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I have the TimeStamp and I need to convert it to Data type object that should match this pattern - "2016-11-16T18:42:33.049Z". How can I do that?

neckobik
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2 Answers2

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Date d = new Date((long)timestamp*1000);

will create a Date instance. Displaying it later is another thing.

I think it's what you want:

DateFormat f = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.mmm'Z'");
System.out.println(f.format(date));

Test:

import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Date;

public class Main {
    public static void main(String[] args) throws Exception {

        Date d = new Date((long)1481723817*1000);
        DateFormat f = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.mmm'Z'");
        System.out.println(f.format(d));
    }
}

>>2016-12-14T14:56:57.056Z
xenteros
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  • I need date to perfectly match the pattern cause I'm using it for query to dynamoDB... Do you know how to achieve that? – neckobik Dec 14 '16 at 13:54
  • You should keep timestamp in DynamoDB. Date is Date and don't worry abut the display format. – xenteros Dec 14 '16 at 13:55
  • @neckobik try the above code. I looks like it works. – xenteros Dec 14 '16 at 13:59
  • actually I need it to make condition for querying secondary index in aws dynamoDB. that is why i need exact same representation as i mentioned before. Case query will compare everything as Strings (Date attribute in Dynamo db is a String value). But ty for your answer any ways. I will test it now. – neckobik Dec 14 '16 at 14:00
  • I think you intended your format to be `"yyyy-MM-dd'T'HH:mm:ss.SSS'Z'"` ending in milliseconds: https://docs.oracle.com/javase/9/docs/api/java/text/SimpleDateFormat.html – Kyr Apr 03 '18 at 17:11
3

Convert like this. 

String date = new java.text.SimpleDateFormat("dd/MM/yyyy HH:mm:ss").format(new java.util.Date (epoch*1000));

you can also use the data object in the manner u want.

xenteros
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Vishesh
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