在 Ruby on Rails 中，有一个功能允许您获取任何日期并打印出它是“很久以前”的时间。
例如：
8 minutes ago
8 hours ago
8 days ago
8 months ago
8 years ago
有没有一种简单的方法可以在 Java 中做到这一点？
问问题
110839 次
33 回答
188
看看PrettyTime库。
使用起来非常简单：
import org.ocpsoft.prettytime.PrettyTime;
PrettyTime p = new PrettyTime();
System.out.println(p.format(new Date()));
// prints "moments ago"
您还可以为国际化消息传递语言环境：
PrettyTime p = new PrettyTime(new Locale("fr"));
System.out.println(p.format(new Date()));
// prints "à l'instant"
如评论中所述，Android 已将此功能内置到android.text.format.DateUtils该类中。
于 2010-10-04T21:32:38.173 回答
71
您是否考虑过TimeUnit枚举？这对这种事情可能非常有用
try {
SimpleDateFormat format = new SimpleDateFormat("dd/MM/yyyy");
Date past = format.parse("01/10/2010");
Date now = new Date();
System.out.println(TimeUnit.MILLISECONDS.toMillis(now.getTime() - past.getTime()) + " milliseconds ago");
System.out.println(TimeUnit.MILLISECONDS.toMinutes(now.getTime() - past.getTime()) + " minutes ago");
System.out.println(TimeUnit.MILLISECONDS.toHours(now.getTime() - past.getTime()) + " hours ago");
System.out.println(TimeUnit.MILLISECONDS.toDays(now.getTime() - past.getTime()) + " days ago");
}
catch (Exception j){
j.printStackTrace();
}
于 2010-10-04T22:30:28.403 回答
55
我采用 RealHowTo 和 Ben J 的答案并制作自己的版本：
public class TimeAgo {
public static final List<Long> times = Arrays.asList(
TimeUnit.DAYS.toMillis(365),
TimeUnit.DAYS.toMillis(30),
TimeUnit.DAYS.toMillis(1),
TimeUnit.HOURS.toMillis(1),
TimeUnit.MINUTES.toMillis(1),
TimeUnit.SECONDS.toMillis(1) );
public static final List<String> timesString = Arrays.asList("year","month","day","hour","minute","second");
public static String toDuration(long duration) {
StringBuffer res = new StringBuffer();
for(int i=0;i< TimeAgo.times.size(); i++) {
Long current = TimeAgo.times.get(i);
long temp = duration/current;
if(temp>0) {
res.append(temp).append(" ").append( TimeAgo.timesString.get(i) ).append(temp != 1 ? "s" : "").append(" ago");
break;
}
}
if("".equals(res.toString()))
return "0 seconds ago";
else
return res.toString();
}
public static void main(String args[]) {
System.out.println(toDuration(123));
System.out.println(toDuration(1230));
System.out.println(toDuration(12300));
System.out.println(toDuration(123000));
System.out.println(toDuration(1230000));
System.out.println(toDuration(12300000));
System.out.println(toDuration(123000000));
System.out.println(toDuration(1230000000));
System.out.println(toDuration(12300000000L));
System.out.println(toDuration(123000000000L));
}}
这将打印以下内容
0 second ago
1 second ago
12 seconds ago
2 minutes ago
20 minutes ago
3 hours ago
1 day ago
14 days ago
4 months ago
3 years ago
于 2014-04-22T09:08:19.300 回答
44
public class TimeUtils {
public final static long ONE_SECOND = 1000;
public final static long SECONDS = 60;
public final static long ONE_MINUTE = ONE_SECOND * 60;
public final static long MINUTES = 60;
public final static long ONE_HOUR = ONE_MINUTE * 60;
public final static long HOURS = 24;
public final static long ONE_DAY = ONE_HOUR * 24;
private TimeUtils() {
}
/**
* converts time (in milliseconds) to human-readable format
* "<w> days, <x> hours, <y> minutes and (z) seconds"
*/
public static String millisToLongDHMS(long duration) {
StringBuffer res = new StringBuffer();
long temp = 0;
if (duration >= ONE_SECOND) {
temp = duration / ONE_DAY;
if (temp > 0) {
duration -= temp * ONE_DAY;
res.append(temp).append(" day").append(temp > 1 ? "s" : "")
.append(duration >= ONE_MINUTE ? ", " : "");
}
temp = duration / ONE_HOUR;
if (temp > 0) {
duration -= temp * ONE_HOUR;
res.append(temp).append(" hour").append(temp > 1 ? "s" : "")
.append(duration >= ONE_MINUTE ? ", " : "");
}
temp = duration / ONE_MINUTE;
if (temp > 0) {
duration -= temp * ONE_MINUTE;
res.append(temp).append(" minute").append(temp > 1 ? "s" : "");
}
if (!res.toString().equals("") && duration >= ONE_SECOND) {
res.append(" and ");
}
temp = duration / ONE_SECOND;
if (temp > 0) {
res.append(temp).append(" second").append(temp > 1 ? "s" : "");
}
return res.toString();
} else {
return "0 second";
}
}
public static void main(String args[]) {
System.out.println(millisToLongDHMS(123));
System.out.println(millisToLongDHMS((5 * ONE_SECOND) + 123));
System.out.println(millisToLongDHMS(ONE_DAY + ONE_HOUR));
System.out.println(millisToLongDHMS(ONE_DAY + 2 * ONE_SECOND));
System.out.println(millisToLongDHMS(ONE_DAY + ONE_HOUR + (2 * ONE_MINUTE)));
System.out.println(millisToLongDHMS((4 * ONE_DAY) + (3 * ONE_HOUR)
+ (2 * ONE_MINUTE) + ONE_SECOND));
System.out.println(millisToLongDHMS((5 * ONE_DAY) + (4 * ONE_HOUR)
+ ONE_MINUTE + (23 * ONE_SECOND) + 123));
System.out.println(millisToLongDHMS(42 * ONE_DAY));
/*
output :
0 second
5 seconds
1 day, 1 hour
1 day and 2 seconds
1 day, 1 hour, 2 minutes
4 days, 3 hours, 2 minutes and 1 second
5 days, 4 hours, 1 minute and 23 seconds
42 days
*/
}
}
于 2010-10-05T01:25:26.160 回答
11
这是基于 RealHowTo 的回答，所以如果你喜欢它，也请给他/她一些爱。
这个清理后的版本允许您指定您可能感兴趣的时间范围。
它对“和”部分的处理也略有不同。我经常发现在使用分隔符连接字符串时，跳过复杂的逻辑并在完成后删除最后一个分隔符通常更容易。
import java.util.concurrent.TimeUnit;
import static java.util.concurrent.TimeUnit.MILLISECONDS;
public class TimeUtils {
/**
* Converts time to a human readable format within the specified range
*
* @param duration the time in milliseconds to be converted
* @param max the highest time unit of interest
* @param min the lowest time unit of interest
*/
public static String formatMillis(long duration, TimeUnit max, TimeUnit min) {
StringBuilder res = new StringBuilder();
TimeUnit current = max;
while (duration > 0) {
long temp = current.convert(duration, MILLISECONDS);
if (temp > 0) {
duration -= current.toMillis(temp);
res.append(temp).append(" ").append(current.name().toLowerCase());
if (temp < 2) res.deleteCharAt(res.length() - 1);
res.append(", ");
}
if (current == min) break;
current = TimeUnit.values()[current.ordinal() - 1];
}
// clean up our formatting....
// we never got a hit, the time is lower than we care about
if (res.lastIndexOf(", ") < 0) return "0 " + min.name().toLowerCase();
// yank trailing ", "
res.deleteCharAt(res.length() - 2);
// convert last ", " to " and"
int i = res.lastIndexOf(", ");
if (i > 0) {
res.deleteCharAt(i);
res.insert(i, " and");
}
return res.toString();
}
}
小代码可以试一试：
import static java.util.concurrent.TimeUnit.*;
public class Main {
public static void main(String args[]) {
long[] durations = new long[]{
123,
SECONDS.toMillis(5) + 123,
DAYS.toMillis(1) + HOURS.toMillis(1),
DAYS.toMillis(1) + SECONDS.toMillis(2),
DAYS.toMillis(1) + HOURS.toMillis(1) + MINUTES.toMillis(2),
DAYS.toMillis(4) + HOURS.toMillis(3) + MINUTES.toMillis(2) + SECONDS.toMillis(1),
DAYS.toMillis(5) + HOURS.toMillis(4) + MINUTES.toMillis(1) + SECONDS.toMillis(23) + 123,
DAYS.toMillis(42)
};
for (long duration : durations) {
System.out.println(TimeUtils.formatMillis(duration, DAYS, SECONDS));
}
System.out.println("\nAgain in only hours and minutes\n");
for (long duration : durations) {
System.out.println(TimeUtils.formatMillis(duration, HOURS, MINUTES));
}
}
}
这将输出以下内容：
0 seconds
5 seconds
1 day and 1 hour
1 day and 2 seconds
1 day, 1 hour and 2 minutes
4 days, 3 hours, 2 minutes and 1 second
5 days, 4 hours, 1 minute and 23 seconds
42 days
Again in only hours and minutes
0 minutes
0 minutes
25 hours
24 hours
25 hours and 2 minutes
99 hours and 2 minutes
124 hours and 1 minute
1008 hours
如果有人需要它，这里有一个类可以将上面的任何字符串转换回毫秒。它对于允许人们在可读文本中指定各种事物的超时非常有用。
于 2011-02-21T05:53:24.097 回答
10
有一个简单的方法可以做到这一点：
假设您想要 20 分钟前的时间：
Long minutesAgo = new Long(20);
Date date = new Date();
Date dateIn_X_MinAgo = new Date (date.getTime() - minutesAgo*60*1000);
就是这样..
于 2013-02-14T09:56:10.157 回答
7
关于内置解决方案：
Java 没有对格式化相对时间的任何内置支持，Java-8 及其新包也没有java.time。如果您只需要英语而不需要其他任何东西，那么手工制作的解决方案可能是可以接受的 - 请参阅@RealHowTo 的答案（尽管它有一个很大的缺点是不考虑将即时增量转换为当地时间的时区单位！）。无论如何，如果你想避免本土的复杂解决方法，尤其是对于其他语言环境，那么你需要一个外部库。
在后一种情况下，我建议使用我的库Time4J（或 Android 上的 Time4A）。它提供了最大的灵活性和最大的 i18n-power。类net.time4j.PrettyTime有七个printRelativeTime...(...)用于此目的的方法。使用测试时钟作为时间源的示例：
TimeSource<?> clock = () -> PlainTimestamp.of(2015, 8, 1, 10, 24, 5).atUTC();
Moment moment = PlainTimestamp.of(2015, 8, 1, 17, 0).atUTC(); // our input
String durationInDays =
PrettyTime.of(Locale.GERMAN).withReferenceClock(clock).printRelative(
moment,
Timezone.of(EUROPE.BERLIN),
TimeUnit.DAYS); // controlling the precision
System.out.println(durationInDays); // heute (german word for today)
另一个java.time.Instant用作输入的示例：
String relativeTime =
PrettyTime.of(Locale.ENGLISH)
.printRelativeInStdTimezone(Moment.from(Instant.EPOCH));
System.out.println(relativeTime); // 45 years ago
该库通过其最新版本 (v4.17) 支持80 种语言以及一些特定于国家/地区的语言环境（尤其是西班牙语、英语、阿拉伯语、法语）。i18n-data 主要基于最新的 CLDR-version v29。使用这个库的其他重要原因是对复数规则的良好支持（在其他语言环境中通常与英语不同）、缩写格式样式（例如：“1 sec ago”）和考虑时区的表达方式。Time4J 甚至在相对时间的计算中意识到诸如闰秒之类的奇特细节（不是很重要，但它形成了与期望范围相关的消息）。与 Java-8的兼容性由于易于使用的类型（如java.time.Instant或）的转换方法而存在java.time.Period。
有什么缺点吗？只有两个。
- 这个库并不小（也因为它的 i18n 数据存储库很大）。
- 该 API 并不广为人知，因此社区知识和支持尚不可用，否则提供的文档非常详细和全面。
（紧凑）替代品：
如果您正在寻找更小的解决方案并且不需要这么多功能并且愿意容忍与 i18n-data 相关的可能的质量问题，那么：
我会推荐ocpsoft/PrettyTime（实际上支持 32 种语言（很快 34 种？）只适合使用
java.util.Date- 请参阅@ataylor 的答案）。不幸的是，具有庞大社区背景的行业标准 CLDR（来自 Unicode 联盟）不是 i18n 数据的基础，因此数据的进一步增强或改进可能需要一段时间......如果您在 Android 上，那么帮助类android.text.format.DateUtils是一个苗条的内置替代品（请参阅此处的其他评论和答案，缺点是它不支持多年和几个月。我相信只有很少有人喜欢这个帮助类的 API 风格。
如果您是Joda-Time的粉丝，那么您可以查看它的类PeriodFormat（在 v2.9.4 版本中支持 14 种语言，另一方面：Joda-Time 肯定也不紧凑，所以我在这里提到它只是为了完整性）。这个库不是一个真正的答案，因为根本不支持相对时间。您至少需要附加文字“之前”（并从生成的列表格式中手动剥离所有较低的单位 - 尴尬）。与 Time4J 或 Android-DateUtils 不同，它没有特别支持缩写或从相对时间自动切换到绝对时间表示。与 PrettyTime 一样，它完全依赖于 Java 社区的私有成员对其 i18n 数据的未经证实的贡献。
于 2015-10-22T07:15:40.170 回答
7
基于这里的一堆答案，我为我的用例创建了以下内容。
示例用法：
String relativeDate = String.valueOf(
TimeUtils.getRelativeTime( 1000L * myTimeInMillis() ));
import java.util.Arrays;
import java.util.List;
import static java.util.concurrent.TimeUnit.DAYS;
import static java.util.concurrent.TimeUnit.HOURS;
import static java.util.concurrent.TimeUnit.MINUTES;
import static java.util.concurrent.TimeUnit.SECONDS;
/**
* Utilities for dealing with dates and times
*/
public class TimeUtils {
public static final List<Long> times = Arrays.asList(
DAYS.toMillis(365),
DAYS.toMillis(30),
DAYS.toMillis(7),
DAYS.toMillis(1),
HOURS.toMillis(1),
MINUTES.toMillis(1),
SECONDS.toMillis(1)
);
public static final List<String> timesString = Arrays.asList(
"yr", "mo", "wk", "day", "hr", "min", "sec"
);
/**
* Get relative time ago for date
*
* NOTE:
* if (duration > WEEK_IN_MILLIS) getRelativeTimeSpanString prints the date.
*
* ALT:
* return getRelativeTimeSpanString(date, now, SECOND_IN_MILLIS, FORMAT_ABBREV_RELATIVE);
*
* @param date String.valueOf(TimeUtils.getRelativeTime(1000L * Date/Time in Millis)
* @return relative time
*/
public static CharSequence getRelativeTime(final long date) {
return toDuration( Math.abs(System.currentTimeMillis() - date) );
}
private static String toDuration(long duration) {
StringBuilder sb = new StringBuilder();
for(int i=0;i< times.size(); i++) {
Long current = times.get(i);
long temp = duration / current;
if (temp > 0) {
sb.append(temp)
.append(" ")
.append(timesString.get(i))
.append(temp > 1 ? "s" : "")
.append(" ago");
break;
}
}
return sb.toString().isEmpty() ? "now" : sb.toString();
}
}
于 2016-05-05T04:16:44.043 回答
6
如果您正在寻找简单的“今天”、“昨天”或“x 天前”。
private String getDaysAgo(Date date){
long days = (new Date().getTime() - date.getTime()) / 86400000;
if(days == 0) return "Today";
else if(days == 1) return "Yesterday";
else return days + " days ago";
}
于 2013-05-28T16:56:19.780 回答
6
于 2019-06-04T20:08:12.847 回答
5
java.time
使用Java 8 及更高版本中内置的java.time框架。
LocalDateTime t1 = LocalDateTime.of(2015, 1, 1, 0, 0, 0);
LocalDateTime t2 = LocalDateTime.now();
Period period = Period.between(t1.toLocalDate(), t2.toLocalDate());
Duration duration = Duration.between(t1, t2);
System.out.println("First January 2015 is " + period.getYears() + " years ago");
System.out.println("First January 2015 is " + period.getMonths() + " months ago");
System.out.println("First January 2015 is " + period.getDays() + " days ago");
System.out.println("First January 2015 is " + duration.toHours() + " hours ago");
System.out.println("First January 2015 is " + duration.toMinutes() + " minutes ago");
于 2015-06-21T12:32:02.240 回答
4
我为jquery-timeago插件创建了一个简单的Java timeago端口，可以满足您的要求。
TimeAgo time = new TimeAgo();
String minutes = time.timeAgo(System.currentTimeMillis() - (15*60*1000)); // returns "15 minutes ago"
于 2011-04-01T16:47:07.750 回答
4
如果您正在为 Android 开发应用程序，它会为所有此类要求提供实用程序类DateUtils 。查看DateUtils#getRelativeTimeSpanString()实用程序方法。
从文档中
CharSequence getRelativeTimeSpanString（很长时间，现在很长时间，很长时间minResolution）
返回一个字符串，将 'time' 描述为相对于 'now' 的时间。过去的时间跨度格式为“42 分钟前”。未来时间跨度的格式类似于“42 分钟内”。
你会像现在一样度过你timestamp的时光。允许您指定要报告的最短时间跨度。System.currentTimeMillis()minResolution
例如，如果设置为 MINUTE_IN_MILLIS，则过去 3 秒的时间将报告为“0 分钟前”。通过 0、MINUTE_IN_MILLIS、HOUR_IN_MILLIS、DAY_IN_MILLIS、WEEK_IN_MILLIS 等之一。
于 2015-10-19T07:40:05.560 回答
4
您可以使用此功能计算时间前
private String timeAgo(long time_ago) {
long cur_time = (Calendar.getInstance().getTimeInMillis()) / 1000;
long time_elapsed = cur_time - time_ago;
long seconds = time_elapsed;
int minutes = Math.round(time_elapsed / 60);
int hours = Math.round(time_elapsed / 3600);
int days = Math.round(time_elapsed / 86400);
int weeks = Math.round(time_elapsed / 604800);
int months = Math.round(time_elapsed / 2600640);
int years = Math.round(time_elapsed / 31207680);
// Seconds
if (seconds <= 60) {
return "just now";
}
//Minutes
else if (minutes <= 60) {
if (minutes == 1) {
return "one minute ago";
} else {
return minutes + " minutes ago";
}
}
//Hours
else if (hours <= 24) {
if (hours == 1) {
return "an hour ago";
} else {
return hours + " hrs ago";
}
}
//Days
else if (days <= 7) {
if (days == 1) {
return "yesterday";
} else {
return days + " days ago";
}
}
//Weeks
else if (weeks <= 4.3) {
if (weeks == 1) {
return "a week ago";
} else {
return weeks + " weeks ago";
}
}
//Months
else if (months <= 12) {
if (months == 1) {
return "a month ago";
} else {
return months + " months ago";
}
}
//Years
else {
if (years == 1) {
return "one year ago";
} else {
return years + " years ago";
}
}
}
1) 这里 time_ago 的单位是微秒
于 2015-12-03T04:02:09.777 回答
3
（来自评论）
如果您在 Android 上需要它，只需使用以下命令：
public static String getTimeAgoFormat(long timestamp) {
return android.text.format.DateUtils.getRelativeTimeSpanString(timestamp).toString();
}
于 2021-09-27T14:12:20.207 回答
2
它不漂亮......但我能想到的最接近的是使用 Joda-Time（如这篇文章中所述：How to calculate elapsed time from now with Joda Time?
于 2010-10-04T21:34:46.233 回答
2
如果我们考虑性能，这是一个更好的代码。它减少了计算次数。 原因 分钟仅在秒数大于 60 时计算，小时仅在分钟数大于 60 时计算，依此类推...
class timeAgo {
static String getTimeAgo(long time_ago) {
time_ago=time_ago/1000;
long cur_time = (Calendar.getInstance().getTimeInMillis())/1000 ;
long time_elapsed = cur_time - time_ago;
long seconds = time_elapsed;
// Seconds
if (seconds <= 60) {
return "Just now";
}
//Minutes
else{
int minutes = Math.round(time_elapsed / 60);
if (minutes <= 60) {
if (minutes == 1) {
return "a minute ago";
} else {
return minutes + " minutes ago";
}
}
//Hours
else {
int hours = Math.round(time_elapsed / 3600);
if (hours <= 24) {
if (hours == 1) {
return "An hour ago";
} else {
return hours + " hrs ago";
}
}
//Days
else {
int days = Math.round(time_elapsed / 86400);
if (days <= 7) {
if (days == 1) {
return "Yesterday";
} else {
return days + " days ago";
}
}
//Weeks
else {
int weeks = Math.round(time_elapsed / 604800);
if (weeks <= 4.3) {
if (weeks == 1) {
return "A week ago";
} else {
return weeks + " weeks ago";
}
}
//Months
else {
int months = Math.round(time_elapsed / 2600640);
if (months <= 12) {
if (months == 1) {
return "A month ago";
} else {
return months + " months ago";
}
}
//Years
else {
int years = Math.round(time_elapsed / 31207680);
if (years == 1) {
return "One year ago";
} else {
return years + " years ago";
}
}
}
}
}
}
}
}
}
于 2018-01-27T10:38:51.140 回答
2
private const val SECOND_MILLIS = 1
private const val MINUTE_MILLIS = 60 * SECOND_MILLIS
private const val HOUR_MILLIS = 60 * MINUTE_MILLIS
private const val DAY_MILLIS = 24 * HOUR_MILLIS
object TimeAgo {
fun timeAgo(time: Int): String {
val now = TimeUnit.MILLISECONDS.toSeconds(System.currentTimeMillis())
if (time > now || time <= 0) {
return "in the future"
}
val diff = now - time
return when {
diff < MINUTE_MILLIS -> "Just now"
diff < 2 * MINUTE_MILLIS -> "a minute ago"
diff < 60 * MINUTE_MILLIS -> "${diff / MINUTE_MILLIS} minutes ago"
diff < 2 * HOUR_MILLIS -> "an hour ago"
diff < 24 * HOUR_MILLIS -> "${diff / HOUR_MILLIS} hours ago"
diff < 48 * HOUR_MILLIS -> "yesterday"
else -> "${diff / DAY_MILLIS} days ago"
}
}
}
称呼
val String = timeAgo(unixTimeStamp)
在 Kotlin 获得时间
于 2020-04-02T15:57:29.467 回答
1
经过长时间的研究，我发现了这一点。
public class GetTimeLapse {
public static String getlongtoago(long createdAt) {
DateFormat userDateFormat = new SimpleDateFormat("E MMM dd HH:mm:ss Z yyyy");
DateFormat dateFormatNeeded = new SimpleDateFormat("MM/dd/yyyy HH:MM:SS");
Date date = null;
date = new Date(createdAt);
String crdate1 = dateFormatNeeded.format(date);
// Date Calculation
DateFormat dateFormat = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");
crdate1 = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss").format(date);
// get current date time with Calendar()
Calendar cal = Calendar.getInstance();
String currenttime = dateFormat.format(cal.getTime());
Date CreatedAt = null;
Date current = null;
try {
CreatedAt = dateFormat.parse(crdate1);
current = dateFormat.parse(currenttime);
} catch (java.text.ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
// Get msec from each, and subtract.
long diff = current.getTime() - CreatedAt.getTime();
long diffSeconds = diff / 1000;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000) % 24;
long diffDays = diff / (24 * 60 * 60 * 1000);
String time = null;
if (diffDays > 0) {
if (diffDays == 1) {
time = diffDays + "day ago ";
} else {
time = diffDays + "days ago ";
}
} else {
if (diffHours > 0) {
if (diffHours == 1) {
time = diffHours + "hr ago";
} else {
time = diffHours + "hrs ago";
}
} else {
if (diffMinutes > 0) {
if (diffMinutes == 1) {
time = diffMinutes + "min ago";
} else {
time = diffMinutes + "mins ago";
}
} else {
if (diffSeconds > 0) {
time = diffSeconds + "secs ago";
}
}
}
}
return time;
}
}
于 2015-09-11T07:07:51.757 回答
1
对于 Android 就像 Ravi 所说的那样，但是由于很多人只想复制粘贴这里的东西。
try {
SimpleDateFormat formatter = new SimpleDateFormat("EEE, dd MMM yyyy HH:mm:ss Z");
Date dt = formatter.parse(date_from_server);
CharSequence output = DateUtils.getRelativeTimeSpanString (dt.getTime());
your_textview.setText(output.toString());
} catch (Exception ex) {
ex.printStackTrace();
your_textview.setText("");
}
有更多时间的人的解释
- 您从某个地方获取数据。首先，你必须弄清楚它的格式。
前任。我从服务器获取数据，格式为 Wed, 27 Jan 2016 09:32:35 GMT [这可能不是你的情况]
这被翻译成
SimpleDateFormat formatter = new SimpleDateFormat("EEE, dd MMM yyyy HH:mm:ss Z");
我怎么知道？在此处阅读文档。
然后在我解析它之后，我得到一个日期。我输入 getRelativeTimeSpanString 的那个日期（没有任何额外的参数对我来说很好，默认为分钟）
如果您没有弄清楚正确的解析 String ，您将得到一个异常，类似于字符 5 的异常。查看字符 5，并更正您的初始解析字符串。. 您可能会遇到另一个例外，重复这些步骤，直到您有正确的公式。
于 2016-01-27T20:00:56.717 回答
1
您可以使用 Java 的 Library RelativeDateTimeFormatter，它正是这样做的：
RelativeDateTimeFormatter fmt = RelativeDateTimeFormatter.getInstance();
fmt.format(1, Direction.NEXT, RelativeUnit.DAYS); // "in 1 day"
fmt.format(3, Direction.NEXT, RelativeUnit.DAYS); // "in 3 days"
fmt.format(3.2, Direction.LAST, RelativeUnit.YEARS); // "3.2 years ago"
fmt.format(Direction.LAST, AbsoluteUnit.SUNDAY); // "last Sunday"
fmt.format(Direction.THIS, AbsoluteUnit.SUNDAY); // "this Sunday"
fmt.format(Direction.NEXT, AbsoluteUnit.SUNDAY); // "next Sunday"
fmt.format(Direction.PLAIN, AbsoluteUnit.SUNDAY); // "Sunday"
fmt.format(Direction.LAST, AbsoluteUnit.DAY); // "yesterday"
fmt.format(Direction.THIS, AbsoluteUnit.DAY); // "today"
fmt.format(Direction.NEXT, AbsoluteUnit.DAY); // "tomorrow"
fmt.format(Direction.PLAIN, AbsoluteUnit.NOW); // "now"
于 2018-12-18T16:39:58.950 回答
1
SQL 时间戳到现在经过的时间。设置自己的时区。
注意1：这将处理单数/复数。
注意 2：这是使用 Joda 时间
String getElapsedTime(String strMysqlTimestamp) {
DateTimeFormatter formatter = DateTimeFormat.forPattern("YYYY-MM-dd HH:mm:ss.S");
DateTime mysqlDate = formatter.parseDateTime(strMysqlTimestamp).
withZone(DateTimeZone.forID("Asia/Kuala_Lumpur"));
DateTime now = new DateTime();
Period period = new Period(mysqlDate, now);
int seconds = period.getSeconds();
int minutes = period.getMinutes();
int hours = period.getHours();
int days = period.getDays();
int weeks = period.getWeeks();
int months = period.getMonths();
int years = period.getYears();
String elapsedTime = "";
if (years != 0)
if (years == 1)
elapsedTime = years + " year ago";
else
elapsedTime = years + " years ago";
else if (months != 0)
if (months == 1)
elapsedTime = months + " month ago";
else
elapsedTime = months + " months ago";
else if (weeks != 0)
if (weeks == 1)
elapsedTime = weeks + " week ago";
else
elapsedTime = weeks + " weeks ago";
else if (days != 0)
if (days == 1)
elapsedTime = days + " day ago";
else
elapsedTime = days + " days ago";
else if (hours != 0)
if (hours == 1)
elapsedTime = hours + " hour ago";
else
elapsedTime = hours + " hours ago";
else if (minutes != 0)
if (minutes == 1)
elapsedTime = minutes + " minute ago";
else
elapsedTime = minutes + " minutes ago";
else if (seconds != 0)
if (seconds == 1)
elapsedTime = seconds + " second ago";
else
elapsedTime = seconds + " seconds ago";
return elapsedTime;
}
于 2021-01-04T17:31:53.877 回答
1
java.time
您可以使用它们java.time.Duration，它们以ISO-8601 标准java.time.Period为模型，并作为JSR-310 实现的一部分与Java-8一起引入。Java-9引入了一些更方便的方法。
- 用于
Duration计算基于时间的数量或时间量。它可以使用基于持续时间的单位进行访问，例如纳秒、秒、分钟和小时。此外，可以使用DAYS单位并将其视为正好等于 24 小时，从而忽略夏令时的影响。 - 用于
Period计算基于日期的时间量。可以使用基于周期的单位（例如天、月和年）来访问它。
演示：
import java.time.Duration;
import java.time.LocalDateTime;
import java.time.Month;
public class Main {
public static void main(String[] args) {
// An arbitrary local date and time
LocalDateTime startDateTime = LocalDateTime.of(2020, Month.DECEMBER, 10, 15, 20, 25);
// Current local date and time
LocalDateTime endDateTime = LocalDateTime.now();
Duration duration = Duration.between(startDateTime, endDateTime);
// Default format
System.out.println(duration);
// Custom format
// ####################################Java-8####################################
String formattedElapsedTime = String.format("%d days, %d hours, %d minutes, %d seconds, %d nanoseconds ago",
duration.toDays(), duration.toHours() % 24, duration.toMinutes() % 60, duration.toSeconds() % 60,
duration.toNanos() % 1000000000);
System.out.println(formattedElapsedTime);
// ##############################################################################
// ####################################Java-9####################################
formattedElapsedTime = String.format("%d days, %d hours, %d minutes, %d seconds, %d nanoseconds ago",
duration.toDaysPart(), duration.toHoursPart(), duration.toMinutesPart(), duration.toSecondsPart(),
duration.toNanosPart());
System.out.println(formattedElapsedTime);
// ##############################################################################
}
}
输出：
PT1395H35M7.355288S
58 days, 3 hours, 35 minutes, 7 seconds, 355288000 nanoseconds ago
58 days, 3 hours, 35 minutes, 7 seconds, 355288000 nanoseconds ago
如果你有两个时刻UTC，你可以使用Instant而不是LocalDateTime例如
import java.time.Duration;
import java.time.Instant;
import java.time.temporal.ChronoUnit;
public class Main {
public static void main(String[] args) {
// Current moment at UTC
Instant now = Instant.now();
// An instant in the past
Instant startDateTime = now.minus(58, ChronoUnit.DAYS)
.minus(2, ChronoUnit.HOURS)
.minus(54, ChronoUnit.MINUTES)
.minus(24, ChronoUnit.SECONDS)
.minus(808624000, ChronoUnit.NANOS);
Duration duration = Duration.between(startDateTime, now);
// Default format
System.out.println(duration);
// Custom format
// ####################################Java-8####################################
String formattedElapsedTime = String.format("%d days, %d hours, %d minutes, %d seconds, %d nanoseconds ago",
duration.toDays(), duration.toHours() % 24, duration.toMinutes() % 60, duration.toSeconds() % 60,
duration.toNanos() % 1000000000);
System.out.println(formattedElapsedTime);
// ##############################################################################
// ####################################Java-9####################################
formattedElapsedTime = String.format("%d days, %d hours, %d minutes, %d seconds, %d nanoseconds ago",
duration.toDaysPart(), duration.toHoursPart(), duration.toMinutesPart(), duration.toSecondsPart(),
duration.toNanosPart());
System.out.println(formattedElapsedTime);
// ##############################################################################
}
}
输出：
PT1394H54M24.808624S
58 days, 2 hours, 54 minutes, 24 seconds, 808624000 nanoseconds ago
58 days, 2 hours, 54 minutes, 24 seconds, 808624000 nanoseconds ago
演示Period：
import java.time.LocalDate;
import java.time.Period;
import java.time.ZoneId;
public class Main {
public static void main(String[] args) {
// Replace ZoneId.systemDefault() with the applicable timezone ID e.g.
// ZoneId.of("Europe/London"). For LocalDate in the JVM's timezone, simply use
// LocalDate.now()
LocalDate endDate = LocalDate.now(ZoneId.systemDefault());
// Let's assume the start date is 1 year, 2 months, and 3 days ago
LocalDate startDate = endDate.minusYears(1).minusMonths(2).minusDays(3);
Period period = Period.between(startDate, endDate);
// Default format
System.out.println(period);
// Custom format
String formattedElapsedPeriod = String.format("%d years, %d months, %d days ago", period.getYears(),
period.getMonths(), period.getDays());
System.out.println(formattedElapsedPeriod);
}
}
输出：
P1Y2M3D
1 years, 2 months, 3 days ago
从Trail: Date Time了解现代日期时间 API 。
- 出于任何原因，如果您必须坚持使用 Java 6 或 Java 7，则可以使用ThreeTen-Backport，它将大部分java.time功能向后移植到 Java 6 和 7。
- 如果您正在为一个 Android 项目工作并且您的 Android API 级别仍然不符合 Java-8，请通过 desugaring和How to use ThreeTenABP in Android Project检查Java 8+ APIs available。
于 2021-02-06T19:00:22.747 回答
0
这是我的Java实现
public static String relativeDate(Date date){
Date now=new Date();
if(date.before(now)){
int days_passed=(int) TimeUnit.MILLISECONDS.toDays(now.getTime() - date.getTime());
if(days_passed>1)return days_passed+" days ago";
else{
int hours_passed=(int) TimeUnit.MILLISECONDS.toHours(now.getTime() - date.getTime());
if(hours_passed>1)return days_passed+" hours ago";
else{
int minutes_passed=(int) TimeUnit.MILLISECONDS.toMinutes(now.getTime() - date.getTime());
if(minutes_passed>1)return minutes_passed+" minutes ago";
else{
int seconds_passed=(int) TimeUnit.MILLISECONDS.toSeconds(now.getTime() - date.getTime());
return seconds_passed +" seconds ago";
}
}
}
}
else
{
return new SimpleDateFormat("HH:mm:ss MM/dd/yyyy").format(date).toString();
}
}
于 2015-12-15T08:05:31.503 回答
0
这个对我有用
public class TimeDifference {
int years;
int months;
int days;
int hours;
int minutes;
int seconds;
String differenceString;
public TimeDifference(@NonNull Date curdate, @NonNull Date olddate) {
float diff = curdate.getTime() - olddate.getTime();
if (diff >= 0) {
int yearDiff = Math.round((diff / (AppConstant.aLong * AppConstant.aFloat)) >= 1 ? (diff / (AppConstant.aLong * AppConstant.aFloat)) : 0);
if (yearDiff > 0) {
years = yearDiff;
setDifferenceString(years + (years == 1 ? " year" : " years") + " ago");
} else {
int monthDiff = Math.round((diff / AppConstant.aFloat) >= 1 ? (diff / AppConstant.aFloat) : 0);
if (monthDiff > 0) {
if (monthDiff > AppConstant.ELEVEN) {
monthDiff = AppConstant.ELEVEN;
}
months = monthDiff;
setDifferenceString(months + (months == 1 ? " month" : " months") + " ago");
} else {
int dayDiff = Math.round((diff / (AppConstant.bFloat)) >= 1 ? (diff / (AppConstant.bFloat)) : 0);
if (dayDiff > 0) {
days = dayDiff;
if (days == AppConstant.THIRTY) {
days = AppConstant.TWENTYNINE;
}
setDifferenceString(days + (days == 1 ? " day" : " days") + " ago");
} else {
int hourDiff = Math.round((diff / (AppConstant.cFloat)) >= 1 ? (diff / (AppConstant.cFloat)) : 0);
if (hourDiff > 0) {
hours = hourDiff;
setDifferenceString(hours + (hours == 1 ? " hour" : " hours") + " ago");
} else {
int minuteDiff = Math.round((diff / (AppConstant.dFloat)) >= 1 ? (diff / (AppConstant.dFloat)) : 0);
if (minuteDiff > 0) {
minutes = minuteDiff;
setDifferenceString(minutes + (minutes == 1 ? " minute" : " minutes") + " ago");
} else {
int secondDiff = Math.round((diff / (AppConstant.eFloat)) >= 1 ? (diff / (AppConstant.eFloat)) : 0);
if (secondDiff > 0) {
seconds = secondDiff;
} else {
seconds = 1;
}
setDifferenceString(seconds + (seconds == 1 ? " second" : " seconds") + " ago");
}
}
}
}
}
} else {
setDifferenceString("Just now");
}
}
public String getDifferenceString() {
return differenceString;
}
public void setDifferenceString(String differenceString) {
this.differenceString = differenceString;
}
public int getYears() {
return years;
}
public void setYears(int years) {
this.years = years;
}
public int getMonths() {
return months;
}
public void setMonths(int months) {
this.months = months;
}
public int getDays() {
return days;
}
public void setDays(int days) {
this.days = days;
}
public int getHours() {
return hours;
}
public void setHours(int hours) {
this.hours = hours;
}
public int getMinutes() {
return minutes;
}
public void setMinutes(int minutes) {
this.minutes = minutes;
}
public int getSeconds() {
return seconds;
}
public void setSeconds(int seconds) {
this.seconds = seconds;
} }
于 2017-04-20T12:40:40.047 回答
0
这是非常基本的脚本。它很容易即兴创作。
结果：（XXX 小时前）或（XX 天前/昨天/今天）
<span id='hourpost'></span>
,or
<span id='daypost'></span>
<script>
var postTime = new Date('2017/6/9 00:01');
var now = new Date();
var difference = now.getTime() - postTime.getTime();
var minutes = Math.round(difference/60000);
var hours = Math.round(minutes/60);
var days = Math.round(hours/24);
var result;
if (days < 1) {
result = "Today";
} else if (days < 2) {
result = "Yesterday";
} else {
result = days + " Days ago";
}
document.getElementById("hourpost").innerHTML = hours + "Hours Ago" ;
document.getElementById("daypost").innerHTML = result ;
</script>
于 2017-06-14T17:22:17.317 回答
0
为此，我Just Now, seconds ago, min ago, hrs ago, days ago, weeks ago, months ago, years ago在此示例中已完成此操作，您可以像这样解析日期2018-09-05T06:40:46.183Z或像下面这样的任何其他日期
在string.xml中添加以下值
<string name="lbl_justnow">Just Now</string>
<string name="lbl_seconds_ago">seconds ago</string>
<string name="lbl_min_ago">min ago</string>
<string name="lbl_mins_ago">mins ago</string>
<string name="lbl_hr_ago">hr ago</string>
<string name="lbl_hrs_ago">hrs ago</string>
<string name="lbl_day_ago">day ago</string>
<string name="lbl_days_ago">days ago</string>
<string name="lbl_lstweek_ago">last week</string>
<string name="lbl_week_ago">weeks ago</string>
<string name="lbl_onemonth_ago">1 month ago</string>
<string name="lbl_month_ago">months ago</string>
<string name="lbl_oneyear_ago" >last year</string>
<string name="lbl_year_ago" >years ago</string>
java代码试试下面
public String getFormatDate(String postTime1) {
Calendar cal=Calendar.getInstance();
Date now=cal.getTime();
String disTime="";
try {
Date postTime;
//2018-09-05T06:40:46.183Z
postTime = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS'Z'").parse(postTime1);
long diff=(now.getTime()-postTime.getTime()+18000)/1000;
//for months
Calendar calObj = Calendar.getInstance();
calObj.setTime(postTime);
int m=calObj.get(Calendar.MONTH);
calObj.setTime(now);
SimpleDateFormat monthFormatter = new SimpleDateFormat("MM"); // output month
int mNow = Integer.parseInt(monthFormatter.format(postTime));
diff = diff-19800;
if(diff<15) { //below 15 sec
disTime = getResources().getString(R.string.lbl_justnow);
} else if(diff<60) {
//below 1 min
disTime= diff+" "+getResources().getString(R.string.lbl_seconds_ago);
} else if(diff<3600) {//below 1 hr
// convert min
long temp=diff/60;
if(temp==1) {
disTime= temp + " " +getResources().getString(R.string.lbl_min_ago);
} else {
disTime = temp + " " +getResources().getString(R.string.lbl_mins_ago);
}
} else if(diff<(24*3600)) {// below 1 day
// convert hr
long temp= diff/3600;
System.out.println("hey temp3:"+temp);
if(temp==1) {
disTime = temp + " " +getResources().getString(R.string.lbl_hr_ago);
} else {
disTime = temp + " " +getResources().getString(R.string.lbl_hrs_ago);
}
} else if(diff<(24*3600*7)) {// below week
// convert days
long temp=diff/(3600*24);
if (temp==1) {
// disTime = "\nyesterday";
disTime = temp + " " +getResources().getString(R.string.lbl_day_ago);
} else {
disTime = temp + " " +getResources().getString(R.string.lbl_days_ago);
}
} else if(diff<((24*3600*28))) {// below month
// convert week
long temp=diff/(3600*24*7);
if (temp <= 4) {
if (temp < 1) {
disTime = getResources().getString(R.string.lbl_lstweek_ago);
}else{
disTime = temp + " " + getResources().getString(R.string.lbl_week_ago);
}
} else {
int diffMonth = mNow - m;
Log.e("count : ", String.valueOf(diffMonth));
disTime = diffMonth + " " + getResources().getString(R.string.lbl_month_ago);
}
}else if(diff<((24*3600*365))) {// below year
// convert month
long temp=diff/(3600*24*30);
System.out.println("hey temp2:"+temp);
if (temp <= 12) {
if (temp == 1) {
disTime = getResources().getString(R.string.lbl_onemonth_ago);
}else{
disTime = temp + " " + getResources().getString(R.string.lbl_month_ago);
}
}
}else if(diff>((24*3600*365))) { // above year
// convert year
long temp=diff/(3600*24*30*12);
System.out.println("hey temp8:"+temp);
if (temp == 1) {
disTime = getResources().getString(R.string.lbl_oneyear_ago);
}else{
disTime = temp + " " + getResources().getString(R.string.lbl_year_ago);
}
}
} catch(Exception e) {
e.printStackTrace();
}
return disTime;
}
于 2018-10-28T06:47:49.967 回答
0
我正在使用 Instant、Date 和 DateTimeUtils。数据（日期）以String类型存储在数据库中，然后转换为Instant。
/*
This method is to display ago.
Example: 3 minutes ago.
I already implement the latest which is including the Instant.
Convert from String to Instant and then parse to Date.
*/
public String convertTimeToAgo(String dataDate) {
//Initialize
String conversionTime = null;
String suffix = "Yang Lalu";
Date pastTime;
//Parse from String (which is stored as Instant.now().toString()
//And then convert to become Date
Instant instant = Instant.parse(dataDate);
pastTime = DateTimeUtils.toDate(instant);
//Today date
Date nowTime = new Date();
long dateDiff = nowTime.getTime() - pastTime.getTime();
long second = TimeUnit.MILLISECONDS.toSeconds(dateDiff);
long minute = TimeUnit.MILLISECONDS.toMinutes(dateDiff);
long hour = TimeUnit.MILLISECONDS.toHours(dateDiff);
long day = TimeUnit.MILLISECONDS.toDays(dateDiff);
if (second < 60) {
conversionTime = second + " Saat " + suffix;
} else if (minute < 60) {
conversionTime = minute + " Minit " + suffix;
} else if (hour < 24) {
conversionTime = hour + " Jam " + suffix;
} else if (day >= 7) {
if (day > 30) {
conversionTime = (day / 30) + " Bulan " + suffix;
} else if (day > 360) {
conversionTime = (day / 360) + " Tahun " + suffix;
} else {
conversionTime = (day / 7) + " Minggu " + suffix;
}
} else if (day < 7) {
conversionTime = day + " Hari " + suffix;
}
return conversionTime;
}
于 2019-06-03T14:14:07.260 回答
0
以下解决方案都是纯Java：
选项 1：没有舍入，只有最大的时间容器
下面的函数只会显示最大的时间容器，例如如果真正的经过时间是"1 month 14 days ago"，这个函数只会显示"1 month ago"。此函数也将始终向下舍入，因此等效于的时间"50 days ago"将显示为"1 month"
public String formatTimeAgo(long millis) {
String[] ids = new String[]{"second","minute","hour","day","month","year"};
long seconds = millis / 1000;
long minutes = seconds / 60;
long hours = minutes / 60;
long days = hours / 24;
long months = days / 30;
long years = months / 12;
ArrayList<Long> times = new ArrayList<>(Arrays.asList(years, months, days, hours, minutes, seconds));
for(int i = 0; i < times.size(); i++) {
if(times.get(i) != 0) {
long value = times.get(i).intValue();
return value + " " + ids[ids.length - 1 - i] + (value == 1 ? "" : "s") + " ago";
}
}
return "0 seconds ago";
}
选项 2：四舍五入
只需用 Math.round(...) 语句包装您想要四舍五入的时间容器，因此如果您想要四舍五入50 days，2 months修改long months = days / 30为long months = Math.round(days / 30.0)
于 2019-09-13T22:55:46.793 回答
0
这是我的测试用例，希望对您有所帮助：
val currentCalendar = Calendar.getInstance()
currentCalendar.set(2019, 6, 2, 5, 31, 0)
val targetCalendar = Calendar.getInstance()
targetCalendar.set(2019, 6, 2, 5, 30, 0)
val diffTs = currentCalendar.timeInMillis - targetCalendar.timeInMillis
val diffMins = TimeUnit.MILLISECONDS.toMinutes(diffTs)
val diffHours = TimeUnit.MILLISECONDS.toHours(diffTs)
val diffDays = TimeUnit.MILLISECONDS.toDays(diffTs)
val diffWeeks = TimeUnit.MILLISECONDS.toDays(diffTs) / 7
val diffMonths = TimeUnit.MILLISECONDS.toDays(diffTs) / 30
val diffYears = TimeUnit.MILLISECONDS.toDays(diffTs) / 365
val newTs = when {
diffYears >= 1 -> "Years $diffYears"
diffMonths >= 1 -> "Months $diffMonths"
diffWeeks >= 1 -> "Weeks $diffWeeks"
diffDays >= 1 -> "Days $diffDays"
diffHours >= 1 -> "Hours $diffHours"
diffMins >= 1 -> "Mins $diffMins"
else -> "now"
}
于 2019-10-14T20:50:36.347 回答
0
getrelativeDateTime函数将为您提供在Whatsapp 通知中看到的日期时间。
要获得未来的相对日期时间，请为其添加条件。这是专门为获取日期时间而创建的，例如 Whatsapp 通知。
private static String getRelativeDateTime(long date) {
SimpleDateFormat DateFormat = new SimpleDateFormat("MMM dd, yyyy", Locale.getDefault());
SimpleDateFormat TimeFormat = new SimpleDateFormat(" hh:mm a", Locale.getDefault());
long now = Calendar.getInstance().getTimeInMillis();
long startOfDay = StartOfDay(Calendar.getInstance().getTime());
String Day = "";
String Time = "";
long millSecInADay = 86400000;
long oneHour = millSecInADay / 24;
long differenceFromNow = now - date;
if (date > startOfDay) {
if (differenceFromNow < (oneHour)) {
int minute = (int) (differenceFromNow / (60000));
if (minute == 0) {
int sec = (int) differenceFromNow / 1000;
if (sec == 0) {
Time = "Just Now";
} else if (sec == 1) {
Time = sec + " second ago";
} else {
Time = sec + " seconds ago";
}
} else if (minute == 1) {
Time = minute + " minute ago";
} else if (minute < 60) {
Time = minute + " minutes ago";
}
} else {
Day = "Today, ";
}
} else if (date > (startOfDay - millSecInADay)) {
Day = "Yesterday, ";
} else if (date > (startOfDay - millSecInADay * 7)) {
int days = (int) (differenceFromNow / millSecInADay);
Day = days + " Days ago, ";
} else {
Day = DateFormat.format(date);
}
if (Time.isEmpty()) {
Time = TimeFormat.format(date);
}
return Day + Time;
}
public static long StartOfDay(Date date) {
Calendar calendar = Calendar.getInstance();
calendar.setTime(date);
calendar.set(Calendar.HOUR_OF_DAY, 0);
calendar.set(Calendar.MINUTE, 0);
calendar.set(Calendar.SECOND, 0);
calendar.set(Calendar.MILLISECOND, 0);
return calendar.getTimeInMillis();
}
于 2020-05-30T04:51:52.333 回答
0
由于缺乏简单性和更新的响应，请遵循更新的 Java 8 及更高版本
import java.time.*;
import java.time.temporal.*;
public class Time {
public static void main(String[] args) {
System.out.println(LocalTime.now().minus(8, ChronoUnit.MINUTES));
System.out.println(LocalTime.now().minus(8, ChronoUnit.HOURS));
System.out.println(LocalDateTime.now().minus(8, ChronoUnit.DAYS));
System.out.println(LocalDateTime.now().minus(8, ChronoUnit.MONTHS));
}
}
这是使用 Java Time API 的版本，它试图解决过去处理日期和时间的问题。
Javadoc
版本 8 https://docs.oracle.com/javase/8/docs/api/index.html?java/time/package-summary.html
版本 11 https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/time/package-summary.html
W3Schools 教程 - https://www.w3schools.com/java/java_date.asp
于 2020-07-29T21:47:03.907 回答