ajax提交：
第一步：ajax提交给servlet数据，进过相关的处理后
Stringmsg=
"error"
;
response.getWriter().write(msg);
返回信息！（如果是要转跳进行第三部，否则对返回的信息做相应的处理，如本例子的弹出提示框。）
第三部：通过前台中的js来实现页面跳转（如果是放在web-inf中jsp,jsp要在web.xml中部署后，这样的url才会有效，参考：http://blog.csdn.net/wanghaiping1993/article/details/23510411中关于web-inf中jsp如何访问
）
window.location.href=
"${pageContext.request.contextPath}/main.jsp"
;
form提交：
第一步：写好form表单后，向servlet提交信息
request.getRequestDispatcher("/WEB-INF/jsp/***.jsp").forward(request,response);
<%@
page
language
=
"java"
import
=
"java.util.*"
pageEncoding
=
"utf-8"
%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+
"://"
+request.getServerName()+
":"
+request.getServerPort()+path+
"/"
;
%>
<!
DOCTYPE
HTML
PUBLIC
"-//W3C//DTD HTML 4.01 Transitional//EN">
<
html
>
<
head
>
<
base
href
=
"
<%=
basePath
%>
"
>
<
title
>
My JSP 'login.jsp' starting page
</
title
>
<
script
type
=
"text/javascript"
src
=
"js/jquery-1.8.0.min.js"
></
script
>
<
script
type
=
"text/javascript"
>
function
ajax_submit(){
var
username=document.getElementsByName(
"username"
)[0].value;
password=document.getElementsByName(
"password"
/* //一定要加入jquery.js */
$.ajax({
url:
'${pageContext.request.contextPath }/Userservlet?method=login'
,
type:
'post'
data:
'username='
+username+
'&password='
+password,239)">success:
(msg) {
if
((msg==
"error"
)) {
}
else if
"success"
))
window.location.href=
"${pageContext.request.contextPath}/main.jsp"
;
}
});
}
</
script
>
</
head
>
<
body
>
<
div
align
=
"center"
style
="
margin
:
auto
;"
>
<%-- <form action="" method="post">
--%>
<
table
>
<
tr
>
密码：
<
input
type
=
"password"
name
=
"password"
></
tr
><
br
>
<
tr
>
<
input
type
=
"submit"
onclick
=
"ajax_submit()"
value
=
"提交"
></
tr
>
</
table
>
</
div
>
<%-- </form>
--%>
</
body
>
</
html
>
