解析数据 org.json.JSONException 时出错：在字符 10 处输入结束

Question

我目前遇到此错误“解析数据时出错 org.json.JSONException: End of input at character 10 of”。我用 Chrome 的 Plugin Postman 测试了我的 PHP，服务器端似乎没问题。请查看下面的代码，感谢您的帮助。

PHP 代码：

public function login($alias, $password){
$user_info = $this->getUserFromDatabase($alias, $password);
if ($user_info != false){
$response["success"] = "true";
$response["user_id"] = $user_info["userID"];
$response["userFirstName"] = $user_info["userFirstName"];
$response["userRank"] = $user_info["userRank"];
echo json_encode($response);
}else{
$response["success"] = "false";
$response["error"] = "true";
echo json_encode($response);
}
}

public function getUserFromDatabase($android_alias, $android_password) {
$db_query = mysql_query("SELECT userID, userFirstName, userRank FROM capUserTable 
WHERE userAlias = '$android_alias' AND userPassword = '$android_password'") 
or die(mysql_error());

$query_results = mysql_fetch_assoc($db_query);
return $query_results;
}

邮递员结果：

{
"success": "true",
"user_id": "1",
"userFirstName": "username",
"userRank": "99"
}

JAVA代码：

private InputStream inputStream = null;
private JSONObject jObject = null;
private String json = "";

public JSONparser() {
}

public JSONObject getJSONFromURL(String URL, List<NameValuePair> params){
try{
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(URL);
httpPost.setEntity(new UrlEncodedFormEntity(params)); 
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
inputStream = httpEntity.getContent();
}catch(UnsupportedEncodingException e){
e.printStackTrace();
Log.e("UnsupportedEncodingException", "Unsupported Encoding Exception" + e.toString());
}catch(ClientProtocolException e){
e.printStackTrace();
}catch(IOException e){
e.printStackTrace();
}catch(Exception e){
e.printStackTrace();
}

try{
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"), 10);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null){
sb.append(line + "\n");
}
inputStream.close();
json = sb.toString();
}catch(Exception e){
Log.e("Buffer Error", "Error converting result " + e.toString());
}

try{
jObject = new JSONObject(json); //this is where the problem occurs
}catch(JSONException e){
Log.e("JSON Parser", "Error parsing data " + e.toString()); 
}
return jObject;
}

Answer 1

其实我之前也遇到过同样的问题.... 我通过更改我的 jason 类来清除错误...

 // function get json from url
  // by making HTTP POST or GET mehtod
  public JSONObject makeHttpRequest(String url, String method,
                  List<NameValuePair> params) {

          // Making HTTP request
          try {

                  // check for request method
                  if(method == "POST"){
                          // request method is POST
                          // 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();

                  }else if(method == "GET"){

                          // request method is GET
                          DefaultHttpClient httpClient = new DefaultHttpClient();
                          String paramString = URLEncodedUtils.format(params, "utf-8");
                          url += "?" + paramString;
                          HttpGet httpGet = new HttpGet(url);

                          HttpResponse httpResponse = httpClient.execute(httpGet);
                          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();
          } catch (Exception e) {
                  Log.e("Buffer Error", "Error converting result " + e.toString());
          }

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

          // return JSON String
          return jObj;

  } 

希望对你有帮助

Answer 2

您好，我已经检查了您的回复得到一个问题

String response = "{success\":\"true\",\"user_id\": \"1\",\"userFirstName\":\"username\", \"userRank\": \"99\"}";

它在 json 验证中显示有效但在成功键中不是 " 开始所以它创建 json 对象，如

{"userFirstName":"username","user_id":"1","success\"":"true","userRank":"99"}

所以它无法从成功中获取值

像这样写

String response = "{\"success\":\"true\",\"user_id\": \"1\",\"userFirstName\":\"username\", \"userRank\": \"99\"}";

它应该像这样解析

         try {
            JSONObject  jobj  = new JSONObject(response);
            String succes = jobj.getString("success");
            String userFirstName = jobj.getString("userFirstName");
            String user_id = jobj.getString("user_id");
            String userRank = jobj.getString("userRank");
            String user_id = jobj.getString("user_id"); 

        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }