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Home/ Questions/Q 6873761
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T04:06:36+00:00 2026-05-27T04:06:36+00:00

Trying to learn how to use JSON and parse the json data.i am using

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Trying to learn how to use JSON and parse the json data.i am using google-gson API to parse my json data.

i am getting my JSON data in the following format

{"guid":{"uri":"http://social.yahooapis.com/v1/me/guid","value":"123456789"}}

and here is my parsing code using google-gson

Gson gson=new Gson();
MyGuid myGuid=new MyGuid();
myGuid=gson.fromJson(response.getBody(), MyGuid.class);

public class MyGuid {

public MyGuid() {

}

private String value;

public String getValue() {
    return value;
}

public void setValue(String value) {
    this.value = value;
}

}

but i am getting the value of guid as null.
I know i am doing wrong but being new to JSON format is what making me more confused.

Any help in this regard will be helpful.

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  1. Editorial Team
    Editorial Team
    2026-05-27T04:06:36+00:00Added an answer on May 27, 2026 at 4:06 am

    Your data structures should reflect the data you are trying to de-serialize.

    For example, you could use something of this form:

    public class Data {
      private Map<String, String> guid;
    
      public Map<String, String> getGuid() {
        return guid;
      }
    
      public void setGuid(Map<String, String> guid) {
        this.guid = guid;
      }
    
      public static void main(String[] args) {
        String json =
            "{\"guid\":{" + "\"uri\":\"http://social.yahooapis.com/v1/me/guid\","
                + "\"value\":\"123456789\"}}";
        Data data = new Gson().fromJson(json, Data.class);
        System.out.println(data.getGuid()
            .get("uri"));
      }
    }
    

    The guid property of the JSON is mapped to the guid property of the Data type. The properties of that object are placed in a Map as strings.

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