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
你可以试试这个。我想它会正常工作的。
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";
}
其他回答
简单且100%的工作解决方案。
处理过去和将来的时间。。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;
}
你可以试试这个。我想它会正常工作的。
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";
}
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 => $"{x.TotalDays / 365:F0} years"}
};
public static string ToRelativeDate(this DateTime input)
{
TimeSpan x = DateTime.Now - input;
string Suffix = x.TotalMinutes > 0 ? " ago" : " from now";
x = new TimeSpan(Math.Abs(x.Ticks));
return offsets.First(n => x.TotalMinutes < n.Key).Value(x) + Suffix;
}
在PHP中，我是这样做的：
<?php
function timesince($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";
} ?>
通过在客户端执行此逻辑，可以减少服务器端负载。在一些Digg页面上查看源代码以供参考。它们让服务器发出一个由Javascript处理的历元时间值。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ToString（）。
推荐文章
- 实体框架核心:在上一个操作完成之前，在此上下文中开始的第二个操作
- 如何为构造函数定制Visual Studio的私有字段生成快捷方式?
- 如何使用JSON确保字符串是有效的JSON。网
- AppSettings从.config文件中获取值
- 通过HttpClient向REST API发布一个空体
- 如何检查IEnumerable是否为空或空?
- 自动化invokerrequired代码模式
- 解析日期字符串并更改格式
- 在c#代码中设置WPF文本框的背景颜色
- 在c#中，什么是单子?
- c#和Java中的泛型有什么不同?和模板在c++ ?
- c#线程安全快速(est)计数器
- 如何将此foreach代码转换为Parallel.ForEach?
- 如何分裂()一个分隔字符串到一个列表<字符串>
- 如何转换列表<字符串>列表<int>?
