java - 如何从http请求中获取json响应？

Question

当尝试从我的 api 检索响应 json 时，我只能获取一个字符串值，但无法将其转换为 jsonobject 来使用。

应以 json 格式响应 {"result":"success","client[id]":"1"}

但我只能检索字符串“result=success,client[id]=1”

public void createWHMCSUser(final String emailID, final String random, final String uid) {

    Thread thread = new Thread(new Runnable() {
        @Override
        public void run() {
            try {
                String e = URLEncoder.encode(emailID, "utf-8");
                String r = URLEncoder.encode(random, "utf-8");
                URL url = new URL("http://create.user.com/includes/api.php?action=AddClient&username=abc123&password=abc123&accesskey=abc123&firstname=User&lastname=Name&email=" + e +"&address1=na&city=na&state=na&poscode=00000&country=US&phonenumber=0000000000&password2=" + r + "&reponsetype=json");
                HttpURLConnection conn = (HttpURLConnection) url.openConnection();
                conn.setRequestMethod("POST");
                conn.setRequestProperty("Content-Type", "application/json");
                conn.setRequestProperty("Accept","application/json");
                conn.setDoOutput(true);
                conn.setDoInput(true);



                Log.i("STATUS", String.valueOf(conn.getResponseCode()));
                Log.i("MSG", conn.getResponseMessage());

                if(conn.getResponseCode() == 200){
                    BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));

                    String readAPIResponse = " ";
                    StringBuilder jsonString = new StringBuilder();
                    while((readAPIResponse = in.readLine()) != null){
                        jsonString.append(readAPIResponse);

                        JSONObject obj = new JSONObject(jsonString.toString());
                        int aJsonString = obj.getInt("Client[id]");
                        SetClientID(uid,aJsonString);



                    }

                    in.close();


                }

                conn.disconnect();
            } catch (Exception e) {
                e.printStackTrace();
            }

        }
    });

    thread.start();

}

Answer 1

这样使用起来更方便快捷有关更多示例和构建库，请检查以下内容: enter link description here

 AndroidNetworking.get("URL")
                     .setTag("test")
                     .setPriority(Priority.MEDIUM)
                     .build()
                     .getAsJSONObject(new JSONObjectRequestListener() {
                        @Override
                        public void onResponse(JSONObject response) {
                          // handle your response here 

                        }
                        @Override
                        public void onError(ANError error) {
                          // handle error
                        }
                    });