php - 从 php 到 flash 的返回变量未定义

Question

当我运行脚本时，除了返回到闪存的变量未定义之外，一切正常。我知道以前曾有人问过这个问题，但我查看了我能找到的有关此问题的每个论坛，但没有一个解决方案有效。当我跟踪变量时，它们都返回“未定义”

ActionScript 3:

function CheckRegistration(event:Event):void{
    var ServerCheck:URLRequest = new URLRequest("http:mywebsite/phpfile.php");
    var CrossCheck:URLVariables = new URLVariables;
    var ServerReply:URLLoader = new URLLoader;
        ServerCheck.data = CrossCheck;
        ServerCheck.method = URLRequestMethod.POST;
        ServerReply.dataFormat = URLLoaderDataFormat.VARIABLES;
        ServerReply.addEventListener(Event.COMPLETE, CheckResult);
        CrossCheck.Requester = String("request");
        CrossCheck.InFirstName = String(FirstName);
        CrossCheck.InLastName = String(LastName);
        CrossCheck.InDeviceID = String(DeviceID);
        ServerReply.load(ServerCheck);
        count++
    if(count >= 3){
        this.removeEventListener(Event.ENTER_FRAME, CheckRegistration);
    }
    trace("Checking Registration")
}

function CheckResult(event:Event):void{
    var CheckVar = String(event.target.data.registration);
    if(CheckVar == "confirmed"){
        trace(CheckVar);
        RegisteredDevice = true;
        this.gotoAndStop(2);
    }else{
        trace(CheckVar);
        this.gotoAndStop(3);
    }
}

PHP:

<?php 
$link = mysql_connect('myserver', 'my_DB', 'mypassword'); 
if (!$link) { 
    die('Connection Failure, Contact FS support' . mysql_error()); 
} 
echo 'Connection Secure'; 
mysql_select_db(my_DB);

if($_POST['Requester'] == "request"){
    $DeviceID = $_POST['InDeviceID'];
    $LastName = $_POST['LastName'];
    $FirstName = $_POST['FirstName'];
    $success = "registration=confirmed";
    $failure = "registration=failure";

    $results = mysql_query("SELECT * FROM myInfo");
    while($row = mysql_fetch_array($results)){
         $RecFirstName = $row['FirstName'];
         $RecLastName = $row['LastName'];
         $RecDeviceID = $row['DeviceID'];
    }

    if(($RecFirstName == $FirstName) || ($RecLastName == $LastName) || ($RecDeviceID == $DeviceID)){
        print "registration=confirmed";
    }else{
        print "registration=failure";   
    }
    mysql_close();
}else{
    print "registration=failure";   
}
?>

Answer 1

我找出问题所在了。我跟踪了 event.target.data 并发现 echo 'Connection Secure'; 添加到了返回 URL 中。我删除了这个并且它起作用了。