Convert DateTime object java to Json string

Question

I want to convert date time object in java to json string in format as below:

{"date":"/Date(18000000+0000)/"}

I did like this but it didn't give me the format i desired:

JSONObject object = new JSONObject();
object.put("date",new Date());

and the result of object.toString()

{"date":"Fri May 04 11:22:32 GMT+07:00 2012"}

i want the string "Fri May 04 11:22:32 GMT+07:00 2012" transform to "/Date(18000000+0000)/" (18000000+0000 here is just a example).

Thanks for your help.

Probably repost [SimpleDateFormat](http://stackoverflow.com/questions/3056703/simpledateformat) — Crazenezz, May 04 '12 at 04:32
@Crazenezz can you give me more detail for my case. i want to format datetime in format "18000000+0000"? — MichaelP, May 04 '12 at 04:35
What are the two portions in your date string? (ie what is 18000000 and what is +0000 ? ) — Greg Kopff, May 04 '12 at 04:40
actually i see the string {"date":"/Date(18000000+0000)/"} is generated from iphone app. i dont know what format for 18000000+0000 is. but i need to generate the same string as iphone app when i get datetime obj in java. — MichaelP, May 04 '12 at 04:44
i dont know, i just want to ask everyone that if i get a datetime obj in java, then how can i convert this obj into format string like this "/Date(18000000+0000)/" — MichaelP, May 04 '12 at 04:51
You don't know what the format is - and you don't know the information needed to try and *derive* what the parts are. How can we help? — Greg Kopff, May 04 '12 at 04:53
@MichaelP: Can you elaborate what format for "18000000+0000"? — Crazenezz, May 04 '12 at 05:56
i just dont know what format "18000000+0000" means, and i want to ask everyone about this format, that's it. — MichaelP, May 04 '12 at 06:59

score 3 · Answer 1 · edited Apr 06 '13 at 18:47

3

Here is my solution, although it is not a good way, but I finally find a workable solution.

SimpleDateFormat format = new SimpleDateFormat("Z");
Date date = new Date();
JSONObject object = new JSONObject();
object.put("date", "/Date(" + String.valueOf(date.getTime()) + format.format(date) + ")/");

edited Apr 06 '13 at 18:47

Waqas Raja

10,802
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answered May 04 '12 at 07:27

MichaelP

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"18000000+0000" means that Unix timestamp + timezone – MichaelP May 04 '12 at 07:31
This is my solution too...worked with .net WCF web service. :) – Waqas Raja Apr 06 '13 at 18:44

score 1 · Answer 2 · answered Mar 20 '15 at 12:44

 public static String convertToJsonDateTime(String javaDate)

  {
     SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-mm-dd HH:mm:ss");
        Date currentDate = null;
        try {
            currentDate = dateFormat.parse(javaDate);
        } catch (ParseException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        long time =currentDate.getTime();
    return  "\\/Date("+time+"+0000)\\/";
}

score 1 · Answer 3 · answered Apr 16 '17 at 17:32

1

My solution : I think this is the simplest Way

DateFormat dateFormat=new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSZ");
yourJsonObject.accumulate("yourDateVarible",dateFormat.format(new Date()));

answered Apr 16 '17 at 17:32

SwapnilKumbhar

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score 0 · Answer 4 · answered May 04 '12 at 04:39

From json.org's description of JSONObject:

The values can be any of these types: Boolean, JSONArray, JSONObject, Number, and String, or the JSONObject.NULL object.

This library doesn't support the "complex" mapping of a Date to a JSON representation. You'll have to do it yourself. You may find something like SimpleDateFormat helpful.

score 0 · Accepted Answer · answered May 04 '12 at 06:12

The date format that you want is /Date(<epoch time><Time Zone>)/.
You can get the epoch time in java using long epoch = System.currentTimeMillis()/1000;(found at this link) and the time zone you can get by using the date and time patteren as Z. Then combine all the strings into one and store it to the Json Object.
Other possibility is that the time you are getting from iOS device may be of the pattern yyMMddHHmmssZ as got from here. Check the output on the iOS device at different times and identify the correct pattern.

score 0 · Answer 6 · answered May 04 '12 at 06:14

0

If your format like this -> "DDmmYYYY+HHMM"

DD -> Day (2 Digit)
mm -> Month (2 Digit)
YYYY -> Year (4 Digit)
HH -> Hour
MM -> Minute

Than you can do like this:

SimpleDateFormat format = new SimpleDateFormat("DDmmYYYY+HHMM");
JSONObject object = new JSONObject();
object.put("date", "/Date(" + format.format(new Date()) + ")/");

answered May 04 '12 at 06:14

Crazenezz

3,416
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that's great, but i not sure "18000000+0000" equivalent to format "DDmmYYYY+HHMM". ok i will research more about this format. – MichaelP May 04 '12 at 07:01

score 0 · Answer 7 · answered May 04 '12 at 06:41

I suppose that's a representation of ISO international date-time format.

YYYYMMDD+HHMM

I think now you will be able to create that string

may be like,

Date d=new Date(); String tmp=""; tmp="/Date("d.getYear()+""+d.getMonth()+""+d.getDate()+"+"+d.getHours()+""+d.getMinutes()+")/";

score 0 · Answer 8 · answered Mar 24 '16 at 13:21

0

Upgrade one of the previous answer:

 public static String convertToJsonDateTime(Date dateToConvert) {
        SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-mm-dd HH:mm:ss");
        long time = dateToConvert.getTime();
        return "/Date(" + time + "+0000)/";
    }

answered Mar 24 '16 at 13:21

Ivan Ponomarenko

19
4

You never use dateFormat after you create it – tomcheney Mar 14 '19 at 16:12

Convert DateTime object java to Json string

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