JSON解析器无法转换我的字符串
问题描述:
我正在从一个PHP MysQl服务器收集某些信息的Android应用程序,但我面临JSON解析器的问题,有时它有效它不起作用,它给了我这些错误:JSON解析器无法转换我的字符串
E/JSON Parser﹕ Error parsing data org.json.JSONException: Value Failed of type java.lang.String cannot be converted to JSONObject
Caused by: java.lang.NullPointerException
和代码这就是产生错误:
try {
// Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("username", username));
params.add(new BasicNameValuePair("password", password));
params.add(new BasicNameValuePair("fname", fname));
params.add(new BasicNameValuePair("lname", lname));
params.add(new BasicNameValuePair("email", email));
params.add(new BasicNameValuePair("city", city));
params.add(new BasicNameValuePair("nrc", nrc));
params.add(new BasicNameValuePair("dob", dob));
params.add(new BasicNameValuePair("cell", cell));
Log.d("username", username);
Log.d("password", password);
Log.d("fname", fname);
Log.d("lname",lname);
Log.d("email",email);
Log.d("city",city);
Log.d("nrc",nrc);
Log.d("dob",dob);
Log.d("cell",cell);
Log.d("request!", "starting");
//Posting user data to script
JSONObject json = jsonParser.makeHttpRequest(
REGISTER_URL, "POST", params);
// full json response
Log.d("Registering attempt", json.toString());
// json success element
success = json.getInt(TAG_SUCCESS);
if (success == 1) {
Log.d("User Profile Created!", json.toString());
的NullException正在由线路原因引起:
Log.d("Registering attempt", json.toString());
尝试将json头添加到您要调用的url,并交叉检查有效的json是否正在返回。点击这里查看..http://jsonviewer.stack.hu/ –
我不明白你的意思,我是新来的Json请你自己解释更多 – Chrome