Question

<html>
<head>
<link type="text/css" rel="stylesheet" href="stylesheet.css"/>
<title>Introduction Process</title>
</head>

<?PHP
$window_width=0;

$window_width= '<script>window.screen.availWidth</script>';

if $window_width <= 500 {
    $Link = "SignUp.php";
}
else{
    $Link = "1stPage.php";
}
?>

<Body>
<center><img src="Alex.png" alt="" align="middle"></center>
<center><div>Hello, I am Alex. I am here to walk you through an 11 question quiz which will help calculate your resilience level.</div></center>

<FORM NAME ="form1" METHOD ="POST" ACTION =<?PHP echo $Link; ?>>

<center><INPUT TYPE = "Submit" class=myButton Name = "Submit1"  VALUE = "Start"></center>
</FORM>


</body>
</html>

我正在尝试测试屏幕的大小，所以我知道将它们发送到我的移动网站或全尺寸网站，我收到此错误“解析错误：语法错误，意外”$ window_width'（T_VARIABLE），期待'（'在第12行的C：\ wamp64 \ www \ IntroductionProcess.php中“

Answer 1

首先，您应该研究如何编写JavaScript if语句。至于如何获取屏幕尺寸，screen.height和screen.width应该只使用JavaScript为您提供屏幕尺寸。在搜索的2分钟内，我在this question找到了答案。请在下次遇到问题时进行更多研究，而不是立即提出问题。

Javascript屏幕大小如果Else声明

1 个答案: