2

I am able to convert the complex Java objects like List, ArrayList etc containing huge volume of data into Json efficiently using the Gson library as below code

    List<CusPojo> list=new ArrayList<CusPojo>();
       .
       .
     Gson gson=new Gson();
     String json=gson.toJson(list);

But if if try the same for a String literal or a String Obj, the conversion is not happening 

    String msg="success";
            **or**
    String msg=new String("success");

    Gson gson=new Gson();
    String json=gson.toJson(msg); 
    System.out.println("json data-- "+json);

Here i expect the data in Json format like

    json data-- {"msg":"success"}

but instead success is what i am getting

    json data-- "success"

I couldn't find any explanation regarding this particulary Please help , thank you in advance..

Improve this question
7
  • 1
    If you want your json output to be {"msg": "success"} try using a class, with a string named msg. Commented Sep 25, 2015 at 7:33
  • can please give me and example? Commented Sep 25, 2015 at 7:35
  • Or put the string in a map with the key "msg" Commented Sep 25, 2015 at 7:36
  • 3
    How can gson know that your variable is called "msg"? What would you expect it to print if the parameter were msg + msg? Commented Sep 25, 2015 at 7:39
  • @LionelPort brings up a good point, you can use a map as well, I think. Commented Sep 25, 2015 at 7:43

6 Answers 6

Reset to default
3

Please note, I rarely write java anymore, and I don't have a compiler in front of me, so there could be some obvious bugs here. But try this.

Assuming you have a class for example:

public class Test {
    public String msg;
}

You can use it, rather than a string in your gson example.

public static void main(String args[]) {
    Test test = new Test();
    test.msg = "success"
    Gson gson = new Gson();
    String json = gson.toJson(test);
}
Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

3

For simple cases where you don't want to use a POJO or map you can just create a JsonObject where it's behaviour is close to a map so you can pass the string as value and provide a property name where the JsonObject's toString() will be in JSON format. So you could do the following:

JsonObject jObj = new JsonObject();
jObj.addProperty("msg", msg);
System.out.println(jObj);
Improve this answer

Comments

0

GSon doesn't save you variable name, it saves field name of serialized class

public class StringSerialize{
      private String msg;
      ......
}
Improve this answer

Comments

0

I think you are doing it wrong. your variable msg is just holding the data. try this, let me know if it helps

    Map<String,String> map = new HashMap<>();       
    map.put("msg", "success");
    Gson gson = new GsonBuilder().serializeNulls().create();
    String json = gson.toJson(map);
    System.out.println(json);
Improve this answer

Comments

0 
Map<String, String> m = new HashMap<String, String>();
m.put("msg", "success");
Gson gson = new GsonBuilder().create();
System.out.println(gson.toJson(m));

Result

{"msg":"success"}
Improve this answer

Comments

-1

Try this for your example:
At first create a model class

class Model {
    public String msg;
}

The name of your member is the key in your json

public static void main() {
    Model model = new Model();
    model.msg = "success"
    Gson gson = new Gson();

    String json = gson.toJson(test); // Model to json
    Model result = gson.fromJson("{\"msg\":\"success\"}", Model.class) // json to Model
}

This is a very good gson-tutorial linked by the gson repository on github

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.