{
return isFuture ? "in " + ts.Hours + " hours" : ts.Hours + " hours ago";
}
if (delta < 48 * HOUR)
{
return isFuture ? "tomorrow" : "yesterday";
}
if (delta < 30 * DAY)
{
return isFuture ? "in " + ts.Days + " days" : ts.Days + " days ago";
}
if (delta < 12 * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return isFuture ? "in " + (months <= 1 ? "one month" : months + " months") : months <= 1 ? "one month ago" : months + " months ago";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return isFuture ? "in " + (years <= 1 ? "one year" : years + " years") : years <= 1 ? "one year ago" : years + " years ago";
}
}
}
于 2011-03-25T00:21:04.483 回答
21
有没有一种简单的方法可以在 Java 中做到这一点？java.util.Date课程似乎相当有限。
这是我快速而肮脏的 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;
}
于 2009-02-20T15:17:43.423 回答
21
iPhone Objective-C 版本
+ (NSString *)timeAgoString:(NSDate *)date {
int delta = -(int)[date timeIntervalSinceNow];
if (delta < 60)
{
return delta == 1 ? @"one second ago" : [NSString stringWithFormat:@"%i seconds ago", delta];
}
if (delta < 120)
{
return @"a minute ago";
}
if (delta < 2700)
{
return [NSString stringWithFormat:@"%i minutes ago", delta/60];
}
if (delta < 5400)
{
return @"an hour ago";
}
if (delta < 24 * 3600)
{
return [NSString stringWithFormat:@"%i hours ago", delta/3600];
}
if (delta < 48 * 3600)
{
return @"yesterday";
}
if (delta < 30 * 24 * 3600)
{
return [NSString stringWithFormat:@"%i days ago", delta/(24*3600)];
}
if (delta < 12 * 30 * 24 * 3600)
{
int months = delta/(30*24*3600);
return months <= 1 ? @"one month ago" : [NSString stringWithFormat:@"%i months ago", months];
}
else
{
int years = delta/(12*30*24*3600);
return years <= 1 ? @"one year ago" : [NSString stringWithFormat:@"%i years ago", years];
}
}
于 2010-02-11T12:23:32.800 回答
20
鉴于世界和她的丈夫似乎正在发布代码示例，这是我不久前根据其中几个答案写的。
我特别需要此代码是可本地化的。所以我有两个类—— Grammar，它指定了可本地化的术语，和FuzzyDateExtensions，它包含一堆扩展方法。我不需要处理未来的日期时间，所以没有尝试用这段代码来处理它们。
我在源代码中留下了一些 XMLdoc，但为了简洁起见，删除了大部分（它们很明显）。我也没有在这里包括每个班级成员：
public class Grammar
{
/// <summary> Gets or sets the term for "just now". </summary>
public string JustNow { get; set; }
/// <summary> Gets or sets the term for "X minutes ago". </summary>
/// <remarks>
/// This is a <see cref="String.Format"/> pattern, where <c>{0}</c>
/// is the number of minutes.
/// </remarks>
public string MinutesAgo { get; set; }
public string OneHourAgo { get; set; }
public string HoursAgo { get; set; }
public string Yesterday { get; set; }
public string DaysAgo { get; set; }
public string LastMonth { get; set; }
public string MonthsAgo { get; set; }
public string LastYear { get; set; }
public string YearsAgo { get; set; }
/// <summary> Gets or sets the term for "ages ago". </summary>
public string AgesAgo { get; set; }
/// <summary>
/// Gets or sets the threshold beyond which the fuzzy date should be
/// considered "ages ago".
/// </summary>
public TimeSpan AgesAgoThreshold { get; set; }
/// <summary>
/// Initialises a new <see cref="Grammar"/> instance with the
/// specified properties.
/// </summary>
private void Initialise(string justNow, string minutesAgo,
string oneHourAgo, string hoursAgo, string yesterday, string daysAgo,
string lastMonth, string monthsAgo, string lastYear, string yearsAgo,
string agesAgo, TimeSpan agesAgoThreshold)
{ ... }
}
该类FuzzyDateString包含：
public static class FuzzyDateExtensions
{
public static string ToFuzzyDateString(this TimeSpan timespan)
{
return timespan.ToFuzzyDateString(new Grammar());
}
public static string ToFuzzyDateString(this TimeSpan timespan,
Grammar grammar)
{
return GetFuzzyDateString(timespan, grammar);
}
public static string ToFuzzyDateString(this DateTime datetime)
{
return (DateTime.Now - datetime).ToFuzzyDateString();
}
public static string ToFuzzyDateString(this DateTime datetime,
Grammar grammar)
{
return (DateTime.Now - datetime).ToFuzzyDateString(grammar);
}
private static string GetFuzzyDateString(TimeSpan timespan,
Grammar grammar)
{
timespan = timespan.Duration();
if (timespan >= grammar.AgesAgoThreshold)
{
return grammar.AgesAgo;
}
if (timespan < new TimeSpan(0, 2, 0)) // 2 minutes
{
return grammar.JustNow;
}
if (timespan < new TimeSpan(1, 0, 0)) // 1 hour
{
return String.Format(grammar.MinutesAgo, timespan.Minutes);
}
if (timespan < new TimeSpan(1, 55, 0)) // 1 hour 55 minutes
{
return grammar.OneHourAgo;
}
if (timespan < new TimeSpan(12, 0, 0) // 12 hours
&& (DateTime.Now - timespan).IsToday())
{
return String.Format(grammar.HoursAgo, timespan.RoundedHours());
}
if ((DateTime.Now.AddDays(1) - timespan).IsToday())
{
return grammar.Yesterday;
}
if (timespan < new TimeSpan(32, 0, 0, 0) // 32 days
&& (DateTime.Now - timespan).IsThisMonth())
{
return String.Format(grammar.DaysAgo, timespan.RoundedDays());
}
if ((DateTime.Now.AddMonths(1) - timespan).IsThisMonth())
{
return grammar.LastMonth;
}
if (timespan < new TimeSpan(365, 0, 0, 0, 0) // 365 days
&& (DateTime.Now - timespan).IsThisYear())
{
return String.Format(grammar.MonthsAgo, timespan.RoundedMonths());
}
if ((DateTime.Now - timespan).AddYears(1).IsThisYear())
{
return grammar.LastYear;
}
return String.Format(grammar.YearsAgo, timespan.RoundedYears());
}
}
除了本地化之外，我想要实现的关键目标之一是“今天”仅表示“这个日历日”，因此IsToday, IsThisMonth,IsThisYear方法如下所示：
public static bool IsToday(this DateTime date)
{
return date.DayOfYear == DateTime.Now.DayOfYear && date.IsThisYear();
}
并且舍入方法是这样的（我已经包括在内RoundedMonths，因为这有点不同）：
public static int RoundedDays(this TimeSpan timespan)
{
return (timespan.Hours > 12) ? timespan.Days + 1 : timespan.Days;
}
public static int RoundedMonths(this TimeSpan timespan)
{
DateTime then = DateTime.Now - timespan;
// Number of partial months elapsed since 1 Jan, AD 1 (DateTime.MinValue)
int nowMonthYears = DateTime.Now.Year * 12 + DateTime.Now.Month;
int thenMonthYears = then.Year * 12 + then.Month;
return nowMonthYears - thenMonthYears;
}
我希望人们觉得这有用和/或有趣:o)
于 2011-04-29T18:31:37.443 回答
19
使用流畅的日期时间
var dateTime1 = 2.Hours().Ago();
var dateTime2 = 3.Days().Ago();
var dateTime3 = 1.Months().Ago();
var dateTime4 = 5.Hours().FromNow();
var dateTime5 = 2.Weeks().FromNow();
var dateTime6 = 40.Seconds().FromNow();
于 2009-09-04T13:15:59.820 回答
17
在 PHP 中，我这样做：
<?php
function timesince($original) {
// array of time period chunks
$chunks = array(
array(60 * 60 * 24 * 365 , 'year'),
array(60 * 60 * 24 * 30 , 'month'),
array(60 * 60 * 24 * 7, 'week'),
array(60 * 60 * 24 , 'day'),
array(60 * 60 , 'hour'),
array(60 , 'minute'),
);
$today = time(); /* Current unix time */
$since = $today - $original;
if($since > 604800) {
$print = date("M jS", $original);
if($since > 31536000) {
$print .= ", " . date("Y", $original);
}
return $print;
}
// $j saves performing the count function each time around the loop
for ($i = 0, $j = count($chunks); $i < $j; $i++) {
$seconds = $chunks[$i][0];
$name = $chunks[$i][1];
// finding the biggest chunk (if the chunk fits, break)
if (($count = floor($since / $seconds)) != 0) {
break;
}
}
$print = ($count == 1) ? '1 '.$name : "$count {$name}s";
return $print . " ago";
} ?>
于 2008-08-20T17:26:49.357 回答
15
我想我会使用类和多态性来试一试。我有一个之前的迭代，它使用了子类，最终导致了太多的开销。我已切换到更灵活的委托/公共属性对象模型，该模型要好得多。我的代码稍微准确一点，我希望我能想出一个更好的方法来生成“几个月前”，看起来并不过分设计。
我想我仍然会坚持使用 Jeff 的 if-then 级联，因为它的代码更少而且更简单（确保它按预期工作肯定更容易）。
对于下面的代码PrintRelativeTime.GetRelativeTimeMessage(TimeSpan ago)返回相对时间消息（例如“昨天”）。
public class RelativeTimeRange : IComparable
{
public TimeSpan UpperBound { get; set; }
public delegate string RelativeTimeTextDelegate(TimeSpan timeDelta);
public RelativeTimeTextDelegate MessageCreator { get; set; }
public int CompareTo(object obj)
{
if (!(obj is RelativeTimeRange))
{
return 1;
}
// note that this sorts in reverse order to the way you'd expect,
// this saves having to reverse a list later
return (obj as RelativeTimeRange).UpperBound.CompareTo(UpperBound);
}
}
public class PrintRelativeTime
{
private static List<RelativeTimeRange> timeRanges;
static PrintRelativeTime()
{
timeRanges = new List<RelativeTimeRange>{
new RelativeTimeRange
{
UpperBound = TimeSpan.FromSeconds(1),
MessageCreator = (delta) =>
{ return "one second ago"; }
},
new RelativeTimeRange
{
UpperBound = TimeSpan.FromSeconds(60),
MessageCreator = (delta) =>
{ return delta.Seconds + " seconds ago"; }
},
new RelativeTimeRange
{
UpperBound = TimeSpan.FromMinutes(2),
MessageCreator = (delta) =>
{ return "one minute ago"; }
},
new RelativeTimeRange
{
UpperBound = TimeSpan.FromMinutes(60),
MessageCreator = (delta) =>
{ return delta.Minutes + " minutes ago"; }
},
new RelativeTimeRange
{
UpperBound = TimeSpan.FromHours(2),
MessageCreator = (delta) =>
{ return "one hour ago"; }
},
new RelativeTimeRange
{
UpperBound = TimeSpan.FromHours(24),
MessageCreator = (delta) =>
{ return delta.Hours + " hours ago"; }
},
new RelativeTimeRange
{
UpperBound = TimeSpan.FromDays(2),
MessageCreator = (delta) =>
{ return "yesterday"; }
},
new RelativeTimeRange
{
UpperBound = DateTime.Now.Subtract(DateTime.Now.AddMonths(-1)),
MessageCreator = (delta) =>
{ return delta.Days + " days ago"; }
},
new RelativeTimeRange
{
UpperBound = DateTime.Now.Subtract(DateTime.Now.AddMonths(-2)),
MessageCreator = (delta) =>
{ return "one month ago"; }
},
new RelativeTimeRange
{
UpperBound = DateTime.Now.Subtract(DateTime.Now.AddYears(-1)),
MessageCreator = (delta) =>
{ return (int)Math.Floor(delta.TotalDays / 30) + " months ago"; }
},
new RelativeTimeRange
{
UpperBound = DateTime.Now.Subtract(DateTime.Now.AddYears(-2)),
MessageCreator = (delta) =>
{ return "one year ago"; }
},
new RelativeTimeRange
{
UpperBound = TimeSpan.MaxValue,
MessageCreator = (delta) =>
{ return (int)Math.Floor(delta.TotalDays / 365.24D) + " years ago"; }
}
};
timeRanges.Sort();
}
public static string GetRelativeTimeMessage(TimeSpan ago)
{
RelativeTimeRange postRelativeDateRange = timeRanges[0];
foreach (var timeRange in timeRanges)
{
if (ago.CompareTo(timeRange.UpperBound) <= 0)
{
postRelativeDateRange = timeRange;
}
}
return postRelativeDateRange.MessageCreator(ago);
}
}
于 2008-08-05T00:42:56.853 回答
13
当您知道查看者的时区时，在日尺度上使用日历日可能会更清楚。我不熟悉 .NET 库，所以很遗憾，我不知道你会如何在 C# 中做到这一点。
在消费者网站上，您也可以在一分钟内轻松应对。“不到一分钟前”或“刚刚”就足够了。
于 2008-08-15T22:42:33.870 回答
13
using System;
using System.Collections.Generic;
using System.Linq;
public static class RelativeDateHelper
{
private static Dictionary<double, Func<double, string>> sm_Dict = null;
private static Dictionary<double, Func<double, string>> DictionarySetup()
{
var dict = new Dictionary<double, Func<double, string>>();
dict.Add(0.75, (mins) => "less than a minute");
dict.Add(1.5, (mins) => "about a minute");
dict.Add(45, (mins) => string.Format("{0} minutes", Math.Round(mins)));
dict.Add(90, (mins) => "about an hour");
dict.Add(1440, (mins) => string.Format("about {0} hours", Math.Round(Math.Abs(mins / 60)))); // 60 * 24
dict.Add(2880, (mins) => "a day"); // 60 * 48
dict.Add(43200, (mins) => string.Format("{0} days", Math.Floor(Math.Abs(mins / 1440)))); // 60 * 24 * 30
dict.Add(86400, (mins) => "about a month"); // 60 * 24 * 60
dict.Add(525600, (mins) => string.Format("{0} months", Math.Floor(Math.Abs(mins / 43200)))); // 60 * 24 * 365
dict.Add(1051200, (mins) => "about a year"); // 60 * 24 * 365 * 2
dict.Add(double.MaxValue, (mins) => string.Format("{0} years", Math.Floor(Math.Abs(mins / 525600))));
return dict;
}
public static string ToRelativeDate(this DateTime input)
{
TimeSpan oSpan = DateTime.Now.Subtract(input);
double TotalMinutes = oSpan.TotalMinutes;
string Suffix = " ago";
if (TotalMinutes < 0.0)
{
TotalMinutes = Math.Abs(TotalMinutes);
Suffix = " from now";
}
if (null == sm_Dict)
sm_Dict = DictionarySetup();
return sm_Dict.First(n => TotalMinutes < n.Key).Value.Invoke(TotalMinutes) + Suffix;
}
}
与此问题的另一个答案相同，但作为具有静态字典的扩展方法。
于 2009-07-17T02:47:42.510 回答
11
@杰夫
var ts = new TimeSpan(DateTime.UtcNow.Ticks - dt.Ticks);
反正做减法DateTime返回 a 。TimeSpan
所以你可以做
(DateTime.UtcNow - dt).TotalSeconds
我也很惊讶地看到手动乘以常数，然后在乘法中添加注释。这是一些误导性的优化吗？
于 2008-08-15T14:20:04.133 回答
11
你可以试试这个。我认为它会正常工作。
long delta = new Date().getTime() - date.getTime();
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 < 0L)
{
return "not yet";
}
if (delta < 1L * MINUTE)
{
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2L * MINUTE)
{
return "a minute ago";
}
if (delta < 45L * MINUTE)
{
return ts.Minutes + " minutes ago";
}
if (delta < 90L * MINUTE)
{
return "an hour ago";
}
if (delta < 24L * HOUR)
{
return ts.Hours + " hours ago";
}
if (delta < 48L * HOUR)
{
return "yesterday";
}
if (delta < 30L * DAY)
{
return ts.Days + " days ago";
}
if (delta < 12L * 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";
}
于 2013-10-15T09:36:10.843 回答
11
您可以使用TimeAgo 扩展，如下所示：
public static string TimeAgo(this DateTime dateTime)
{
string result = string.Empty;
var timeSpan = DateTime.Now.Subtract(dateTime);
if (timeSpan <= TimeSpan.FromSeconds(60))
{
result = string.Format("{0} seconds ago", timeSpan.Seconds);
}
else if (timeSpan <= TimeSpan.FromMinutes(60))
{
result = timeSpan.Minutes > 1 ?
String.Format("about {0} minutes ago", timeSpan.Minutes) :
"about a minute ago";
}
else if (timeSpan <= TimeSpan.FromHours(24))
{
result = timeSpan.Hours > 1 ?
String.Format("about {0} hours ago", timeSpan.Hours) :
"about an hour ago";
}
else if (timeSpan <= TimeSpan.FromDays(30))
{
result = timeSpan.Days > 1 ?
String.Format("about {0} days ago", timeSpan.Days) :
"yesterday";
}
else if (timeSpan <= TimeSpan.FromDays(365))
{
result = timeSpan.Days > 30 ?
String.Format("about {0} months ago", timeSpan.Days / 30) :
"about a month ago";
}
else
{
result = timeSpan.Days > 365 ?
String.Format("about {0} years ago", timeSpan.Days / 365) :
"about a year ago";
}
return result;
}
或者使用来自 Timeago 的带有 Razor 扩展的jQuery 插件。
于 2015-09-08T14:02:39.373 回答
10
您可以通过在客户端执行此逻辑来减少服务器端负载。在一些 Digg 页面上查看源代码以供参考。他们让服务器发出一个由 Javascript 处理的纪元时间值。这样您就不需要管理最终用户的时区。新的服务器端代码将类似于：
public string GetRelativeTime(DateTime timeStamp)
{
return string.Format("<script>printdate({0});</script>", timeStamp.ToFileTimeUtc());
}
您甚至可以在那里添加一个 NOSCRIPT 块，然后执行一个 ToString()。
于 2008-08-17T15:56:26.810 回答
9
这是stackoverflow使用的算法，但用perlish伪代码更简洁地重写了错误修复（没有“一小时前”）。该函数需要一个（正）秒数，并返回一个人类友好的字符串，如“3 小时前”或“昨天”。
agoify($delta)
local($y, $mo, $d, $h, $m, $s);
$s = floor($delta);
if($s<=1) return "a second ago";
if($s<60) return "$s seconds ago";
$m = floor($s/60);
if($m==1) return "a minute ago";
if($m<45) return "$m minutes ago";
$h = floor($m/60);
if($h==1) return "an hour ago";
if($h<24) return "$h hours ago";
$d = floor($h/24);
if($d<2) return "yesterday";
if($d<30) return "$d days ago";
$mo = floor($d/30);
if($mo<=1) return "a month ago";
$y = floor($mo/12);
if($y<1) return "$mo months ago";
if($y==1) return "a year ago";
return "$y years ago";
于 2008-09-23T01:09:30.193 回答
9
Java for client-side gwt usage:
import java.util.Date;
public class RelativeDateFormat {
private static final long ONE_MINUTE = 60000L;
private static final long ONE_HOUR = 3600000L;
private static final long ONE_DAY = 86400000L;
private static final long ONE_WEEK = 604800000L;
public static String format(Date date) {
long delta = new Date().getTime() - date.getTime();
if (delta < 1L * ONE_MINUTE) {
return toSeconds(delta) == 1 ? "one second ago" : toSeconds(delta)
+ " seconds ago";
}
if (delta < 2L * ONE_MINUTE) {
return "one minute ago";
}
if (delta < 45L * ONE_MINUTE) {
return toMinutes(delta) + " minutes ago";
}
if (delta < 90L * ONE_MINUTE) {
return "one hour ago";
}
if (delta < 24L * ONE_HOUR) {
return toHours(delta) + " hours ago";
}
if (delta < 48L * ONE_HOUR) {
return "yesterday";
}
if (delta < 30L * ONE_DAY) {
return toDays(delta) + " days ago";
}
if (delta < 12L * 4L * ONE_WEEK) {
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 static long toSeconds(long date) {
return date / 1000L;
}
private static long toMinutes(long date) {
return toSeconds(date) / 60L;
}
private static long toHours(long date) {
return toMinutes(date) / 60L;
}
private static long toDays(long date) {
return toHours(date) / 24L;
}
private static long toMonths(long date) {
return toDays(date) / 30L;
}
private static long toYears(long date) {
return toMonths(date) / 365L;
}
}
于 2009-11-14T18:44:50.030 回答
9
我从比尔盖茨的一个博客中得到了这个答案。我需要在我的浏览器历史记录中找到它，然后我会给你链接。
执行相同操作的 Javascript 代码（根据要求）：
function posted(t) {
var now = new Date();
var diff = parseInt((now.getTime() - Date.parse(t)) / 1000);
if (diff < 60) { return 'less than a minute ago'; }
else if (diff < 120) { return 'about a minute ago'; }
else if (diff < (2700)) { return (parseInt(diff / 60)).toString() + ' minutes ago'; }
else if (diff < (5400)) { return 'about an hour ago'; }
else if (diff < (86400)) { return 'about ' + (parseInt(diff / 3600)).toString() + ' hours ago'; }
else if (diff < (172800)) { return '1 day ago'; }
else {return (parseInt(diff / 86400)).toString() + ' days ago'; }
}
基本上，您以秒为单位工作。
于 2010-02-01T19:51:13.127 回答
8
我认为已经有很多与这篇文章相关的答案，但是可以使用它，它就像插件一样易于使用，并且对于程序员来说也易于阅读。发送您的特定日期，并以字符串形式获取其值：
public string RelativeDateTimeCount(DateTime inputDateTime)
{
string outputDateTime = string.Empty;
TimeSpan ts = DateTime.Now - inputDateTime;
if (ts.Days > 7)
{ outputDateTime = inputDateTime.ToString("MMMM d, yyyy"); }
else if (ts.Days > 0)
{
outputDateTime = ts.Days == 1 ? ("about 1 Day ago") : ("about " + ts.Days.ToString() + " Days ago");
}
else if (ts.Hours > 0)
{
outputDateTime = ts.Hours == 1 ? ("an hour ago") : (ts.Hours.ToString() + " hours ago");
}
else if (ts.Minutes > 0)
{
outputDateTime = ts.Minutes == 1 ? ("1 minute ago") : (ts.Minutes.ToString() + " minutes ago");
}
else outputDateTime = "few seconds ago";
return outputDateTime;
}
于 2013-03-16T06:36:07.017 回答
6
var ts = new TimeSpan(DateTime.Now.Ticks - dt.Ticks);
于 2012-05-27T17:30:29.780 回答
6
如果你想有一个像 的输出"2 days, 4 hours and 12 minutes ago"，你需要一个时间跨度：
TimeSpan timeDiff = DateTime.Now-CreatedDate;
然后您可以访问您喜欢的值：
timeDiff.Days
timeDiff.Hours
ETC...
于 2015-09-08T14:02:16.003 回答
5
I would provide some handy extensions methods for this and make the code more readable. First, couple of extension methods for Int32.
public static class TimeSpanExtensions {
public static TimeSpan Days(this int value) {
return new TimeSpan(value, 0, 0, 0);
}
public static TimeSpan Hours(this int value) {
return new TimeSpan(0, value, 0, 0);
}
public static TimeSpan Minutes(this int value) {
return new TimeSpan(0, 0, value, 0);
}
public static TimeSpan Seconds(this int value) {
return new TimeSpan(0, 0, 0, value);
}
public static TimeSpan Milliseconds(this int value) {
return new TimeSpan(0, 0, 0, 0, value);
}
public static DateTime Ago(this TimeSpan value) {
return DateTime.Now - value;
}
}
Then, one for DateTime.
public static class DateTimeExtensions {
public static DateTime Ago(this DateTime dateTime, TimeSpan delta) {
return dateTime - delta;
}
}
Now, you can do something like below:
var date = DateTime.Now;
date.Ago(2.Days()); // 2 days ago
date.Ago(7.Hours()); // 7 hours ago
date.Ago(567.Milliseconds()); // 567 milliseconds ago
于 2012-09-13T12:15:49.307 回答
5
/**
* {@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";
}
}
于 2014-09-05T01:11:31.120 回答
4
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 Google。
于 2013-08-06T07:54:50.763 回答
4
土耳其语本地化版本的文森特答案。
const int SECOND = 1;
const int MINUTE = 60 * SECOND;
const int HOUR = 60 * MINUTE;
const int DAY = 24 * HOUR;
const int MONTH = 30 * DAY;
var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
if (delta < 1 * MINUTE)
return ts.Seconds + " saniye önce";
if (delta < 45 * MINUTE)
return ts.Minutes + " dakika önce";
if (delta < 24 * HOUR)
return ts.Hours + " saat önce";
if (delta < 48 * HOUR)
return "dün";
if (delta < 30 * DAY)
return ts.Days + " gün önce";
if (delta < 12 * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months + " ay önce";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years + " yıl önce";
}
于 2019-01-03T08:03:07.237 回答
3
当然，摆脱“一小时前”问题的简单解决方法是增加“一小时前”有效的窗口。改变
if (delta < 5400) // 90 * 60
{
return "an hour ago";
}
进入
if (delta < 7200) // 120 * 60
{
return "an hour ago";
}
这意味着 110 分钟前发生的事情将读作“一小时前”——这可能并不完美，但我会说它比“1 小时前”的当前情况要好。
于 2008-08-25T06:12:29.623 回答
3
在某种程度上，您可以DateTime通过以秒到年计算相对时间来执行您的功能，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，但是如果您对其进行一些修改和调试，它可能会完成这项工作。
于 2020-10-13T06:28:46.130 回答
2
// Calculate total days in current year
int daysInYear;
for (var i = 1; i <= 12; i++)
daysInYear += DateTime.DaysInMonth(DateTime.Now.Year, i);
// Past date
DateTime dateToCompare = DateTime.Now.Subtract(TimeSpan.FromMinutes(582));
// Calculate difference between current date and past date
double diff = (DateTime.Now - dateToCompare).TotalMilliseconds;
TimeSpan ts = TimeSpan.FromMilliseconds(diff);
var years = ts.TotalDays / daysInYear; // Years
var months = ts.TotalDays / (daysInYear / (double)12); // Months
var weeks = ts.TotalDays / 7; // Weeks
var days = ts.TotalDays; // Days
var hours = ts.TotalHours; // Hours
var minutes = ts.TotalMinutes; // Minutes
var seconds = ts.TotalSeconds; // Seconds
if (years >= 1)
Console.WriteLine(Math.Round(years, 0) + " year(s) ago");
else if (months >= 1)
Console.WriteLine(Math.Round(months, 0) + " month(s) ago");
else if (weeks >= 1)
Console.WriteLine(Math.Round(weeks, 0) + " week(s) ago");
else if (days >= 1)
Console.WriteLine(Math.Round(days, 0) + " days(s) ago");
else if (hours >= 1)
Console.WriteLine(Math.Round(hours, 0) + " hour(s) ago");
else if (minutes >= 1)
Console.WriteLine(Math.Round(minutes, 0) + " minute(s) ago");
else if (seconds >= 1)
Console.WriteLine(Math.Round(seconds, 0) + " second(s) ago");
Console.ReadLine();
于 2019-05-26T02:31:53.863 回答
2
使用解构和 Linq 获得“n [最大时间单位] 前”的“单线”：
TimeSpan timeSpan = DateTime.Now - new DateTime(1234, 5, 6, 7, 8, 9);
(string unit, int value) = new Dictionary<string, int>
{
{"year(s)", (int)(timeSpan.TotalDays / 365.25)}, //https://en.wikipedia.org/wiki/Year#Intercalation
{"month(s)", (int)(timeSpan.TotalDays / 29.53)}, //https://en.wikipedia.org/wiki/Month
{"day(s)", (int)timeSpan.TotalDays},
{"hour(s)", (int)timeSpan.TotalHours},
{"minute(s)", (int)timeSpan.TotalMinutes},
{"second(s)", (int)timeSpan.TotalSeconds},
{"millisecond(s)", (int)timeSpan.TotalMilliseconds}
}.First(kvp => kvp.Value > 0);
Console.WriteLine($"{value} {unit} ago");
你得到786 year(s) ago
用当前的年份和月份，比如
TimeSpan timeSpan = DateTime.Now - new DateTime(2020, 12, 6, 7, 8, 9);
你得到4 day(s) ago
随着实际日期，如
TimeSpan timeSpan = DateTime.Now - DateTime.Now.Date;
你得到9 hour(s) ago
于 2020-12-10T16:00:48.177 回答
2
这是我的功能，就像一个魅力:)
public static string RelativeDate(DateTime theDate)
{
var span = DateTime.Now - theDate;
if (span.Days > 365)
{
var years = (span.Days / 365);
if (span.Days % 365 != 0)
years += 1;
return $"about {years} {(years == 1 ? "year" : "years")} ago";
}
if (span.Days > 30)
{
var months = (span.Days / 30);
if (span.Days % 31 != 0)
months += 1;
return $"about {months} {(months == 1 ? "month" : "months")} ago";
}
if (span.Days > 0)
return $"about {span.Days} {(span.Days == 1 ? "day" : "days")} ago";
if (span.Hours > 0)
return $"about {span.Hours} {(span.Hours == 1 ? "hour" : "hours")} ago";
if (span.Minutes > 0)
return $"about {span.Minutes} {(span.Minutes == 1 ? "minute" : "minutes")} ago";
if (span.Seconds > 5)
return $"about {span.Seconds} seconds ago";
return span.Seconds <= 5 ? "about 5 seconds ago" : string.Empty;
}
于 2017-09-04T01:27:10.793 回答
-1
My way is much more simpler. You can tweak with the return strings as you want
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;
}
于 2018-03-15T13:03:01.000 回答
-1
简单且 100% 有效的解决方案。
也处理以前和未来的时间..以防万一
public string GetTimeSince(DateTime postDate)
{
string message = "";
DateTime currentDate = DateTime.Now;
TimeSpan timegap = currentDate - postDate;
if (timegap.Days > 365)
{
message = string.Format(L("Ago") + " {0} " + L("Years"), (((timegap.Days) / 30) / 12));
}
else if (timegap.Days > 30)
{
message = string.Format(L("Ago") + " {0} " + L("Months"), timegap.Days/30);
}
else if (timegap.Days > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Days"), timegap.Days);
}
else if (timegap.Hours > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Hours"), timegap.Hours);
}
else if (timegap.Minutes > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Minutes"), timegap.Minutes);
}
else if (timegap.Seconds > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Seconds"), timegap.Seconds);
}
// let's handle future times..just in case
else if (timegap.Days < -365)
{
message = string.Format(L("In") + " {0} " + L("Years"), (((Math.Abs(timegap.Days)) / 30) / 12));
}
else if (timegap.Days < -30)
{
message = string.Format(L("In") + " {0} " + L("Months"), ((Math.Abs(timegap.Days)) / 30));
}
else if (timegap.Days < 0)
{
message = string.Format(L("In") + " {0} " + L("Days"), Math.Abs(timegap.Days));
}
else if (timegap.Hours < 0)
{
message = string.Format(L("In") + " {0} " + L("Hours"), Math.Abs(timegap.Hours));
}
else if (timegap.Minutes < 0)
{
message = string.Format(L("In") + " {0} " + L("Minutes"), Math.Abs(timegap.Minutes));
}
else if (timegap.Seconds < 0)
{
message = string.Format(L("In") + " {0} " + L("Seconds"), Math.Abs(timegap.Seconds));
}
else
{
message = "a bit";
}
return message;
}
于 2021-09-01T21:49:09.617 回答