问题描述
我想在Flash中创建一个用户登录系统,但我需要通过PHP才能这样做的时候到MySQL。我看了看周围的一些教程,但我一直得到错误。
I am trying to create a user log-in system in Flash but I need to communicate to MySQL through PHP in order to do so. I looked around at a few tutorials, but I have been getting errors.
下面是我在Flash制作的界面,我试图用controlpanel.php沟通
Here is the interface I have made in flash, which I am trying to communicate with controlpanel.php
http://i.imgur.com/JrTWm.png?1
下面是code
AS文件
package actions
{
import flash.display.MovieClip;
import flash.events.MouseEvent;
import flash.net.URLRequest;
import flash.net.URLVariables;
import flash.net.URLLoader;
import flash.events.Event;
import flash.net.URLLoaderDataFormat;
import flash.net.URLRequestMethod;
import flash.text.TextFieldAutoSize;
public class main extends MovieClip
{
public function main():void
{
submit_button.buttonMode = true;
submit_button.addEventListener(MouseEvent.MOUSE_DOWN, checkLogin)
username.text = "";
password.text = "";
}
public function checkLogin(e:MouseEvent):void
{
if(username.text==""||password.text=="")
{
if (username.text == "")
{
username.text = "Enter your username";
}
if (password.text == "")
{
password.text="Enter your password";
}
}
else
{
processLogin();
}
}
public function processLogin():void
{
var phpVars:URLVariables = new URLVariables();
var phpFileRequest:URLRequest = new URLRequest("http://mathlympics.cu.cc/php/controlpanel.php");
phpFileRequest.method = URLRequestMethod.POST;
phpFileRequest.data = phpVars;
var phpLoader:URLLoader=new URLLoader();
phpLoader.dataFormat = URLLoaderDataFormat.VARIABLES;
phpVars.systemCall = "checkLogin";
phpVars.username = username.text;
phpVars.password = password.text;
phpLoader.load(phpFileRequest);
phpLoader.addEventListener(Event.COMPLETE, showResult);
}
public function showResult(event:Event):void
{
result_text.autoSize = TextFieldAutoSize.LEFT;
result_text.text = ""+ event.target.data.systemResult;
}
}
}
PHP文件 - 将connect.php
PHP file - connect.php
<?php
$db_username = "censored";
$db_name = "censored";
$db_password = "censored";
$db_host = "mysql2.000webhost.com";
mysql_connect($db_host,$db_username, $db_password, $db_name);
mysql_select_db($db_name) or die (mysql_error());
?>
PHP文件 - controlpanel.php
PHP file - controlpanel.php
<?php
error_reporting(E_ALL);
include_once("connect.php");
$username = "admin";
$password = "password";
$sql = "SELECT * FROM user WHERE username='$username' AND password='$password'";
$query = mysql_query($sql);
$login_counter = mysql_num_rows($query);
if ($login_counter > 0)
{
$data = mysql_fetch_array($query);
$userbio = $data["user_bio"];
echo "systemResult=" . $userbio;
}
else
{
echo "systemResult=Invalid";
}
?>
我没有得到任何错误,但是当我preSS提交按钮,是说不确定的结果文本框中,即使当我输入正确的用户名和密码。
I do not get any error but when I press the submit button, is says undefined in the result text box, even when I type the right username and password.
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为那些你感兴趣的是:这是我的网站这里输入链接的描述
To those of you interested: Here is my website enter link description here
推荐答案
所以,问题最终被你有 SELECT * FROM用户...
时,它应该是 SELECT * FROM用户...
=)
So the problem ended up being you had SELECT * FROM user...
when it should have been SELECT * FROM users...
=)
这篇关于什么是不对的AS3 code与PHP文件进行通信?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!