PHP＆amp; Android自定义登录无法正常工作

Question

Registracion部分工作得很好（不包括在这里）， 但是当我点击登录按钮（bLogin）时，它给了我这个：

org.json.JSONException: No value for name

这是LoginActivity.class：

public class LoginActivity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);

        final EditText etUsername = (EditText) findViewById(R.id.etUsername);
        final EditText etPassword = (EditText) findViewById(R.id.etPassword);
        final TextView tvRegisterLink = (TextView) findViewById(R.id.tvRegisterLink);
        final Button bLogin = (Button) findViewById(R.id.bSignIn);

        tvRegisterLink.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                Intent registerIntent = new Intent(LoginActivity.this, RegisterActivity.class);
                LoginActivity.this.startActivity(registerIntent);
            }
        });

        bLogin.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                final String username = etUsername.getText().toString();
                final String password = etPassword.getText().toString();

                // Response received from the server
                Response.Listener<String> responseListener = new Response.Listener<String>() {
                    @Override
                    public void onResponse(String response) {
                        try {
                            JSONObject jsonResponse = new JSONObject(response);
                            boolean success = jsonResponse.getBoolean("success");

                            if (success) {
                                String name = jsonResponse.getString("name");
                                int age = jsonResponse.getInt("age");

                                Intent intent = new Intent(LoginActivity.this, UserAreaActivity.class);
                                intent.putExtra("name", name);
                                intent.putExtra("age", age);
                                intent.putExtra("username", username);
                                LoginActivity.this.startActivity(intent);
                            } else {
                                AlertDialog.Builder builder = new AlertDialog.Builder(LoginActivity.this);
                                builder.setMessage("Login Failed")
                                        .setNegativeButton("Retry", null)
                                        .create()
                                        .show();
                            }

                        } catch (JSONException e) {
                            e.printStackTrace();
                        }
                    }
                };

                LoginRequest loginRequest = new LoginRequest(username, password, responseListener);
                RequestQueue queue = Volley.newRequestQueue(LoginActivity.this);
                queue.add(loginRequest);
            }
        });
    }
}

我认为错误将在php部分，但我无法找到它的确切位置。 这是login.php：

        <?php
  $con = "Connection works fine.";

$username = $_POST["username"];
$password = $_POST["password"];

$statement = mysqli_prepare($con, "SELECT * FROM user WHERE username = ? AND password = ?");
mysqli_stmt_bind_param($statement, "ss", $username, $password);
mysqli_stmt_execute($statement);

mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $userID, $name, $username, $age,$password);

$response = array();
$response["success"] = false;

while(mysqli_stmt_fetch($statement)){
    $response["success"] = true;
    $response["name"] = $name;
    $response["age"] = $age;
    $response["username"] = $username;
    $response["password"] = $password;
}

echo json_encode($response);
?>