Question

我想通过网络服务访问Android应用。在Web服务中，执行新注册。在Android应用程序中，进行新注册的xml文件。数据在SQL Server数据库中成功保存，并由Web服务正确保存，返回数据以jason字符串形式获取。但是当字符串转换为JSONObject时，它会给出如下错误：

Error parsing data org.json.JSONException: Value [{"userid":105,"created_at":"03-Oct-2013","success":1,"email":"rty@gmail.com","password":"rty12345","name":"rtyu"}] of type org.json.JSONArray cannot be converted to JSONObject

我已将注册活动注册为RegisterActivity.java

              else
              {
                  erName.setText("");
                  erPass.setText("");
                  erEmail.setText("");
                  erCopass.setText("");
                  UserFunction userFunction = new UserFunction();
                  JSONObject json = userFunction.registerUser(name, email, password);

                  // check for login response
                  try {
                      if (json.getString(KEY_SUCCESS) != null) {
                          String res = json.getString(KEY_SUCCESS); 
                          if(Integer.parseInt(res) == 1){
                              // user successfully registred
                              // Store user details in SQLite Database
                              Databasehandler db = new Databasehandler(getApplicationContext());
                              JSONObject json_user = json.getJSONObject("user");

                              // Clear all previous data in database
                              userFunction.logoutUser(getApplicationContext());
                              db.addUser(json_user.getString(KEY_NAME), json_user.getString(KEY_EMAIL),   
                           json.getString(KEY_UID), json_user.getString(KEY_CREATED_AT));                        
                              // Launch Dashboard Screen
                              Intent login = new Intent(getApplicationContext(), LoginActivity.class);
                              // Close all views before launching Dashboard
                              login.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
                              startActivity(login);
                              // Close Registration Screen
                              Toast.makeText(RegisterActivity.this,"You are Registered   successfully",Toast.LENGTH_SHORT).show();
                              finish();
                          }else{
                              // Error in registration
                              Toast.makeText(RegisterActivity.this,"User Allready Registered!!!",Toast.LENGTH_LONG).show();
                          }
                      }
                  } catch (JSONException e) {
                      e.printStackTrace();
                  }
              }
        }
          });

因为错误发生在以下行：

  if (json.getString(KEY_SUCCESS) != null)

在JSONParser类中，jObj获取null值。问题在于：jObj = new JSONObject(json);
JSONParser类的代码：

 public class JSONParser {

static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

// constructor
public JSONParser() {

}

public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {

    // Making HTTP request
    try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);
        httpPost.setEntity(new UrlEncodedFormEntity(params));

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "n");
        }
        is.close();
        json = sb.toString();
        Log.e("JSON", json);
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);            
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

    // return JSON String
    return jObj;

}

}

服务调用的另一类UserFunction：公共类UserFunction {

 private JSONParser jsonParser;

  // Testing in localhost using wamp or xampp 
 // use http://10.0.2.2/ to connect to your localhost ie http://localhost/
    private static String loginURL = "http://192.168.1.120/rvAndroidServices.ashx";
private static String registerURL = "http://192.168.1.120/rvAndroidServices.ashx";
private static String name1 = "http://192.168.1.120/rvAndroidServices.ashx";

private static String login_tag = "login";
private static String register_tag = "register";
private static String name_tag = "name";

// constructor
public UserFunction(){
    jsonParser = new JSONParser();
}

/**
 * function make Login Request
 * @param email
 * @param password
 * */
public JSONObject loginUser(String email, String password){
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", login_tag));
    params.add(new BasicNameValuePair("email", email));
    params.add(new BasicNameValuePair("password", password));
    jsonParser= new JSONParser();
    JSONObject json = jsonParser.getJSONFromUrl(loginURL, params);
    // return json
    // Log.e("JSON", json.toString());
    return json;
}

/**
 * function make Login Request
 * @param name
 * @param email
 * @param password
 * */
public JSONObject registerUser(String name, String email, String password){
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", register_tag));
    params.add(new BasicNameValuePair("name", name));
    params.add(new BasicNameValuePair("email", email));
    params.add(new BasicNameValuePair("password", password));
    jsonParser   = new JSONParser();
    // getting JSON Object
    JSONObject json = jsonParser.getJSONFromUrl(registerURL, params);
    // return json
    return json;
}


/**
 * Function get Login status
 * */
public boolean isUserLoggedIn(Context context){
    Databasehandler db = new Databasehandler(context);
    int count = db.getRowCount();
    if(count > 0){
        // user logged in
        return true;
    }
    return false;
}

public String getAppCategorydetail(Context context){
    Databasehandler db = new Databasehandler(context);
    String count = db.getAppCategorydetail();

        return count;

}
/**
 * Function to logout user
 * Reset Database
 * */
public boolean logoutUser(Context context){
    Databasehandler db = new Databasehandler(context);
    db.resetTables();
    return true;
}

public JSONObject chname(String name) 
{
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", name_tag));
    params.add(new BasicNameValuePair("name", name));
     JSONObject json = jsonParser.getJSONFromUrl(name1, params);
    return json;

}

}

Answer 1

在响应字符串中，你得到jsonArray而不是JsonObject，所以当你说

时

try {
    jObj = new JSONObject(json);            
}

这里json需要在JsonArray中转换而不是JSONObject。之后从json数组中获取第一个对象。

类似的东西：

try {
JSONArray jArr = new JSONArray(json);  
JSONObject jObj = jArr.getJSONObject(0);

}

希望这有帮助！

Answer 2

Error parsing data org.json.JSONException: Value [{"userid":105,"created_at":"03-Oct-2013","success":1,"email":"rty@gmail.com","password":"rty12345","name":"rtyu"}] of type org.json.JSONArray cannot be converted to JSONObject

您的例外解释一切

您的字符串为JSONArray not JSONObject，您需要从JSONObject获取JSONArray。

因此，使用JSONArray获取JSONOBbject 更改为：

jObj  = new JSONArray(json).getJSONObject(0);

Answer 3

试试这个......

JSONArray data = jsonObj.getJSONArray("data");

Answer 4

您的网络服务（apis）可能无法将数据返回为＆＃34; json＆＃34; 如果apis是用php编写的 - 尝试在下面添加一行

标题（＆＃39;内容类型：application / json＆＃39;）;

它应该有用。

错误解析数据 - 无法将字符串转换为JSONObject

4 个答案: