我使用examplep创建了一个数据库"new"。localhost数据库有两个表user和score。我有一个AS3代码，它有2个输入文本字段，用于将用户和分数值插入数据库表中。现在我正在尝试使用用户名从数据库中检索插入的分数。我已经取了另一个文本字段来取用户名和一个按钮，当我写"sarah"时，点击按钮会返回已经插入数据库的sarah的分数。但代码显示了一个错误。我试了很多次，但都无法修复。请帮忙。这是我的代码
AS3代码：
btn2.addEventListener(MouseEvent.MOUSE_DOWN, fetchscore);
function fetchscore(event:MouseEvent)
{
var phpVars:URLVariables = new URLVariables();
var phpFileRequest:URLRequest = new URLRequest('http://localhost/collectscore.php');
phpFileRequest.method = URLRequestMethod.POST;
var phpLoader:URLLoader = new URLLoader();
phpLoader.dataFormat = URLLoaderDataFormat.VARIABLES;
phpLoader.addEventListener(Event.COMPLETE, showResult);
phpVars.systemCall = "processLogin";
phpVars.cname = Name.text;
phpLoader.load(phpFileRequest);
}
function showResult(e:Event)
{
//trace(phpVars.result);
result_text.text = "" + e.target.data.systemResult;
}
fetchscore.php
<?php
include('connect.php');
$username = $_POST['cname'];
if ($_POST['systemCall'] == "processLogin"){
$sqlqry = "SELECT * FROM scoreu WHERE username='$username'";//scoreu is my DBtable with two field user and score
$query = mysqli_query($sqlqry);
$login_counter = mysqli_num_rows($query);
if ($login_counter > 0) {
while ($data = mysqli_fetch_array($query)) {
if (mysqli_query($link), "SELECT score FROM scoreu WHERE user='$username'")) {
$findscore = $data['score'];
print 'result=$findscore';
}
}
} else {
print 'result=The login details dont match names.';
}
}
?>
connect.php
<?php
// connect.php
$db_name = 'new';
$db_username = 'root';
$db_password = '';
$db_host = 'localhost';
$link = mysqli_connect($db_host, $db_username, $db_password, $db_name);
if (mysqli_connect_errno()) {
die('Failed to connect to the server : '.mysqli_connect_error());
}
?>
您的代码中有一些"问题"，让我们从PHP代码开始。
PHP代码：
<?php
// collectscore.php
include('connect.php');
$username = $_POST['cname'];
if ($_POST['systemCall'] == "processLogin"){
// you need only one request to get the score
$query = mysqli_query($link, "SELECT score FROM scoreu WHERE user = '$username'");
// you have to fetch the result into a php var
if ($data = mysqli_fetch_assoc($query)) {
$findscore = $data['score'];
// you should use double quotes when passing vars into a string
// otherwise, if you want use single quotes, you can write it : print 'result='.$findscore;
print "result=$findscore";
} else {
print 'result=The login details dont match names.';
}
}
?>
ActionScript代码：
function fetchscore(event:MouseEvent): void
{
var phpVars:URLVariables = new URLVariables();
phpVars.systemCall = "processLogin";
phpVars.cname = cname.text; // name text field
var phpFileRequest:URLRequest = new URLRequest('http://127.0.0.1/collectscore.php');
phpFileRequest.method = URLRequestMethod.POST;
// you forgot to send your POST vars, for that, we use URLRequest.data
phpFileRequest.data = phpVars;
var phpLoader:URLLoader = new URLLoader();
phpLoader.dataFormat = URLLoaderDataFormat.VARIABLES;
phpLoader.addEventListener(Event.COMPLETE, showResult);
phpLoader.load(phpFileRequest);
}
function showResult(e:Event):void
{
// here the var should be the same as in the PHP script, result in this case : print "result=$findscore";
trace(e.target.data.result);
}
希望这能有所帮助。