Ajax当中如何回传一个JSP？
4.PassBackJsp
例 4.1
<%@ page contentType="text/html; charset=GBK"%>
<%@ page language="java"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<style type="text/css">
table.default {
width:510px;
}
table.default td {
border:1px solid black;
height:23px;
}
table.default td.item {
background:#006589;
color:#fff;
text-align:center;
}
</style>
<script type="text/javascript">
var xmlHttp;
function createXmlHttp() {
if (window.XMLHttpRequest) {
xmlHttp = new XMLHttpRequest(); //FireFox、Opera等浏览器支持的创建方式
} else {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");//IE浏览器支持的创建方式
}
}
function refreshCart() {
createXmlHttp();
xmlHttp.onreadystatechange = showCartIn;
xmlHttp.open("GET", "cart.jsp" , true);
xmlHttp.send(null);
}
function showCartIn() {
if (xmlHttp.readyState == 4) {
/* thus, in this way, it return a jsp,so it is simple.if it return back a Servlet, it is difficult. this is a
very good idea. */
document.getElementById("shoppingcart").innerHTML = xmlHttp.responseText;
}
}
</script>
</head>
<body οnlοad="refreshCart()">
<div id="shoppingcart">
</div>
</body>
</html>
更多内容请见原文，文章转载自：https://blog.csdn.net/qq_43650923/article/details/103051814
