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How can I get the current date and time in below format , date = "Mon, 13 Jul 2020 14:08:30 GMT"

need help in java code to get current date and time in below format :
newdate = "Mon, 13 Jul 2020 14:08:30 GMT"
After this I need to replace current date with earlier one:
vJsonfile1.replace(earlierdate," "+ newdate);

You can use ZonedDateTime and the RFC_1123 format to get the output you need:
DateTimeFormatter dtf = DateTimeFormatter.RFC_1123_DATE_TIME;
ZonedDateTime zdt = ZonedDateTime.now(ZoneId.of("GMT"));
System.out.println(dtf.format(zdt));
Wed, 15 Jul 2020 19:07:37 GMT
Note the 1123_DATE_TIME format doesnt play nice with North American Time zones so itll work as long as its GMT or European time zones otherwise below will suffice too:
DateTimeFormatter dtf = DateTimeFormatter.ofPattern("EEE, dd MMM yyyy HH:mm:ss z");
ZonedDateTime zdt = ZonedDateTime.now();
System.out.println(dtf.format(zdt));
Wed, 15 Jul 2020 14:13:22 CDT
Which will output the current time with the time zone its in.

Here is an example that formats your datetime and then changes the date portion of it:
ZonedDateTime zonedDateTime = ZonedDateTime.now();
System.out.println(zonedDateTime.format(DateTimeFormatter.ofPattern("EEE, dd MMM yyyy HH:mm:ss z")));
zonedDateTime = ZonedDateTime.of(LocalDate.of(2020, 12, 15), zonedDateTime.toLocalTime(), zonedDateTime.getZone());
System.out.println(zonedDateTime.format(DateTimeFormatter.ofPattern("EEE, dd MMM yyyy HH:mm:ss z")));

This code does exactly what you wanted it to do.
Output:
Mi, 15 Jul 2020 08:55:21 MESZ
Code:
SimpleDateFormat formatter = new SimpleDateFormat("EEE, dd MMM yyyy hh:mm:ss z");
Date currentDate = new Date();
System.out.println(formatter.format(currentDate));

Joda: parse string as DateTime with weird timezone format

Is there a valid joda DateTimeFormat for date strings like the following:
Mon, 23 Jul 2018 07:08:26 +0300 GMT
I have tried:
DateTimeFormatter FMT1 = DateTimeFormat.forPattern("E, d MMM yyyy HH:mm:ss Z");
DateTimeFormatter FMT2 = DateTimeFormat.forPattern("E, d MMM yyyy HH:mm:ss z");
DateTimeFormatter FMT3 = DateTimeFormat.forPattern("E, d MMM yyyy HH:mm:ss z Z");
but none of these work.
I had a look in here https://www.joda.org/joda-time/apidocs/org/joda/time/format/DateTimeFormat.html and I cannot figure out a way to parse that date without having to change the string itself first.
Is there a way?

I take it that +0300 is the true offset and GMT is just there to say that +0300 is relative to GMT. Joda-Time supports apostrophes for delimiting constant text that shouldn’t be interpreted:
DateTimeFormatter formatter = DateTimeFormat.forPattern("E, d MMM yyyy HH:mm:ss Z 'GMT'");
DateTime dt = DateTime.parse("Mon, 23 Jul 2018 07:08:26 +0300 GMT", formatter);
System.out.println(dt);
Output on my computer in Europe/Copenhagen time zone:
2018-07-23T06:08:26.000+02:00
As an aside the idea of using a lowercase and an uppercase z wasn’t too bad. It works if you put them in the opposite order:
DateTimeFormatter formatter = DateTimeFormat.forPattern("E, d MMM yyyy HH:mm:ss Z z");
2018-07-23T04:08:26.000Z
I prefer the apostrophes, though. It’s hard for a reader to know what to expect from Z z.

How to Parse Date in java [duplicate]

This question already has answers here:
How can I convert Date.toString back to Date?
(5 answers)
Closed 4 years ago.
Unparseable date: "Tue Jul 03 16:59:51 IST 2018" Exception
String date="Tue Jul 03 16:59:51 IST 2018";
i want to parse it.
My code is
SimpleDateFormat newformat=SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssZ")
Date d=newformat.parse(date);

You are using old and quite problematic classes, especially Date.
For your example, perhaps consider using LocalDateTime
String date = "Tue Jul 03 16:59:51 IST 2018";
DateTimeFormatter format = DateTimeFormatter
.ofPattern("EEE MMM dd HH:mm:ss z yyyy");
LocalDateTime dateTime = LocalDateTime.parse(date, format);

Since your pattern has to match the string you want to parse you need to adjust the pattern as following:
EEE MMM dd HH:mm:ss z yyyy
Infos gathered from the docs.
You should also use the new java.time API introduced with Java 8.
String s = "Tue Jul 03 16:59:51 IST 2018";
//Java 7 way
SimpleDateFormat newformat = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy");
Date d = newformat.parse(s);
//Java 8 way
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("EEE MMM dd HH:mm:ss z yyyy");
LocalDateTime dateTime = LocalDateTime.parse(s, formatter);

How to get Date in the below Format "dd MMM, yyyy z"

How to make the date to have a GMT offset like mentioned here
import java.util.*;
import java.text.*;
import java.lang.*;
class TFTest
{
public static void main(String[] args)
{
SimpleDateFormat sdf = new SimpleDateFormat("dd MMM, yyyy z");
Date dt = new Date();
System.out.println("\n\n\nparsed date:"+sdf.format(dt)+"\n\n");
}
}
the above program outputs the value as
parsed date:02 Aug, 2016 IST.
But I want the value to be parsed date:02 Aug, 2016 GMT +05:30
How to get in the specified format ..?

The pattern that should work is dd MMM, yyyy 'GMT' XXX indeed X is the timezone in ISO 8601 which seems to be what you are looking for.
Output:
parsed date:02 Aug, 2016 GMT +05:30

Try, for more documentation visit simpledateformat
"dd MMM, yyyy 'GTM' XXX"

This pattern "dd MMM, yyyy z ZZZZ" will print in the given format
Format formatter = new SimpleDateFormat("dd MMM, yyyy z ZZZZ");
Result like :
02 Aug, 2016 GMT +0000

Error parsing date string

I need to parse this string as a date:
Mon Jun 10 00:00:00 CEST 2013
Here is what I do:
SimpleDateFormat sdf = new SimpleDateFormat("ccc MMM dd HH:mm:ss z yyyy");
Date date = sdf.parse(dateString);
But I get a ParseException:
Unparseable date: "Wed Oct 02 00:00:00 CEST 2013" (at offset 0)
Any help please?

As others have said, you need EEE instead of ccc - but you should also specify a locale, so that it doesn't try to parse the month and day names (and other things) using your system default locale:
SimpleDateFormat sdf = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy",
Locale.US);

Your format is wrong. You need to use EEE instead of ccc, where E means Day name in week.
SimpleDateFormat sdf = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy");
Have a look at the docs regarding all the valid patterns available for SimpleDateFormat.

Replace ccc with EEE in the pattern to specify the day of the week:
SimpleDateFormat sdf = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy");
Example: https://gist.github.com/kmb385/8781482

Update the format as below :
SimpleDateFormat sdf = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy");

It is a Locale problem. That's because dates are represented differently between Locales, so the JVM fires an exception if the Date is not in the correct format. You can solve it by setting a custom Locale:
String str = "Mon Jun 10 00:00:00 EST 2013";
Locale.setDefault(Locale.US);
SimpleDateFormat sdf = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy");
Date date = sdf.parse(str);
System.out.println(date);
IDEone examples do work because the default locale is Locale.US

java.time
The accepted answer uses SimpleDateFormat which was the correct thing to do in Feb 2014. In Mar 2014, the java.util date-time API and their formatting API, SimpleDateFormat were supplanted by the modern Date-Time API. Since then, it is highly recommended to stop using the legacy date-time API.
Note: Never use date-time formatting/parsing API without a Locale.
Solution using the modern date-time API:
import java.time.ZonedDateTime;
import java.time.format.DateTimeFormatter;
import java.util.Locale;
class Main {
public static void main(String[] args) {
String stdDateTime = "Mon Jun 10 00:00:00 CEST 2013";
DateTimeFormatter parser = DateTimeFormatter.ofPattern("EEE MMM dd HH:mm:ss z uuuu", Locale.ENGLISH);
ZonedDateTime zdt = ZonedDateTime.parse(stdDateTime, parser);
System.out.println(zdt);
}
}
Output:
2013-06-10T00:00+02:00[Europe/Paris]
If for any reason, you need an instance of java.util.Date, you can get it as follow:
Date date = Date.from(zdt.toInstant());
ONLINE DEMO
Learn more about the modern Date-Time API from Trail: Date Time.