Not able to create proper JSON using Java

Question

My situation is crazy. I'm creating a JSON object, which is perfectly fine from server side end. But, this is not getting acceptable due to extra " which is created via server side. My server side code is

String id = "123";
    String hql = "FROM Person E WHERE E.userId = "+id;
    Query query = session.createQuery(hql);
    List<?> results = query.list();     
    JSONArray arr = new JSONArray();
    JSONObject obj = new JSONObject();
    for(int i = 0; i < results.size(); i++) {
        Personalisation p = (Personalisation) results.get(i);
        obj.put("courseId", p.getCourseId());
        obj.put("CourseValue", p.getCourseValue());
    }
    System.out.println(obj);

It is printing

{"CourseValue":"{\"color\": \"green\",\"value\": \"#f00\"}","courseId":"C5"}

This is fine from server side end. You could see extra " before {\"color tag, when I try to parse the same thing on client side, it doesnt accept due to illegal character. What should I need to do?

Here is the fiddle too http://jsfiddle.net/hLkUz/43/

wero · Accepted Answer · 2016-03-22 08:39:45Z

4

p.getCourseValue() seems to return a String containing JSON:

{"color": "green","value": "#f00"}

Now when you put this into JSONObject and serialize the object it will escape the JSON string again.

Instead of putting the course value as JSON tring into the JSONObject you need to build up the course value object yourself.

answered Mar 22, 2016 at 8:39

wero

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1 Comment

Syed Over a year ago

I tried it using new JSONObject(p.getCourseValue()). Still its not accepting. Any info?

RH-indra Poudel · Accepted Answer · 2016-03-22 08:43:50Z

0

as far as I know you cannot parse a JSON object as a string so use

 System.out.println(obj.toString());

instead of

System.out.println(obj);

answered Mar 22, 2016 at 8:43

RH-indra Poudel

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6 Comments

RH-indra Poudel Over a year ago

and use obj.put("CourseValue", new JSONObject(p.getCourseValue())); while inserting

Syed Over a year ago

Its not accepting new JSONObject(p.getCourseValue()). It says remove argument

Syed Over a year ago

and I'm not worried about SOP. I'm worried because its not being accepted by client

RH-indra Poudel Over a year ago

give a try to (JSONObject)new JSONParser().parse(p.getCourseValue());

Syed Over a year ago

Yeah I tried by new JsonParser().parse(p.getCourseValue()).getAsJsonObject()

|

Syed · Accepted Answer · 2016-03-22 09:09:04Z

0

Able to construct by using

new JsonParser().parse(p.getCourseValue()).getAsJsonObject()

answered Mar 22, 2016 at 9:09

Syed

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