php - 如何从 IOS 模拟器访问我的服务器 MySql 以快速检索数据

Question

这段代码可能有什么问题，我正在尝试进入我的已打开的 MAMP 服务器，我在服务器中有一个 php 文件，我正在其中测试连接等等，这个:

<?php
header('Content-type: application/json');
if($_POST) {
    $username   = $_POST['username'];
    $password   = $_POST['password'];
    echo $username
    echo $password

    if($username && $password) {

            $db_name     = 'DBTest';
            $db_user     = 'pedro';
            $db_password = 'pedro';
            $server_url  = 'localhost';

            $mysqli = new mysqli('localhost', $db_user, $db_password, $db_name);

            /* check connection */
            if (mysqli_connect_errno()) {
                error_log("Connect failed: " . mysqli_connect_error());
                echo '{"success":0,"error_message":"' . mysqli_connect_error() . '"}';
            } else {
                if ($stmt = $mysqli->prepare("SELECT username FROM users WHERE username = ? and password = ?")) {

                    $password = md5($password);

                    /* bind parameters for markers */
                    $stmt->bind_param("ss", $username, $password);

                    /* execute query */
                    $stmt->execute();

                    /* bind result variables */
                    $stmt->bind_result($id);

                    /* fetch value */
                    $stmt->fetch();

                    /* close statement */
                    $stmt->close();
                }

                /* close connection */
                $mysqli->close();

                if ($id) {
                    error_log("User $username: password match.");
                    echo '{"success":1}';
                } else {
                    error_log("User $username: password doesn't match.");
                    echo '{"success":0,"error_message":"Invalid Username/Password"}';
                }
            }
    } else {
        echo '{"success":0,"error_message":"Invalid Username/Password."}';
    }
}else {
    echo '{"success":0,"error_message":"Invalid Data."}';
}
?>

而在 Xcode 中，应用程序目前有 3 个 View ，所有 View 都在 swift 中，但重要的是:

//
//  LogInViewController.swift
//  ParkingApp
//
//  Created by Pedro Alonso on 02/06/15.
//  Copyright (c) 2015 Pedro Alonso. All rights reserved.
//

import UIKit

class LogInViewController: UIViewController {

    @IBOutlet weak var loginLabel: UILabel!
    @IBOutlet weak var usernameField: UITextField!
    @IBOutlet weak var passwordField: UITextField!

    override func viewDidLoad() {
        super.viewDidLoad()

        // Do any additional setup after loading the view.
    }

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


    /*
    // MARK: - Navigation

    // In a storyboard-based application, you will often want to do a little preparation before navigation
    override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
        // Get the new view controller using segue.destinationViewController.
        // Pass the selected object to the new view controller.
    }
    */

    @IBAction func enterTapped(sender: UIButton) {

        var username: String = usernameField.text
        var password: String = passwordField.text


        if ( username.isEmpty || password.isEmpty) {

            var alertView: UIAlertView = UIAlertView()
            alertView.title = "Failed"
            alertView.message = "Error in the username or password"
            alertView.delegate = self
            alertView.addButtonWithTitle("Ok")
            alertView.show()


        } else {

            var post: String = "username=\(username)&password=\(password)"

            NSLog("Post data: %@", post)
            println(post)

            var url: NSURL = NSURL(string: "http://localhost:8888/jsonlogin2.php")!

            var postData: NSData = post.dataUsingEncoding(NSASCIIStringEncoding, allowLossyConversion: false)!

            var postLenght: String = String(postData.length)

            var request: NSMutableURLRequest = NSMutableURLRequest(URL: url)

            request.HTTPMethod = "POST"

            request.HTTPBody = postData

            request.setValue(postLenght, forHTTPHeaderField: "Content-Length")

            request.setValue("application/x-form-urlencoded", forHTTPHeaderField: "Content-Type")

            request.setValue("application/json", forHTTPHeaderField: "Accept")

            var responseError: NSError?

            var response: NSURLResponse?

            var urlData: NSData? = NSURLConnection.sendSynchronousRequest(request, returningResponse: &response, error: &responseError)

            if urlData != nil {

                let res = response as! NSHTTPURLResponse!

                println(urlData)

                NSLog("Response code: %ld", res.statusCode)

                if (res.statusCode >= 200 && res.statusCode < 300) {

                    var responseData: NSString = NSString(data: urlData!, encoding: NSUTF8StringEncoding)!

                    NSLog("Response: ==> %@", responseData)

                    var error: NSError?

                    let jsonData: NSDictionary = NSJSONSerialization.JSONObjectWithData(urlData!, options: NSJSONReadingOptions.MutableContainers, error: &error) as! NSDictionary

                    let succes: Int = jsonData.valueForKey("succes") as! Int

                    if succes == 1 {

                        NSLog("Login Success")

                        var prefs: NSUserDefaults = NSUserDefaults.standardUserDefaults()

                        prefs.setObject(username, forKey: "USERNAME")

                        prefs.setInteger(1, forKey: "ISLOGGEDIN")

                        prefs.synchronize()

                        self.dismissViewControllerAnimated(true, completion: nil)


                    } else {

                        var errorMsg: String?

                        if jsonData["error_message"] as? String != nil {

                            errorMsg = jsonData["error_message"] as! String?

                        } else {

                            errorMsg = "Unknown error"
                        }

                        var alertView: UIAlertView = UIAlertView()

                        alertView.title = "Sign in failed"

                        alertView.message = errorMsg
                        alertView.delegate = self
                        alertView.addButtonWithTitle("Ok")
                        alertView.show()


                    }


                } else {

                    var alertView:UIAlertView = UIAlertView()
                    alertView.title = "Sign in Failed!"
                    alertView.message = "Connection Failed"
                    alertView.delegate = self
                    alertView.addButtonWithTitle("OK")
                    alertView.show()
                }


            } else {

                var alertView:UIAlertView = UIAlertView()
                alertView.title = "Sign in Failed!"
                alertView.message = "Connection Failure"
                if let error = responseError {
                    alertView.message = (error.localizedDescription)
                }
                alertView.delegate = self
                alertView.addButtonWithTitle("OK")
                alertView.show()
            }

        }

    }
}

.php 文件在/Applications/MAMP/htdocs 中，不清楚的是为什么给我响应代码 500，我不知道为什么会这样。有帮助吗？？谢谢。

编辑:响应:

<NSHTTPURLResponse: 0x7fc42149ac60> { URL: http://localhost:8888/jsonlogin2.php } { status code: 500, headers {
    Connection = close;
    "Content-Length" = 0;
    "Content-Type" = "text/html; charset=UTF-8";
    Date = "Thu, 04 Jun 2015 12:11:35 GMT";
    Server = "Apache/2.2.29 (Unix) mod_wsgi/3.4 Python/2.7.8 PHP/5.6.7 mod_ssl/2.2.29 OpenSSL/0.9.8zd DAV/2 mod_fastcgi/2.4.6 mod_perl/2.0.8 Perl/v5.20.0";
    "X-Powered-By" = "PHP/5.6.7";
} }

我在模拟器中safari可以访问到localhost:8888，所以没有连接问题。

EDIT2:所以这显然是请求，因为它告诉我无效数据跳过所有并返回此:

2015-06-04 17:16:11.914 ParkingApp[3777:126598] Response: ==> {"success":0,"error_message":"Invalid Data."}

我完成请求的方式可能有什么问题？

EDIT2:我更改了代码并激活了 mysql 日志以查看查询，但 $stmt->get_result() 或 fetch() 仍然没有执行任何操作，我不知道为什么。我没有在整个 IOS 中都这样做，但这里的简单浏览器是麻烦的部分。

修改部分:

$mysqli = new mysqli('localhost', $db_user, $db_password, $db_name);

            /* check connection */
            if (mysqli_connect_errno()) {
                error_log("Connect failed: " . mysqli_connect_error());
                echo '{"success":0,"error_message":"' . mysqli_connect_error() . '"}';
            } else {
                $query = "SELECT dataOne,password FROM users WHERE username = ? and password = ?";

                if ($stmt = $mysqli->prepare($query)) {

                    //$password = md5($password);

                    /* bind parameters for markers */
                    $stmt->bind_param("ss", $username, $password);

                    /* execute query */
                    $stmt->execute();

                    //$stmt->debugDumpParams();
                    echo $stmt->sqlstate;
                    var_dump($stmt);

                    /* bind result variables */
                    //$stmt->bind_result($dataOne,$password);

                    $result = $stmt->get_result();

                    printf("test: ", $dataOne, $password);

                     //fetch value

                    while($stmt->fetch()) {

                        echo $dataOne;

                    }


                    /* close statement */
                    $stmt->close();
                }

                /* close connection */
                $mysqli->close();

                if ($result != null) {
                    error_log("User $username: password match.");
                    echo '{"success":1, "dataOne:"'.$dataOne.'}';
                } else {
                    error_log("User $username: password doesn't match.");
                    echo '{"success":0,"error_message":"Invalid Username/Password"}';
                }
            }

$stmt 没有在 get_result() 上重新调整任何东西，或者没有进入 while(fetch()) 我只是现在不知道。有帮助吗？

Answer 1

好吧，如果您的 Web 服务器抛出 HTTP 错误代码 500(内部错误)，那是因为您的 PHP 脚本崩溃了。我会尝试阅读 php 日志，并尝试对 php 脚本进行一些调试。

也许您的 iOS 应用程序发布的数据有问题，导致 php 脚本失败？

在这种情况下，从 safari 访问 localhost:8888 也不能证明 php 脚本正在运行，因为它需要您发布任何数据才能执行脚本。 if($_POST) { .只需浏览该脚本，if 语句就永远不会是 true .

编辑:

有时一次验证一个组件会有所帮助。尝试构建一个简单的 html 表单来发布 username和 password针对您的服务器 ( http://localhost:8888/jsonlogin2.php )。当您看到它按预期运行时，请继续确保该应用程序正常运行。这样你就可以判断你的错误是在服务器上(php 脚本)还是在你的应用程序中。

查一下也不错$_POST像这样:

if (!empty($_POST)) {}

这将检查 $_POST是空的。

您的应用也在使用 application/x-form-urlencoded我猜这应该是:application/x-www-form-urlencoded .

但又一次。制作一个本地 html 表单，并确保您的 php 脚本正在运行，然后移至该应用程序。