Android应用程序中的JSON对象请求错误

Question

我正在开发针对android的应用程序，现在我正在执行用户“CreateAccount”和“Login”活动。我能够将用户的信息插入数据库，但我无法从数据库获取用户的信息来执行登录。这是我的Java代码：

String showUrl = "http://192.168.0.11/webapps/showUser.php"; 
public void searchLoginInfo(View view) { 
    JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.POST,showUrl,null, new Response.Listener<JSONObject>() 
     { 
     @Override 
     public void onResponse(JSONObject response) { 
      try { 
       JSONArray users = response.getJSONArray("users"); 
       for (int i = 0; i < users.length(); i++) { 
        JSONObject user = users.getJSONObject(i); 
        String username = user.getString("username").toLowerCase(); 
        String password = user.getString("password").toLowerCase(); 
        String retypedPassword = user.getString("retypedpassword").toLowerCase(); 
        String email = user.getString("passphrase").toLowerCase(); 
       } 
      } catch (JSONException e) { 
       e.printStackTrace(); 
      } 
     } 
    }, new Response.ErrorListener() 
     { 
     @Override 
     public void onErrorResponse(VolleyError error) { 
      Log.e(TAG, "onErrorResponse: ERORR!!!!!!!!!!!"); 
      Log.e(TAG, "onErrorResponse: ERORR!!!!!!!!!!!"+error.toString()); 
     } 
    }); 
    requestQueue.add(jsonObjectRequest); 
} 

的错误，我得到的是：E/MyActivity：onErrorResponse：ERORR !!!!!!!!!!! E/MyActivity：onErrorResponse：ERORR !!!!!!!!!!! com.android.volley.ParseError：org.json.JSONException：Value java.lang.String类型的连接无法转换为JSONObject。 这是我showUser.php脚本：

<?php 
if($_SERVER["REQUEST_METHOD"]=="POST"){ 
include 'connection.php'; 
showUser(); 
} 
function showUser(){ 
global $connect; 
$query="SELECT* FROM user;"; 
$result=mysqli_query($connect,$query); 
$number_of_rows=mysqli_num_rows($result); 
$temp_array=array(); 
if($number_of_rows>0){ 
    while($row=mysqli_fetch_assoc($result)){ 
     $temp_array[]=$row; 
    } 
} 
header('Content-Type:application/json'); 
echo json_encode(array("users"=>$temp_array)); 
mysqli_close($connect); 
} 
?> 

从我有红色的在互联网上有可能是与归类的问题？我的表排序和列是utfmb4_general_ci和我的服务器连接排序规则是utfmb4_unicode_ci PrintScreen

我也使用POSTMAN来查看服务器的实际响应，它对我来说看起来不错。

{ 
    "users": [ 
{ 
    "username": "charlie", 
    "password": "harris", 
    "retypedpassword": "harris", 
    "email": "charlie@yahoo.com" 
}, 
{ 
    "username": "tomas", 
    "password": "kovtun", 
    "retypedpassword": "kovtun", 
    "email": "tkovtu@yahoo.com" 
} 
] 
} 

这个问题真的让我失望，所以任何帮助将不胜感激！谢谢！

Answer 1

使用此：

public void searchLoginInfo(View view) { 
StringRequest stringRequest = new StringRequest(Request.Method.POST,showUrl, new Response.Listener<String>() 
    { 
    @Override 
    public void onResponse(String response) { 
     try { 
      response = response.replaceAll("[^\\p{ASCII}]", ""); 
      JSONObject jsonObject = new JSONObject(response); 
      JSONArray users = jsonObject.getJSONArray("users"); 
      for (int i = 0; i < users.length(); i++) { 
       JSONObject user = users.getJSONObject(i); 
       String username = user.getString("username").toLowerCase(); 
       String password = user.getString("password").toLowerCase(); 
       String retypedPassword = user.getString("retypedpassword").toLowerCase(); 
       String email = user.getString("passphrase").toLowerCase(); 
      } 
     } catch (JSONException e) { 
      e.printStackTrace(); 
     } 
    } 
}, new Response.ErrorListener() 
    { 
    @Override 
    public void onErrorResponse(VolleyError error) { 
     Log.e(TAG, "onErrorResponse: ERORR!!!!!!!!!!!"); 
     Log.e(TAG, "onErrorResponse: ERORR!!!!!!!!!!!"+error.toString()); 
    } 
}); 
requestQueue.add(stringRequest); 
}