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java 两个时间相减
先转成Date 再获取毫秒数来减
java.text.SimpleDateFormat sdf = new SimpleDateFormat(“yyyy-MM-dd HH:mm:ss”);
java.util.Calendar c = Calendar.getInstance();
String date1 = “2014-10-10 00:00:00”;
String date2 = “2015-10-10 00:00:00”;
c.setTime(sdf.parse(date1));
int y1 = c.get(Calendar.YEAR);
int m1 = c.get(Calendar.MONTH);
c.setTime(sdf.parse(date2));
int y2 = c.get(Calendar.YEAR);
int m2 = c.get(Calendar.MONTH);
int y = Math.abs(y2 – y1);
int m = y * 12 + Math.abs(m1-m2);
long d1 = sdf.parse(date1).getTime();
long d2 = sdf.parse(date2).getTime();
int d = (int) (Math.abs(d2-d1) / (1000 * 60 * 60 * 24));
System.out.println(“相差年: ” + y);
System.out.println(“相差月: ” + m);
System.out.println(“相差天: ” + d);
JAVA时间相减
用下面这种方法吧：
Calendar date = Calendar.getInstance();//today
//int k = 5;//add 5 days
int k = -5;//sub 5 days
date.add( Calendar.DATE, k );
String format = “yyyy-MM-dd HH:mm:ss”;
SimpleDateFormat simpleDate = new SimpleDateFormat( format );
String newDate = simpleDate.format( date.getTime() );
System.out.println(“newDate is “+newDate);
java实现两个时间相减得到年月
大概判断了一下，比较粗糙，仅供参考。
private static void function17() throws Exception {
Scanner sc=new Scanner(System.in);
SimpleDateFormat sdf=new SimpleDateFormat(“yyyy-MM-dd”);
System.out.println(“请输入起始时间”);
Date date1=sdf.parse(sc.next());
System.out.println(“请输入截止时间”);
Date date2=sdf.parse(sc.next());
long time1=date1.getTime();
long time2=date2.getTime();
long time=(time2-time1)/1000;
long year=time/(24*3600*365);
long month=time%(24*3600*365)/(24*3600*30);
if(year0){
System.out.println(year+”年”+month+”月”);
}else{
System.out.println(month+”月”);
}
}
java 中 日期如何相加减？？？
这个东西很简单。
现在是2004-03-26 13：31：40
过去是：2004-01-02 11：30：24
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：
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”);
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