从 Flash AS3 向 PHP 发送和接收数据 [英] Sending and receiving data from Flash AS3 to PHP

问题描述

我知道这是经常被问到的问题，但我已经浏览了整个互联网，以找出我在用于从 AS3 向 PHP 发送和接收数据的代码中犯的错误，反之亦然.你能找出错误吗?这是我的代码:

I know this is frequently asked, but I have looked all over the internet to find the mistake I'm making with the code I've used to send and receive data from AS3 to PHP and viceversa. Can you find the mistake? Here is my code:

AS3:

import flash.events.MouseEvent;
import flash.net.URLLoader;
import flash.net.URLRequest;
import flash.net.URLVariables;
import flash.net.URLLoaderDataFormat;
import flash.net.URLRequestMethod;
import flash.events.Event;

submitbtn.addEventListener(MouseEvent.CLICK, sendData)

function sendData(event:MouseEvent):void
{
    var loader : URLLoader = new URLLoader;
    var urlreq:URLRequest = new URLRequest("http://[mydomain]/test.php");
    var urlvars: URLVariables = new URLVariables;
    loader.dataFormat = URLLoaderDataFormat.VARIABLES;
    urlreq.method = URLRequestMethod.POST;
    urlvars.uname = nametxt.text;
    urlvars.apellido = aptxt.text;
    urlvars.email = emtxt.text;
    urlvars.cedula = cctxt.text;
    urlvars.score = scoretxt.text;
    urlreq.data = urlvars;
    loader.addEventListener(Event.COMPLETE, completed);
    loader.load(urlreq);
}

function completed(event:Event): void
{
    var loader2: URLLoader = URLLoader(event.target);
    trace(loader2.data.done);
    resptxt.text = event.target.data.done;
}

[域]/test.php 中的 PHP:

<?php
    $username = $_POST["uname"];
    $apellido = $_POST["apellido"];
    $cedula = $_POST["cedula"];
    $email = $_POST["email"];
    $score = $_POST["score"];
    print_r($_POST);
    if (!($link=mysql_connect(databasemanager,username,password))) 
       { 
          echo "Error conectando a la base de datos."; 
          exit(); 
       } 
       if (!mysql_select_db(database,$link)) 
       { 
          echo "Error seleccionando la base de datos."; 
          exit(); 
       }
       try
       {
           mysql_query("insert into scores(name,lastName,email,document,score) values('$username','$apellido','$email','$cedula','$score')",$link);                
           print "done=true";          
       }
       catch(Exception $e)
       {
           print "done=$e->getMessage()";          
       }
       echo "done=true";    
?>

感谢您的回答.

推荐答案

您的 AS 代码似乎是正确的.所以问题可能出在 PHP 上.请先用这个 PHP 文件进行测试:

Your AS code seems to be right. So the problem might be in PHP. Please test first with this PHP file:

<?php
       echo "test=1&done=true";    
?>

这应该让您的电影跟踪 "true".然后您应该调试您的 PHP.print_r($_POST); 当然会破坏你的输出.可能你忘记删除这个调试语句了:-)

This should then let your movie trace "true". You then should debug your PHP. print_r($_POST); destroys your output of course. May be you did forget to remove this debugging statement :-)

@Jesse 和 @Ascension Systems，检查 URLVariables 的文档:http:///livedocs.adobe.com/flash/9.0_de/ActionScriptLangRefV3/flash/net/URLVariables.html

@Jesse and @Ascension Systems, check the docs for URLVariables: http://livedocs.adobe.com/flash/9.0_de/ActionScriptLangRefV3/flash/net/URLVariables.html

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