用OkHttp从响应中创建一个新的JsonObject

Question

我正在尝试使用OKHttp和https://github.com/hongyangAndroid/okhttp-utils来实现登录功能，但是我不知道如何使用Callback中的Response创建一个新的JsonObject。

  OkHttpUtils.post()
            .url(url)
            .addParams("phone", "18681873411")
            .addParams("password", "18681873411")
            .build()
            .execute(new Callback() {
                @Override
                public Object parseNetworkResponse(Response response) throws Exception {
                    Log.i("ws","---->>parseNetworkResponse" + response.body().string());


                    JSONObject jsonObj = null;
                    try {
                        jsonObj = new JSONObject(response.body().string());
                    } catch (JSONException e) {
                        Log.i("ws","---->>JSONException");
                        e.printStackTrace();
                    } catch (IOException e) {
                        Log.i("ws","---->>IOException");
                        e.printStackTrace();
                    }
                    Log.i("ws","---->>111 :" );
                    String opCode = jsonObj.getString("OpCode");
                    String message = jsonObj.getString("Message");
                    Log.i("ws","---->>111 opCode:" + opCode);
                    Log.i("ws","---->>111 Message:" + message);
                    JSONObject result = new JSONObject();
                    result.put("qrcode",opCode);
                    result.put("message", message);
                    return result;

                }

                @Override
                public void onError(Call call, Exception e) {
                    Log.i("ws","---->>onError" + e);
                }

                @Override
                public void onResponse(Object response) {
                    Log.i("ws","---->>onResponse" + response.toString());
                }
            });

以及错误信息：

parseNetworkResponse{"OpCode":0,"Message":"登陆成功","Other":{"id":1,"name":null,"phone":"18681873411","QQuid":null,"weixinuid":null,"gender":null,"age":null,"credits":null,"address":null,"signature":null,"imageurl":null,"password":"18681873411","created_at":"2016-05-19T17:05:02.000Z","updated_at":"2016-05-19T17:05:02.000Z"}}
05-20 11:29:01.953 13496-13621/com.smartspace.magicmirror I/ws: ---->>IOException
05-20 11:29:01.954 13496-13621/com.smartspace.magicmirror I/ws: ---->>111 :
05-20 11:29:01.955 13496-13496/com.smartspace.magicmirror I/ws: ---->>onErrorjava.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String org.json.JSONObject.getString(java.lang.String)' on a null object reference

- Shuai Wang

---

你确定你的 jsonObj 包含了正确解析的值吗？你可以尝试将代码 String opCode = jsonObj.getString("OpCode"); String message = jsonObj.getString("Message"); 放在你的 try 语句内，就在 jsonObj = new JSONObject(response.body().string()); 下面。 - AL.

Answer 1

由于调用response.body().string()会从返回流中读取内容，然后关闭它，因此第二次调用它时将不返回任何内容：\

更好的方式：

if (!response.isSuccessful()) throw new IOException("Unexpected code " + response); 
    String responseData = response.body().string();
    Log.e(TAG, "onResponse: " + responseData);
    try {
        JSONObject jsonObject = new JSONObject(responseData);
        // Do something here 
        } catch (JSONException e) {
...}

- toantran