时差计划

我正在使用以下函数来计算时差。 它没有显示正确的输出。 1个月的时差后，它显示出2分钟的差异。

我的计划有什么问题？

public String TimestampDiff(Timestamp t) { long t1 = t.getTime(); String st = null; long diff; java.util.Date date = new java.util.Date(); long currT = date.getTime(); System.out.println(); System.out.println(" current timesstamp is " + currT); diff = (currT - t1) / 60; int years = (int) Math.floor(diff / (1000 * 60 * 60 * 24 * 365)); double remainder = Math.floor(diff % (1000 * 60 * 60 * 24 * 365)); int days = (int) Math.floor(remainder / (1000 * 60 * 60 * 24)); remainder = Math.floor(remainder % (1000 * 60 * 60 * 24)); int hours = (int) Math.floor(remainder / (1000 * 60 * 60)); remainder = Math.floor(remainder % (1000 * 60 * 60)); int minutes = (int) Math.floor(remainder / (1000 * 60)); remainder = Math.floor(remainder % (1000 * 60)); int seconds = (int) Math.floor(remainder / (1000)); System.out.println("\nyr:Ds:hh:mm:ss " + years + ":" + days + ":" + hours + ":" + minutes + ":" + seconds); if (years == 0 && days == 0 && hours == 0 && minutes == 0) { st = "few seconds ago"; } else if (years == 0 && days == 0 && hours == 0) { st = minutes + " minuts ago"; } else if (years == 0 && days == 0) { st = hours + " hours ago"; } else if (years == 0 && days == 1) { st = new SimpleDateFormat("'yesterday at' hh:mm a").format(t1); } else if (years == 0 && days > 1) { st = new SimpleDateFormat(" MMM d 'at' hh:mm a").format(t1); } else if (years > 0) { st = new SimpleDateFormat("MMM d ''yy 'at' hh:mm a").format(t1); } st = st.replace("AM", "am").replace("PM", "pm"); return st; } 

Answer 1

 import org.apache.commons.lang.time.DateUtils; import java.text.SimpleDateFormat; @Test public void testDate() throws Exception { long t1 = new SimpleDateFormat("dd.MM.yyyy").parse("20.03.2013").getTime(); long now = System.currentTimeMillis(); String result = null; long diff = Math.abs(t1-now); if(diff < DateUtils.MILLIS_PER_MINUTE){ result = "few seconds ago"; }else if(diff < DateUtils.MILLIS_PER_HOUR){ result = (int)(diff/DateUtils.MILLIS_PER_MINUTE) + " minuts ago"; }else if(diff < DateUtils.MILLIS_PER_DAY){ result = (int)(diff/DateUtils.MILLIS_PER_HOUR) + " hours ago"; }else if(diff < DateUtils.MILLIS_PER_DAY * 2){ result = new SimpleDateFormat("'yesterday at' hh:mm a").format(t1); }else if(diff < DateUtils.MILLIS_PER_DAY * 365){ result = new SimpleDateFormat(" MMM d 'at' hh:mm a").format(t1); } else{ result = new SimpleDateFormat("MMM d ''yy 'at' hh:mm a").format(t1); } result = result.replace("AM", "am").replace("PM", "pm"); System.out.println(result); } 

Answer 2

我建议看看Joda Time ，注意到：

Joda-Time是Java SE 8之前Java事实上的标准日期和时间库。现在要求用户迁移到java.time（JSR-310）。

安装

• 对于基于Debian的系统： libjoda-time-java 。 jar将在/usr/share/java作为joda-time.jar
• 对于其他人： 下载最新的jar，例如joda-time-2.2-dist.zip，其中包括joda-time-2.2.jar

使用Eclipse时，将其添加到Java Build路径（Project> Properties> Java Build Path> Add External Jar）

相关的JavaDoc

• 约会时间
• 期
• PeriodFormatter
• PeriodFormatterBuilder

示例代码

 import java.sql.Timestamp; import java.util.Date; import org.joda.time.DateTime; import org.joda.time.Period; import org.joda.time.format.PeriodFormatter; import org.joda.time.format.PeriodFormatterBuilder; public class MinimalWorkingExample { static Date date = new Date(1990, 4, 28, 12, 59); public static String getTimestampDiff(Timestamp t) { final DateTime start = new DateTime(date.getTime()); final DateTime end = new DateTime(t); Period p = new Period(start, end); PeriodFormatter formatter = new PeriodFormatterBuilder() .printZeroAlways().minimumPrintedDigits(2).appendYears() .appendSuffix(" year", " years").appendSeparator(", ") .appendMonths().appendSuffix(" month", " months") .appendSeparator(", ").appendDays() .appendSuffix(" day", " days").appendSeparator(" and ") .appendHours().appendLiteral(":").appendMinutes() .appendLiteral(":").appendSeconds().toFormatter(); return p.toString(formatter); } public static void main(String[] args) { String diff = getTimestampDiff(new Timestamp(2013, 3, 20, 7, 51, 0, 0)); System.out.println(diff); } } 

输出：

 22 years, 10 months, 01 day and 18:52:00 

为什么我推荐新的解决方案

• 它更短（与1665个字符/ 41行相比，726个字符/ 14行）
• 它更容易理解
• 它更容易调整
• 代码和表示的分离更加清晰
• 我不想修复你的代码

Answer 3

不要long为int ，因为它将失去其精度。 将所有int更改为long ，并查看差异。

希望这个帮助