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
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
用于客户端gwt的Java:
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;
}
}
其他回答
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];
}
}
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
这是stackoverflow使用的算法,但使用了错误修复(没有“一小时前”)的perlish伪代码进行了更简洁的重写。该函数在秒前取一个(正数),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";
@杰夫
我知道你的有点长。然而,随着对“昨天”和“几年”的支持,它似乎确实更为有力。但根据我的经验,当使用此选项时,用户最有可能在前30天内查看内容。只有真正的铁杆人才会在这之后出现。所以,我通常选择保持简短。
这是我目前在我的一个网站上使用的方法。这只返回相对的日期、小时和时间。然后用户必须在输出中加上“ago”。
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();
}
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,我会给你链接。
执行相同操作的Javascript代码(按要求):
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,你是以秒为单位工作的。