最近因为开发一个AR的项目，客户要求限制账号密码登陆，所以研究了一下远程服务器的登陆，本来对程序了解的不是很多，在这里总结一下这两天的工作。

工具：

1.unity

2.xampp(注：写的PHP文档需要放在C:\xampp\htdocs文件中)

先搭建和服务器的连接,

PHP如下

$myData=mysqli_connect( "localhost" ,"root" ,"" );//localhost本地服务器 登陆账户为root 密码为空

if ( mysqli_connect_errno())

{

echo "eron";

return;

}

// 选择数据库

mysqli_query($myData,"set names utf8") ;

mysqli_select_db($myData ,"Cool_yt_rgb" );//服务器数据库名字cool_yt_rgb

// 寻

$userID = $_POST['usernamePost'];//$_POST 后面的名字是在unity里面需要调用的

$userPWD = $_POST['passwordPost'];

//从cool_yt_rgb数据库建立的表中选择userinfo表的username和passport

$sql = "SELECT username FROM userinfo where username='".$userID."' and password ='".$userPWD."' ";

$result = mysqli_query($myData,$sql)or die("SQL error!");

$num_results = mysqli_num_rows($result);

echo $num_results;

?>

UNITY脚本如下

using System.Collections;

using System.Collections.Generic;

using UnityEngine;

using UnityEngine.UI;

public class LoginURL : MonoBehaviour

{

public InputField userNameField;//登陆用户名 public InputField passWordField;//登录密码 public Text feedbackmsg;//实时返回状态文字

string LoginsURL = "http://localhost/cool_yt_rgb/LoginURL.php";//服务器连接

public void isLoad()

{

string idText = userNameField.text;

string pwdText = passWordField.text;

StartCoroutine(LoginToDB(idText, pwdText));

}

IEnumerator LoginToDB (string username, string password)

{

WWWForm form = new WWWForm();

form.AddField("usernamePost", username);

form.AddField("passwordPost", password);

WWW www = new WWW(LoginsURL, form);

yield return www;

if (www.text == "1")

{

Application.LoadLevel("111");

}

else

{

feedbackmsg.CrossFadeAlpha(100f, 0f, false);

feedbackmsg.color = Color.red;

feedbackmsg.text = "Login Error";

}

}

}

然后做好登录界面

关于把SQL上传服务器的工作还没做 暂时先到这一步

关于给服务器增加数据可以做一个注册的连接，PHP如下

//Variable for the connection

$servername = "localhost";

$server_username = "root";

$server_password = "";

$dbName = "Cool_yt_rgb";

//Variable from the user

$username = 'nihap';//$_POST["usernamePost"];//在数据库中增加表的名字

$password = '456789';//$_POST["passwordPost"];//在数据库中增加表的密码

//Make Connection

$conn = new mysqli($servername, $server_username, $server_password, $dbName);

//Check Connection

if(!$conn){

die("Connection Failed. ". mysqli_connect_error());

}

$sql = "INSERT INTO userinfo (username,password)

VALUES ('".$username."','".$password."')";

$result = mysqli_query($conn ,$sql);

if(!result) echo "threre was an error";

else echo "Everything ok";

?>