[스크립트] 해상도에 따른 페이지 열기(팝업)
페이지 정보
작성자 관리자 작성일15-06-24 21:48 조회2,624회 댓글0건본문
<html>
<head>
<title>..</title>
<meta http-equiv="Content-Type" content="text/html; charset=euc-kr">
<script language="JavaScript">
<!--
function redirectPage() {
var url800x600 = "main1.html"; //800*600 에서 열릴문서
var url1024x768 = "main2.html"; //1024*768 에서 열릴문서
var url1152x864 = "main3.html"; //1152*864 에서 열릴문서
if ((screen.width == 800) && (screen.height == 600))
window.location.href= url800x600;
else if ((screen.width == 1024) && (screen.height == 768))
window.location.href= url1024x768;
else if ((screen.width == 1152) && (screen.height == 864))
window.location.href= url1152x864;
else window.location.href= url800x600;
}
//-->
</script>
</head>
<body Onload="redirectPage()">
</body>
</html>
<head>
<title>..</title>
<meta http-equiv="Content-Type" content="text/html; charset=euc-kr">
<script language="JavaScript">
<!--
function redirectPage() {
var url800x600 = "main1.html"; //800*600 에서 열릴문서
var url1024x768 = "main2.html"; //1024*768 에서 열릴문서
var url1152x864 = "main3.html"; //1152*864 에서 열릴문서
if ((screen.width == 800) && (screen.height == 600))
window.location.href= url800x600;
else if ((screen.width == 1024) && (screen.height == 768))
window.location.href= url1024x768;
else if ((screen.width == 1152) && (screen.height == 864))
window.location.href= url1152x864;
else window.location.href= url800x600;
}
//-->
</script>
</head>
<body Onload="redirectPage()">
</body>
</html>
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