Convert a date format in epoch

JavaDatetimeDateEpoch

Java Problem Overview


I have a string with a date format such as

Jun 13 2003 23:11:52.454 UTC

containing millisec... which I want to convert in epoch. Is there an utility in Java I can use to do this conversion?

Java Solutions


Solution 1 - Java

This code shows how to use a java.text.SimpleDateFormat to parse a java.util.Date from a String:

String str = "Jun 13 2003 23:11:52.454 UTC";
SimpleDateFormat df = new SimpleDateFormat("MMM dd yyyy HH:mm:ss.SSS zzz");
Date date = df.parse(str);
long epoch = date.getTime();
System.out.println(epoch); // 1055545912454

Date.getTime() returns the epoch time in milliseconds.

Solution 2 - Java

You can also use the Java 8 API

import java.time.ZonedDateTime;
import java.time.format.DateTimeFormatter;

public class StackoverflowTest{
    public static void main(String args[]){
        String strDate = "Jun 13 2003 23:11:52.454 UTC";
        DateTimeFormatter dtf  = DateTimeFormatter.ofPattern("MMM dd yyyy HH:mm:ss.SSS zzz");
        ZonedDateTime     zdt  = ZonedDateTime.parse(strDate,dtf);        
        System.out.println(zdt.toInstant().toEpochMilli());  // 1055545912454  
    }
}

The DateTimeFormatter class replaces the old SimpleDateFormat. You can then create a ZonedDateTime from which you can extract the desired epoch time.

The main advantage is that you are now thread safe.

Thanks to Basil Bourque for his remarks and suggestions. Read his answer for full details.

Solution 3 - Java

tl;dr

ZonedDateTime.parse( 
                        "Jun 13 2003 23:11:52.454 UTC" , 
                        DateTimeFormatter.ofPattern ( "MMM d uuuu HH:mm:ss.SSS z" ) 
                    )
              .toInstant()
              .toEpochMilli()

>1055545912454

java.time

This Answer expands on the Answer by Lockni.

DateTimeFormatter

First define a formatting pattern to match your input string by creating a DateTimeFormatter object.

String input = "Jun 13 2003 23:11:52.454 UTC";
DateTimeFormatter f = DateTimeFormatter.ofPattern ( "MMM d uuuu HH:mm:ss.SSS z" );

ZonedDateTime

Parse the string as a ZonedDateTime. You can think of that class as: ( Instant + ZoneId ).

ZonedDateTime zdt = ZonedDateTime.parse ( "Jun 13 2003 23:11:52.454 UTC" , f );

>zdt.toString(): 2003-06-13T23:11:52.454Z[UTC]

https://i.stack.imgur.com/MZe55.png" alt="Table of types of date-time classes in modern java.time versus legacy." />

Count-from-epoch

I do not recommend tracking date-time values as a count-from-epoch. Doing so makes debugging tricky as humans cannot discern a meaningful date-time from a number so invalid/unexpected values may slip by. Also such counts are ambiguous, in granularity (whole seconds, milli, micro, nano, etc.) and in epoch (at least two dozen in by various computer systems).

But if you insist you can get a count of milliseconds from the epoch of first moment of 1970 in UTC (1970-01-01T00:00:00) through the Instant class. Be aware this means data-loss as you are truncating any nanoseconds to milliseconds.

Instant instant = zdt.toInstant ();

>instant.toString(): 2003-06-13T23:11:52.454Z

long millisSinceEpoch = instant.toEpochMilli() ; 

>1055545912454


About java.time

The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.

To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.

The Joda-Time project, now in maintenance mode, advises migration to the java.time classes.

You may exchange java.time objects directly with your database. Use a JDBC driver compliant with JDBC 4.2 or later. No need for strings, no need for java.sql.* classes. Hibernate 5 & JPA 2.2 support java.time.

Where to obtain the java.time classes?

Solution 4 - Java

Create Common Method to Convert String to Date format

public static void main(String[] args) throws Exception {
    long test = ConvertStringToDate("May 26 10:41:23", "MMM dd hh:mm:ss");
    long test2 = ConvertStringToDate("Tue, Jun 06 2017, 12:30 AM", "EEE, MMM dd yyyy, hh:mm a");
    long test3 = ConvertStringToDate("Jun 13 2003 23:11:52.454 UTC", "MMM dd yyyy HH:mm:ss.SSS zzz");
}

private static long ConvertStringToDate(String dateString, String format) {
    try {
        return new SimpleDateFormat(format).parse(dateString).getTime();
    } catch (ParseException e) {}
    return 0;
}

Solution 5 - Java

  String dateTime="15-3-2019 09:50 AM" //time should be two digit like 08,09,10 
   DateTimeFormatter dtf  = DateTimeFormatter.ofPattern("dd-MM-yyyy hh:mm a");
        LocalDateTime zdt  = LocalDateTime.parse(dateTime,dtf);
        LocalDateTime now = LocalDateTime.now();
        ZoneId zone = ZoneId.of("Asia/Kolkata");
        ZoneOffset zoneOffSet = zone.getRules().getOffset(now);
        long a= zdt.toInstant(zoneOffSet).toEpochMilli();
        Log.d("time","---"+a);

you can get zone id form this a link!

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionuser804979View Question on Stackoverflow
Solution 1 - JavaBohemianView Answer on Stackoverflow
Solution 2 - JavaOrtomala LokniView Answer on Stackoverflow
Solution 3 - JavaBasil BourqueView Answer on Stackoverflow
Solution 4 - Javauser1960808View Answer on Stackoverflow
Solution 5 - JavaFlashView Answer on Stackoverflow