使用json格式android解析Java中的Json麻烦

Question

How can I get the information from my .json file with the following format?:

[{"name":"My name","country":"Country name"}]

I'm getting with the following:

Json file:

{"name":"My name","country":"Country name"}

Java file:

@Override
        protected JSONObject doInBackground(String... args) {

            JsonParser jParser = new JsonParser();
            JSONObject json;
            json = jParser.getJSONFromUrl("http://example.com/myfile.json");

            System.out.println("JSON: " + json);

            if(json != null) {

                try {

                    name = json.getString("name");
                    country = json.getString("country");

                } catch (JSONException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }

            }

            return null;
        }

Answer 1

You are not getting JSONObject in your response, you are getting JSONArray that holds one object. Therefore these lines are wrong

JSONObject json;
json = jParser.getJSONFromUrl("http://example.com/myfile.json");

And you should replace it with

JSONArray json;

Then you get your 0th object

JSONObject wholeObject = json.getJSONObject(0);

And get strings from it

name = wholeObject.getString("name");
country = wholeObject.getString("country");

Answer 2

Can you try this:

 protected JSONObject doInBackground(String... args) {

        JsonParser jParser = new JsonParser();
        JSONObject json;
        json = jParser.getJSONFromUrl("http://example.com/myfile.json");

        System.out.println("JSON: " + json);

        if(json != null) {

            try {
                JSONArray jsonArray = new JSONArray(json.toString());
                JSONObject newJSON = jsonArray.get(0); //first index
                name = newJSON.getString("name");
                country = newJSON.getString("country");

            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

        }

        return null;
    }