menembus batas impian


isi nya belum dicoba. sekedar catatan. sumber dr mana malah terlanjur hilang (maaf buat penulis asli ya)

The following procedures are introduced for time in JAVA in comparison, was the time difference. For example, if you want to implement a function, can be submitted within 30 minutes, not more than 30 minutes after submission. Then: calculate the time difference ** / SimpleDateFormat sdf = new SimpleDateFormat (‘yyyy-MM-dd HH: mm: ss’); S. ..

The following procedures are introduced for time in JAVA in comparison, was the time difference. For example, if you want to implement a function, can be submitted within 30 minutes, not more than 30 minutes after submission. Then:

Calculate the time difference ** /
SimpleDateFormat sdf = new SimpleDateFormat (“yyyy-MM-dd HH: mm: ss”);
String systemTime = sdf. Format (new Date ()). ToString ();
The interception to the time string time format 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; / / divided 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: Now is 2004-03-26 13:31:40
Past :2004-01-02 11:30:24
I now have two dates is poor, poor form: XX day of XX hours XX min XX sec
Method 1:
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 2:
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 3:
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; / / divided 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 days, careful examination revealed the format of time
Here dateA, dateB format with yyyy-MM-dd line! Remember!
/ **
* Based on two dates, to obtain the number of days apart
* 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;
)

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s

%d blogger menyukai ini: