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我想從PHP傳遞一些變量閃爍,使用此ActionScript代碼IM:回聲不返回PHP的動作連接變量值
public function gameOver(score:Number)
{
totalScore.text = score.toString();
var scriptVars:URLVariables = new URLVariables();
scriptVars.score = score;
var scriptLoader:URLLoader = new URLLoader();
var scriptRequest:URLRequest = new URLRequest("checkScores.php");
scriptRequest.method = URLRequestMethod.POST;
scriptRequest.data = scriptVars;
scriptLoader.load(scriptRequest);
scriptLoader.addEventListener(Event.COMPLETE, handleLoadSuccessful);
scriptLoader.addEventListener(IOErrorEvent.IO_ERROR, handleLoadError);
}
function handleLoadSuccessful(e:Event):void
{
trace("Scores Loaded");
var vars:URLVariables = new URLVariables(e.target.data);
nickname1.text = vars.nickname;
score1.text = vars.score;
}
function handleLoadError($evt:IOErrorEvent):void
{
trace("Load failed.");
nickname1.text ="error";
}
這PHP代碼:
<?php
... some code for the mysql connection and select sentence ...
$topScores = mysqli_query($con, $topScores);
$topScores = mysqli_fetch_array($topScores);
echo "&nickname=$topScores[nickname]&score=$topScores[score]";
?>
兩個運行沒有錯誤,問題是,我得到的閃存不是變量值,但變量的名稱,換句話說,我得到的vars.nickname是
$topScores[nickname]
和vars.score
$topScores[score]
如果我單獨運行PHP時得到這樣的:
&nickname=jonny&score=100
這是我想要得到實際的變量值,任何幫助將不勝讚賞。
它的工作!但是我收到了一個新錯誤,錯誤#2101:傳遞給URLVariables.decode()的字符串必須是包含名稱/值對的URL編碼查詢字符串。我修正了改變我的php echo sintax來回顯「nickname =」。urlencode($ topScores ['nickname'])。「&score =」。$ topScores ['score'];非常感謝! – FruitDealer