Android Volley字符串响应未转换为JsonObject

时间:2018-06-29 12:12:55

标签: php android json string android-volley

我试图从服务器的 String Response 中获取数据(在PHP上运行服务器后端并发送JsonObject响应),然后将其转换为 JsonObject 以获取数据正确地。但是 Android Volley 显示以下错误:

org.json.JSONException:类型java.lang.String的值db无法转换为JSONObject

我的代码如下

     public void loginFunction(String a, String b) {

        progressDialog.setMessage("Please Wait, We are Inserting Your Data on Server");
        progressDialog.show();

        final String user_email_insert = a;
        final String user_password_insert = b;
        String HttpUrl = getString(R.string.server_name)+getString(R.string.log_in_api);

        StringRequest stringRequest = new StringRequest(Request.Method.POST, HttpUrl,
                new Response.Listener<String>() {
                    @Override
                    public void onResponse(String ServerResponse) {

                        progressDialog.dismiss();
                        try {

                            JSONObject person = new JSONObject(ServerResponse);
                            String name = person.getString("user_full_name");
                            String email = person.getString("user_email");
                            Log.d("1111",person.toString());

                            Toast.makeText(getApplicationContext(), name+"\n"+email, Toast.LENGTH_LONG).show();

                        } catch (JSONException e) {
                            e.printStackTrace();
                            Toast.makeText(getApplicationContext(), "Error: " + e.getMessage(), Toast.LENGTH_LONG).show();
                        }


                    }
                },
                new Response.ErrorListener() {
                    @Override
                    public void onErrorResponse(VolleyError volleyError) {


                        progressDialog.dismiss();

                        Toast.makeText(LogIn.this, volleyError.toString(), Toast.LENGTH_LONG).show();
                    }
                }) {
            @Override
            protected Map<String, String> getParams() {


                Map<String, String> params = new HashMap<String, String>();

                params.put("user_email", user_email_insert);
                params.put("user_password", user_password_insert);


                return params;
            }

        };


        RequestQueue requestQueue = Volley.newRequestQueue(LogIn.this);
        requestQueue.add(stringRequest);

    } 

PHP代码是:

        <?php 



    include "db_connection.php";

   $user_email = $_POST["user_email"];

   $user_password = $_POST["user_password"];


   $sql = "SELECT user_password,user_full_name,user_email FROM user_info WHERE user_email ='" .$user_email. "'";
   mysqli_query($con,'SET CHARACTER SET utf8');
   mysqli_query($con,"SET SESSION collation_connection ='utf8_general_ci'");

   $res = mysqli_query($con,$sql);

   $row = mysqli_fetch_row($res);

   if($user_password == $row[0]){

   // echo "Login successfull".$row[1];

$myObj->user_full_name = $row[1];
$myObj->user_email = $row[2];


$myJSON = json_encode($myObj,JSON_UNESCAPED_UNICODE);

echo $myJSON;



     //echo "Login successfull";



   }else{

     echo "Wrong password";
     //header( 'Location: login_failed.php' ) ;

   }



   mysqli_close($con);




  ?> 

0 个答案:

没有答案