错误解析数据-字符串无法转换为JSONObject

问题描述 投票：3回答：4

我想通过Web服务访问Android应用。在Web服务中，执行新的注册。在android应用中，进行新注册的xml文件。数据成功保存在SQL Server数据库中，并通过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，它给我这样的错误：

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

我已经进行了注册为RegisterActivity.java的活动

              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();
                  }
              }
        }
          });

由于在行上发生错误：

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

在JSONParser类中，jObj获得空值。问题出在下面：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;

}

}

UserFunction的另一类，其中服务调用： 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;

}

}
android json jsonobject
4个回答
2
投票
Error parsing data org.json.JSONException: Value [{"userid":105,"created_at":"03-Oct-2013","success":1,"email":"[email protected]","password":"rty12345","name":"rtyu"}] of type org.json.JSONArray cannot be converted to JSONObject

您的例外向您解释了一切

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

因此，请使用JSONArray获得JSONOBbject 更改为：

jObj  = new JSONArray(json).getJSONObject(0);
3
投票

在响应字符串中，您获取的是jsonArray而不是JsonObject，所以当您说

try {
    jObj = new JSONObject(json);            
}

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

类似：

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

希望这会有所帮助！

0
投票

尝试一下...

JSONArray data = jsonObj.getJSONArray("data");
0
投票

[您的网络服务（apis）可能不会以“ json”的形式返回数据如果API是用php编写的-请尝试在下面添加行

header（'Content-Type：application / json'）;

它应该工作。

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