Converting Java String object into Json using Gson library

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..

• jabgibson: If you want your json output to be {"msg": "success"} try using a class, with a string named msg.
• user4939493: can please give me and example?
• Lionel Port: Or put the string in a map with the key "msg"
• Andy Turner: How can gson know that your variable is called "msg"? What would you expect it to print if the parameter were msg + msg?
• jabgibson: @LionelPort brings up a good point, you can use a map as well, I think.

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