日期在java中将时间戳转换为特定格式（年、月、周、天、时间、小时、分钟和秒）

Question

这是我的密码

例如，2年、2个月、1周、2天、1小时、2分钟、35秒

    String stdate ="01/01/2014 09:30:30";
    String endate ="09/11/2015 11:30:30";
    SimpleDateFormat df = new SimpleDateFormat("dd/MM/yyyy hh:mm:ss");
    Date d1 = new Date();
    Date d2 = new Date();
    long year =(1000*60*60*24*365l);
    long month =(1000*60*60*24*30l);
    long weeks =(1000*60*60*24*7l);
    long days =(1000*60*60*24l);

    try{

        d1 = df.parse(stdate);
        d2 = df.parse(endate);
        long diff = d2.getTime()-d1.getTime();
        long diffYear = diff/(1000*60*60*24*365l);
        long diffMonth = (diff-(diffYear*year))/month;
        long diffWeeks = ((diff%month))/weeks;
        long diffDays = ((diff%weeks))/days;

        System.out.println(diffYear+" years  ");
        System.out.println(diffMonth+" months ");
        System.out.println(diffWeeks+" week ");
        System.out.println(diffDays+" days "); // wrong output,
    }catch(Exception e){

    }

o/p:

1年
10个月
2周
5天

我不想利用乔达的时间。它应该在java中。util.*

请回答，提前谢谢

Answer 1

你可以尝试从Joda-time使用Period

String stdate = "01/01/2014 09:30:30";
String endate = "09/11/2015 11:30:30";
SimpleDateFormat df = new SimpleDateFormat("dd/MM/yyyy hh:mm:ss");
Date d1 = df.parse(stdate);
Date d2 = df.parse(endate);;

DateTime startTime = new DateTime(d1), endTime = new DateTime(d2);
Period p = new Period(startTime, endTime);
System.out.printf("%-8s %d %n","years:",p.getYears());
System.out.printf("%-8s %d %n","months:",p.getMonths());
System.out.printf("%-8s %d %n","weeks:",p.getWeeks());
System.out.printf("%-8s %d %n","days:",p.getDays());
System.out.printf("%-8s %d %n","hours:",p.getHours());
System.out.printf("%-8s %d %n","minutes:",p.getMinutes());
System.out.printf("%-8s %d %n","second:",p.getSeconds());

输出：

years:   1 
months:  10 
weeks:   1 
days:    1 
hours:   2 
minutes: 0 
second:  0 

---

更新：

回答你最初的问题：就像你从diff中减去已经属于year的天数来计算月份一样，你需要减去

• 在year和month上用于计算weeks的天数之和
• 如果要计算days，则已属于年、月、周的天数总和

所以你的代码看起来像

long diffYear = diff / year;
long diffMonth = (diff - (diffYear * year)) / month;
//long diffWeeks= ((diff % month)) / weeks;
long diffWeeks = (diff - (diffYear * year + diffMonth * month)) / weeks;
//long diffDays = ((diff % weeks)) / days;
long diffDays = (diff - (diffYear * year + diffMonth * month + diffWeeks*weeks)) / days;//((diff % weeks)) / days;

警告：这种计算天数或周数的方法将每个月视为30天，这并不总是正确的，因为也可能有28、29、30、31天的月份。这就是为什么你会看到5而不是1天

警告2：不要使用1000*60*60*24*365l，对于较大的数字，它可能会导致整数溢出（使用的第一个数字是整数），而应该使用1000L*60*60*24*365。所以

• 将l更改为L，因为l看起来像1，所以可能会让人困惑
• 开始与long相乘

为了让事情变得更简单，你甚至可以把它写成

long seconds = 1000L;
long minutes = seconds * 60;
long hours = minutes * 60;
long days = hours * 24;
long weeks = days * 7;
long month = days * 30;
long year = days * 365;