在安卓设备上显示我的 php 回声

我想在我的数据库中已经存在用户名或电子邮件时显示我的php回显

$sql = "SELECT * FROM user WHERE username='$username' OR email='$email'";
    $check = mysqli_fetch_array(mysqli_query($con,$sql));
        if(isset($check)){
            echo "username or email already exist";
        }else {
            $statement = mysqli_prepare($con, "INSERT INTO User (username, email, age, password) VALUES (?, ?, ?, ?)");
            mysqli_stmt_bind_param($statement, "ssis", $username, $email, $age, $password);
        }
mysqli_stmt_execute($statement);
mysqli_stmt_close($statement);
mysqli_close($con);

当电子邮件或用户名已存在时，它不会创建帐户，但它不会显示我的回声并退出我的注册活动。

这是我的服务器请求.java

 @Override
    protected Void doInBackground(Void... params) {
        ArrayList<NameValuePair> dataToSend = new ArrayList<>();
        dataToSend.add(new BasicNameValuePair("username", user.username));
        dataToSend.add(new BasicNameValuePair("email", user.email));
        dataToSend.add(new BasicNameValuePair("password", user.password));
        dataToSend.add(new BasicNameValuePair("age", user.age + ""));
        HttpParams httpRequestParams = getHttpRequestParams();
        HttpClient client = new DefaultHttpClient(httpRequestParams);
        HttpPost post = new HttpPost(SERVER_ADDRESS
                + "Register.php");

        try {
            post.setEntity(new UrlEncodedFormEntity(dataToSend));
            client.execute(post);
        } catch (Exception e) {
            e.printStackTrace();
        }
        return null;
    }
    private HttpParams getHttpRequestParams() {
        HttpParams httpRequestParams = new BasicHttpParams();
        HttpConnectionParams.setConnectionTimeout(httpRequestParams,
                CONNECTION_TIMEOUT);
        HttpConnectionParams.setSoTimeout(httpRequestParams,
                CONNECTION_TIMEOUT);
        return httpRequestParams;
    }
    @Override
    protected void onPostExecute(Void result) {
        super.onPostExecute(result);
        progressDialog.dismiss();
        userCallBack.done(null);
    }