ios - 如何用swift连接mysql？

Question

我有一个网络应用程序，我想制作一个 iOS 应用程序，我不想使用 HTTP 请求，我的网站有自己的数据库(这是一个 MySQL 数据库)。我在谷歌上搜索了很多，但找不到适合我的解决方案。你们以前有人做过吗？

Answer 1

将 swift 连接到 mysql 和 php 非常容易。首先，您需要一个 REST API。您可以使用任何可用的框架来创建 rest api。您还可以仅使用 PHP 编写 Web 服务代码。所以在这里我将展示任何php框架的使用。

所以首先创建一个文件来存储你的数据库常量。

<?php
/**
 * Created by PhpStorm.
 * User: Belal
 * Date: 12/08/16
 * Time: 7:58 PM
 */

define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');
define('DB_NAME', 'iphone');

然后创建另一个php文件来创建数据库连接。

<?php

class DbConnect
{
    private $conn;

    function __construct()
    {
    }

    /**
     * Establishing database connection
     * @return database connection handler
     */
    function connect()
    {
        require_once 'Config.php';

        // Connecting to mysql database
        $this->conn = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);

        // Check for database connection error
        if (mysqli_connect_errno()) {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }

        // returing connection resource
        return $this->conn;
    }
}

现在您还需要一个文件来处理您的数据库操作。

<?php

class DbOperation
{
    private $conn;

    //Constructor
    function __construct()
    {
        require_once dirname(__FILE__) . '/Config.php';
        require_once dirname(__FILE__) . '/DbConnect.php';
        // opening db connection
        $db = new DbConnect();
        $this->conn = $db->connect();
    }

    //Function to create a new user
    public function createTeam($name, $memberCount)
    {
        $stmt = $this->conn->prepare("INSERT INTO team(name, member) values(?, ?)");
        $stmt->bind_param("si", $name, $memberCount);
        $result = $stmt->execute();
        $stmt->close();
        if ($result) {
            return true;
        } else {
            return false;
        }
    }

}

最后，您需要创建将处理您的 http 请求的 php 文件。

<?php

//creating response array
$response = array();

if($_SERVER['REQUEST_METHOD']=='POST'){

    //getting values
    $teamName = $_POST['name'];
    $memberCount = $_POST['member'];

    //including the db operation file
    require_once '../includes/DbOperation.php';

    $db = new DbOperation();

    //inserting values 
    if($db->createTeam($teamName,$memberCount)){
        $response['error']=false;
        $response['message']='Team added successfully';
    }else{

        $response['error']=true;
        $response['message']='Could not add team';
    }

}else{
    $response['error']=true;
    $response['message']='You are not authorized';
}
echo json_encode($response);

现在只需在您的 iOS 应用程序上创建 View ，然后在单击按钮时向您的 php 文件发送请求。代码如下。

//
//  ViewController.swift
//  SwiftPHPMySQL
//
//  Created by Belal Khan on 12/08/16.
//  Copyright © 2016 Belal Khan. All rights reserved.
//

import UIKit

class ViewController: UIViewController {

    //URL to our web service
    let URL_SAVE_TEAM = "http://192.168.1.103/MyWebService/api/createteam.php"


    //TextFields declarations
    @IBOutlet weak var textFieldName: UITextField!
    @IBOutlet weak var textFieldMember: UITextField!



    //Button action method
    @IBAction func buttonSave(sender: UIButton) {

        //created NSURL
        let requestURL = NSURL(string: URL_SAVE_TEAM)

        //creating NSMutableURLRequest
        let request = NSMutableURLRequest(URL: requestURL!)

        //setting the method to post
        request.HTTPMethod = "POST"

        //getting values from text fields
        let teamName=textFieldName.text
        let memberCount = textFieldMember.text

        //creating the post parameter by concatenating the keys and values from text field
        let postParameters = "name="+teamName!+"&member="+memberCount!;

        //adding the parameters to request body
        request.HTTPBody = postParameters.dataUsingEncoding(NSUTF8StringEncoding)


        //creating a task to send the post request
        let task = NSURLSession.sharedSession().dataTaskWithRequest(request){
            data, response, error in

            if error != nil{
                print("error is \(error)")
                return;
            }

            //parsing the response
            do {
                //converting resonse to NSDictionary
                let myJSON =  try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary

                //parsing the json
                if let parseJSON = myJSON {

                    //creating a string
                    var msg : String!

                    //getting the json response
                    msg = parseJSON["message"] as! String?

                    //printing the response
                    print(msg)

                }
            } catch {
                print(error)
            }

        }
        //executing the task
        task.resume()

    }


    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }


}

您需要做的另一件事是在您的 Info.plist 文件中添加以下行，这是因为默认情况下您不能将请求发送到不安全的 url，所以因为我们有 http 我们必须做最后一件事。

<!-- add from here -->
    <key>NSAppTransportSecurity</key>
    <dict>
        <key>NSAllowsArbitraryLoads</key>
        <true/>
        <key>NSExceptionDomains</key>
        <dict>
            <key>yourdomain.com</key>
            <dict>
                <key>NSIncludesSubdomains</key>
                <true/>
                <key>NSThirdPartyExceptionRequiresForwardSecrecy</key>
                <false/>
            </dict>
        </dict>
    </dict>
    <!-- end of the code -->

来源: iOS MySQL Database Tutorial