Reputation: 53
I want to parse an String to Date but the date obtained is incorrect. my code is like :
SimpleDateFormat df = new SimpleDateFormat("dd-MMM-yy hh.mm.ss.S a");
date1 = df.parse("17-DEC-19 05.40.39.364000000 PM");
but date1 is: Sat Dec 21 22:47:19 IRST 2019
I need to date: * Dec 17 17:40:39 IRST 2019
Upvotes: 3
Views: 87
Reputation: 79075
The SimpleDateFormat
does not have precision beyond milliseconds (.SSS
).
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
import java.util.Locale;
public class Main {
public static void main(String[] args) throws ParseException {
SimpleDateFormat df = new SimpleDateFormat("dd-MMM-yy hh.mm.ss.SSS a", Locale.ENGLISH);
Date date1 = df.parse("17-DEC-19 05.40.39.364 PM");
System.out.println(date1);
}
}
Output:
Tue Dec 17 17:40:39 GMT 2019
Note that the java.util
date-time API and their formatting API, SimpleDateFormat
are outdated and error-prone. It is recommended to stop using them completely and switch to the modern date-time API* .
Using modern date-time API:
import java.time.LocalDateTime;
import java.time.format.DateTimeFormatter;
import java.time.format.DateTimeFormatterBuilder;
import java.util.Locale;
public class Main {
public static void main(String[] args) {
DateTimeFormatter df = new DateTimeFormatterBuilder()
.parseCaseInsensitive() // For case-insensitive (e.g. AM/am) parsing
.appendPattern("dd-MMM-yy hh.mm.ss.n a")
.toFormatter(Locale.ENGLISH);
LocalDateTime ldt = LocalDateTime.parse("17-DEC-19 05.40.39.364000000 PM", df);
System.out.println(ldt);
}
}
Output:
2019-12-17T17:40:39.364
Learn more about the modern date-time API from Trail: Date Time.
* For any reason, if you have to stick to Java 6 or Java 7, you can use ThreeTen-Backport which backports most of the java.time functionality to Java 6 & 7. If you are working for an Android project and your Android API level is still not compliant with Java-8, check Java 8+ APIs available through desugaring and How to use ThreeTenABP in Android Project.
Upvotes: 2