无法使用swift连接到MYSQL数据库

Question

这是我的快捷代码：

import UIKit

class SignUpViewController: UIViewController {

@IBOutlet weak var emailField: UITextField!
@IBOutlet weak var nameField: UITextField!
@IBOutlet weak var pwdField: UITextField!
@IBOutlet weak var pwdAgain: UITextField!

@IBAction func SignUpTapped(_ sender: UIButton) {
    if pwdField.text == pwdAgain.text {
    var request = URLRequest(url: NSURL(string: "sftp://....compute.amazonaws.com/var/www/html/register.php") as! URL) //... is hide my user name
    request.httpMethod = "POST"
    let postString = "email=\(emailField.text!)&name=\(nameField.text)&password=\(pwdField.text)"
    request.httpBody = postString.data(using: String.Encoding.utf8)

    let task = URLSession.shared.dataTask(with: request) { (data, response, error) in
        if error != nil{
            print("response = \(error)")
            return
        }
        print("response = \(response)")

        let responseString = NSString(data: data!, encoding: String.Encoding.utf8.rawValue)
        print("responseString = \(responseString)")
    }
    task.resume()
    }else if pwdAgain.text == "" || pwdField.text == "" || emailField.text == "" || nameField.text == ""{
        let alert = UIAlertController(title: "Notification", message: "Please fill all block!", preferredStyle: UIAlertControllerStyle.alert)
        self.present(alert, animated: true, completion: nil)

    }else{
        let alert = UIAlertController(title: "Notification", message: "Please enter same password twice!", preferredStyle: UIAlertControllerStyle.alert)
        self.present(alert, animated: true, completion: nil)
    }
}

override func viewDidLoad() {
    super.viewDidLoad()

    }
}

和我的php代码：register.php

<?php
include 'dbConnect.php';

$email = $_POST['email'];
$name = $_POST['name'];
$pwd = $_POST['password'];
$id = date('mdHis');

$checkEmail = false;
$checkName = false;
$dbInsert = false;

if ($conn->query("SELECT * FROM fyp.blogger WHERE email='$email'")-   >num_rows == 0) {
    $checkEmail = true;
}

if ($conn->query("SELECT * FROM fyp.blogger WHERE name='$name'")->num_rows == 0) {
    $checkName = true;
}

if ($checkName && $checkEmail) {
    $dbInsert = $conn->query("INSERT INTO fyp.blogger (blogger_id,email,name,pwd) VALUES ($id, '$email','$name', '$pwd')");
}
$conn->close();
?>

问题是我尝试测试连接，就像阻止我一样。

以下错误：

有关于我使用sftp的事吗？我只想将一些数据保存到MYSQL数据库 并读取数据。 我希望任何人都可以解决问题。