程序
<?php
if(!$_GET["screenX"]) {
echo '
<script>
location = location.href+"?screenX="+screen.width+"&screenY="+screen.height;
</script>
';
exit;
}
$screenX = $_GET["screenX"];
$screenY = $_GET["screenY"];
?>
以下是其他内容
hmVHTML5中文学习网 - HTML5先行者学习网hmVHTML5中文学习网 - HTML5先行者学习网
