簡體   English   中英

JSON解析返回空值

[英]Json parsing return null value

在我的Android應用程序中,我正在使用json解析登錄。json解析網址包含瀏覽器上的json數據,如下所示

{
    posts: [{
        status: 1,
        user_id: 10,
        access_token: "BRBL4JJXB9",
        image: "images/users/.PNG",
        msg: "Login Successful."
    }]
}

但是當調用相同的URL進行JSON解析時,它返回null。我的解析代碼如下

 @Override
 protected Void doInBackground(Void... params) {
     String url = "";
     url = URLS.BASEURL + "mobile_api.php?action=userLogin&user_name=" +plusname+"&image="+plusimage+"&email="+URLEncoder.encode(gmail) +"&gcm_id="+deviceid;

     JSONReader reader = new JSONReader();
     json = getJsonGET(url);

     if (json != null) {

         try {                 
             JSONObject response = new JSONObject(json);
             JSONArray arr = response.getJSONArray("posts");

             for (int index = 0; index < arr.length(); index++) {

                 JSONObject jobj = arr.getJSONObject(index);
                 status = jobj.getString("status");

                 if (status.equalsIgnoreCase("1")) {
                     userid = jobj.getString("user_id");
                     accesstoken = jobj.getString("access_token");

                     SharedPreferences.Editor edit = m_pref.edit();
                     edit.putString("userid", userid);
                     edit.putString("accesstoken", accesstoken);

                     edit.commit();

                 } else {

                        }
             }
             } catch (JSONException e) {
                    e.printStackTrace();
               }
     }
            return null;

}


private String getJsonGET(String url) {

    String contentAsString;

    StringBuilder total = new StringBuilder();
    try {

         InputStream is = null;

         try {
              URL url2 = new URL(url);
              HttpURLConnection conn = (HttpURLConnection)                                  url2.openConnection();
              conn.setReadTimeout(100000 /* milliseconds */);
              conn.setConnectTimeout(150000 /* milliseconds */);
              conn.setRequestMethod("GET");
              conn.setDoInput(true);
              conn.connect();
              int response = conn.getResponseCode();
              Log.d("", "The response is: " + response);
              is = conn.getInputStream();

              // Convert the InputStream into a string

              Reader reader = null;
              reader = new InputStreamReader(is, "UTF-8");
              BufferedReader r = new BufferedReader(new InputStreamReader(is));

              String line;
              while ((line = r.readLine()) != null) {
                  total.append(line);
              }

              char[] buffer = new char[total.length()];
              reader.read(buffer);

              contentAsString = new String(buffer);

          } finally {
                     if (is != null) {
                        is.close();
                     }

            }
      } catch (Exception e) {
          return null;
        }

  return total.toString();

}

變量“ json”返回null。請幫助我找到解決方案。此處r.readLine()為null

您的json輸出應為..

{"posts": [{"status": 1,"user_id": 10,"access_token": "BRBL4JJXB9","image": "images/users/.PNG","msg": "Login Successful."}]} 

其次,您將statususer_id解析為string但是它是int因此您的語法和條件應類似於。

int status = jsonObject.getInt("status");
if (status == 1)) {
     int userid = jobj.getInt("user_id");
     String accesstoken = jobj.getString("access_token");

     SharedPreferences.Editor edit = m_pref.edit();
     edit.putInt("userid", userid);
     edit.putString("accesstoken", accesstoken);
     edit.commit();
}

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM