8

How can I parse json using gson? I have a json array with multiple object types, and I don't know, what kind of object I need to create to save this structure. I cannot change the json data format (I don't control the server). Can I use gson or other library parse this json array, how should I do?

This is the json code block:

[
  {
    "type": 1,
    "object": {
      "title1": "title1",
      "title2": "title2"
    }
  },
  {
    "type": 2,
    "object": [
      "string",
      "string",
      "string"
    ]
  },
  {
    "type": 3,
    "object": [
      {
        "url": "url",
        "text": "text",
        "width": 600,
        "height": 600
      },
      {
        "url": "url",
        "text": "text",
        "width": 600,
        "height": 600
      }
    ]
  },
  {
    "type": 4,
    "object": {
      "id": 337203,
      "type": 1,
      "city": "1"
    }
  }
]
1
  • Hi! Have you read and tried my answer yet? Commented Sep 9, 2015 at 10:19

3 Answers 3

8

This json structure is inherently gson-unfriendly. i.e You cannot model this cleanly in java because the "object" key refers to a dynamic type. The best you can do with this structure is model it like so:

    public class Models extends ArrayList<Models.Container> {

    public class Container {
        public int type;
        public Object object;
    }

    public class Type1Object {
        public String title1;
        public String title2;
    }

    public class Type3Object {
        public String url;
        public String text;
        public int width;
        public int height;
    }

    public class Type4Object {
        public int id;
        public int type;
        public int city;
    }

}

Then do some awkward switch on type and the object field:

String json = "{ ... json string ... }";
Gson gson = new Gson();
Models model = gson.fromJson(json, Models.class);


for (Models.Container container : model) {

    String innerJson = gson.toJson(container.object);

    switch(container.type){
        case 1:
            Models.Type1Object type1Object = gson.fromJson(innerJson, Models.Type1Object.class);
            // do something with type 1 object...                                
            break;
        case 2:
            String[] type2Object = gson.fromJson(innerJson, String[].class);
            // do something with type 2 object...
            break;
        case 3:
            Models.Type3Object[] type3Object = gson.fromJson(innerJson, Models.Type3Object[].class);
            // do something with type 3 object...
            break;
        case 4:
            Models.Type4Object type4Object = gson.fromJson(innerJson, Models.Type4Object.class);
            // do something with type 4 object...
            break;

    }
}

Ultimately the best solution is to get the json structure changed to something more compatible with java.

E.g:

[
  {
    "type": 1,
    "type1Object": {
      "title1": "title1",
      "title2": "title2"
    }
  },
  {
    "type": 2,
    "type2Object": [
      "string",
      "string",
      "string"
    ]
  },
  {
    "type": 3,
    "type3Object": [
      {
        "url": "url",
        "text": "text",
        "width": 600,
        "height": 600
      },
      {
        "url": "url",
        "text": "text",
        "width": 600,
        "height": 600
      }
    ]
  },
  {
    "type": 4,
    "type4Object": {
      "id": 337203,
      "type": 1,
      "city": "1"
    }
  }
]
Sign up to request clarification or add additional context in comments.

Comments

4

This may be a bit late for the original poster, but hopefully it will help someone else.

I am using Gson in Android. I have seen everyone use custom classes and long way round solutions. Mine is basic.

I have an ArrayList of many different Object types (Models for my database) - Profile is one of them. I get the item using mContactList.get(i) which returns:

{"profile": 
    {"name":"Josh",
     "position":"Programmer",
     "profile_id":1,
     "profile_image_id":10,
     "user_id":1472934469
    },
 "user":
    {"email":"[email protected]",
     "phone_numbers":[],
     "user_id":1,
     "user_type_id":1
    },
 "follower":
    {"follower_id":3,
     "following_date":1.4729345E9,
     "referred_by_id":2,
     "user_from_id":1,
     "user_to_id":2
    },
 "media":
    {"link":"uploads/profiles/profile-photos/originals/1-G9FSkRCzikP4QFY.png",
     "media_description":"",
     "media_id":10,
     "media_name":"",
     "media_slug":"",
     "medium_link":"uploads/profiles/profile-photos/thumbs-medium/1-G9FSkRCzikP4QFY.png",
     "thumbnail_link":"uploads/profiles/profile-photos/thumbs-small/1-G9FSkRCzikP4QFY.png",
     "uploader_id":1
    }
}

Now I create the Gson object:

Gson gson = new Gson();
// this creates the JSON string you see above with all of the objects
String str_obj = new Gson().toJson(mContactList.get(i)); 

Now instead of creating a custom class - just pass it through as a JsonObject using the following code:

JsonObject obj = gson.fromJson(str_obj, JsonObject.class);

And now, you can call the object like so:

JsonObject profile = obj.getAsJsonObject("profile");

1 Comment

How can I get the name property of profile? @Haring10
0

You can set the methods in your model class very easily. Just make a StringRequest. Below is a snippet:

List<YourModelClass> inpList;
StringRequest greq = new StringRequest(Request.Method.POST, yourURL, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                try {
                        Log.d("response array===>  ", response.toString());

                        Type type = new TypeToken<List<YourModelClass>>(){}.getType();
                        inpList = new Gson().fromJson(response, type);

                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                error.printStackTrace();

            }
        }){
            @Override
            protected Map<String, String> getParams() throws AuthFailureError {
                Map<String, String> params = new HashMap<String, String>();
                //return params back to server, if any
            }
        };

        yourVolley.getRequestQueue().add(greq);

I have used this to generate your model class from you json. Your model class will look something like this:

 package com.example;

import javax.annotation.Generated;
import com.google.gson.annotations.Expose;

@Generated("org.jsonschema2pojo")
public class YourModelClass {

@Expose
private Integer type;
@Expose
private Object object;

/**
* 
* @return
* The type
*/
public Integer getType() {
return type;
}

/**
* 
* @param type
* The type
*/
public void setType(Integer type) {
this.type = type;
}

/**
* 
* @return
* The object
*/
public Object getObject() {
return object;
}

/**
* 
* @param object
* The object
*/
public void setObject(Object object) {
this.object = object;
}

}
-----------------------------------com.example.Object.java-----------------------------------

package com.example;

import javax.annotation.Generated;
import com.google.gson.annotations.Expose;

@Generated("org.jsonschema2pojo")
public class Object {

@Expose
private Integer id;
@Expose
private Integer type;
@Expose
private String city;

/**
* 
* @return
* The id
*/
public Integer getId() {
return id;
}

/**
* 
* @param id
* The id
*/
public void setId(Integer id) {
this.id = id;
}

/**
* 
* @return
* The type
*/
public Integer getType() {
return type;
}

/**
* 
* @param type
* The type
*/
public void setType(Integer type) {
this.type = type;
}

/**
* 
* @return
* The city
*/
public String getCity() {
return city;
}

/**
* 
* @param city
* The city
*/
public void setCity(String city) {
this.city = city;
}

}

2 Comments

but the object has four types, how to deal with it
You can change the model class and define object as a list of "object" class. i,e you will make another class object and define its getters and setters separately. But yes i think object value as array should have a different name than object value as an object

Your Answer

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

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.