java calculate the time difference
Core Tip: The following procedure describes the time in JAVA in comparison, to obtain the time difference. For example, if you want to implement a function, can be submitted within 30 minutes, not more than 30 minutes after the submission. Then: calculate the time difference ** / SimpleDateFormat sdf = new SimpleDateFormat ('yyyy-MM-dd HH: mm: ss'); S. ..
The following procedure describes the time in JAVA in comparison, to obtain the time difference. For example, if you want to implement a function, can be submitted within 30 minutes, not more than 30 minutes after the submission. Then:
Calculate the time difference ** /
SimpleDateFormat sdf = new SimpleDateFormat ("yyyy-MM-dd HH: mm: ss");
String systemTime = sdf. Format (new Date ()). ToString ();
To the time of the interception time format string into a string ** /
Date begin = sdf.parse (2008-03-28 11:55:30);
Date end = sdf.parse (systemTime);
long between = (end.getTime ()-begin.getTime ()) / 1000; / / divide by 1000 to convert seconds
long day = between / (24 * 3600);
long hour = between% (24 * 3600) / 3600;
long minute = between% 3600/60;
long second = between% 60/60;
if ((hour == 0) & & (day == 0) & & (minute <= 30)) {/****/}
For example: It is now 2004-03-26 13:31:40
Past :2004-01-02 11:30:24
I now get two dates is poor, poor form: XX XX days XX hours XX minutes seconds
Method One:
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)
{
}
Method Two:
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 + "days" + hour + "hours" + min + "minutes" + s + "seconds");
Method three:
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; / / divide by 1000 to convert seconds
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 + "days" + hour1 + "hours" + minute1 + "sub" + second1 + "seconds");
I am here out of a mistake, not the number of days to double check that the format of time
Here dateA, dateB format with the same yyyy-MM-dd! Remember!
/ **
* Based on the two dates, the number of days apart to obtain
* Method Name:
* @ Param dateA
* @ Param dateB
* @ Return
* /
public static int getBetweenDayNumber (String dateA, String dateB) {
long dayNumber = 0;
long DAY = 24L * 60L * 60L * 1000L;
SimpleDateFormat df = new SimpleDateFormat ("yyyy-MM-dd");
try {
java.util.Date d1 = df.parse (dateA);
java.util.Date d2 = df.parse (dateB);
dayNumber = (d2.getTime () - d1.getTime ()) / DAY;
} Catch (Exception e) {
e.printStackTrace ();
}
return (int) dayNumber;
}
