我寫了一些JavaScript代碼來顯示頁面加載後的模式,然後單擊按鈕重定向另一個URL。這工作得很好。但現在我需要使用外部JavaScript文件來完成它。提前致謝。這是我做的:如何使用外部js文件工作模態?
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title> Pop Up</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap-theme.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#myModal").modal('show');
});
</script>
</head>
<body>
<div id="myModal" class="modal fade">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title">Here is your message Title</h4>
</div>
<div class="modal-body">
<p>Here is your message details</p>
<a href="https://www.fiverr.com/" type="button" class="btn btn-default">Exit</a>
</div>
</div>
</div>
</div>
<script>
$(".modal").on("hidden.bs.modal", function() {
window.location = "https://www.google.com/";
});
</script>
</body>
</html>
我該如何做到這一點與外部JavaScript文件?
$(「。modal」)將不會顯示任何內容,因爲文件在頭部 – charlietfl