使用Gson库将Java String对象转换为Json

Question

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..

Answer 1

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);
}

Answer 2

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);

Answer 3

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

public class StringSerialize{
      private String msg;
      ......
}

Answer 4

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);

Answer 5

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"}

Answer 6

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