Question

java计算时间差方法？

Answer 1

比如乱闹：现在是2004-03-26 13：31：裤陪信40

过去是：2004-01-02 11：30：24

我现在要获得两个日胡轮期差，差的形式为：XX天XX小时XX分XX秒

方法一：

DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

try{ Date d1 = df.parse("2004-03-26 13:31:40");

Date d2 = df.parse("2004-01-02 11:30:24");

long diff = d1.getTime() - d2.getTime();

long days = diff / (1000 * 60 * 60 * 24)}

catch (Exception e){}

方法二：

SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

java.util.Date now = df.parse("2004-03-26 13:31:40");

java.util.Date date=df.parse("2004-01-02 11:30:24");

long l=now.getTime()-date.getTime();

long day=l/(24*60*60*1000);

long hour=(l/(60*60*1000)-day*24);

long min=((l/(60*1000))-day*24*60-hour*60);

long s=(l/1000-day*24*60*60-hour*60*60-min*60);

System.out.println(""+day+"天"+hour+"小时"+min+"分"+s+"秒");

方法三：

SimpleDateFormat dfs = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

java.util.Date begin=dfs.parse("2004-01-02 11:30:24");

java.util.Date end = dfs.parse("2004-03-26 13:31:40");

long between=(end.getTime()-begin.getTime())/1000;//除以1000是为了转换成秒

long day1=between/(24*3600);

long hour1=between%(24*3600)/3600;

long minute1=between%3600/60;

long second1=between%60/60;

System.out.println(""+day1+"天"+hour1+"小时"+minute1+"分"+second1+"秒");

====================================================

java 比较时间大小

String s1="2008-01-25 09:12:09";

String s2="2008-01-29 09:12:11";

java.text.DateFormat df=new java.text.SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

java.util.Calendar c1=java.util.Calendar.getInstance();

java.util.Calendar c2=java.util.Calendar.getInstance();

try{c1.setTime(df.parse(s1));c2.setTime(df.parse(s2));}catch(java.text.ParseException e){

System.err.println("格式不正确");}

int result=c1.compareTo(c2);

if(result==0)

System.out.println("c1相等c2");

else if(result<0)

System.out.println("c1小于c2");

else

Answer 2

要获得两个日期差，差的形式为：XX天XX小时XX分XX秒

方法一：

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DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");try{  Date d1 = df.parse("2004-03-26 13:31:40");  Date d2 = df.parse("2004-01-02 11:30:24"); 巧漏 long diff = d1.getTime() - d2.getTime();  long days = diff / (1000 * 60 * 60 * 24);}catch (Exception e){}

方法二：

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SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");java.util.Date now = df.parse("2004-03-26 13:31:40");java.util.Date date=df.parse("2004-01-02 11:30:24");long l=now.getTime()-date.getTime();long day=l/(24*60*60*1000);long hour=(l/(60*60*1000)-day*24);long min=((l/(60*1000))-day*24*60-hour*60);long s=(l/1000-day*24*60*60-hour*60*60-min*60);System.out.println(""+day+"天"+hour+"小时"+min+"族察分"+s+"秒");

方孝穗烂法三：

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SimpleDateFormat dfs = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");java.util.Date begin=dfs.parse("2004-01-02 11:30:24");java.util.Date end = dfs.parse("2004-03-26 13:31:40");long between=(end.getTime()-begin.getTime())/1000;//除以1000是为了转换成秒long day1=between/(24*3600);long hour1=between%(24*3600)/3600;long minute1=between%3600/60;long second1=between%60/60;System.out.println(""+day1+"天"+hour1+"小时"+minute1+"分"+second1+"秒");

希望本文所述对大家的java程序设计有所帮助。