1. jQuery.post( url, [data], [callback], [type] ) ：使用POST方式来进行异步请求
<body> <div id="main"> <button id="myBut">Ajax获取数据</button> <div id="container"></div> </div> </body>1.2.2 style代码
<style type="text/css">
#main {
margin: 0 auto;
width: 400px;
}
#container {
width: 400px;
height: 300px;
border: 1px dashed #666;
text-align: center;
line-height: 300px;
}
</style><script type="text/javascript" src="js/jquery-1.8.3.js"></script>
<script type="text/javascript">
$(function() {
$('#myBut').click(function() {
$.post("JqueryAjaxServlet", {
age : 18,
name : "zhang"
}, function(data, textStatus) {
var container = $('#container');
var resultData = $.parseJSON(data);
var age = resultData.age;
var name = resultData.name;
container.html("name:" + name + "," + "age:" + age);
});
});
});
</script>
1.2.4 JqueryAjaxServlet.java代码public class JqueryAjaxServlet extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
String age = request.getParameter("age");
String name = request.getParameter("name");
String personJSON = "{\"name" + "\":\"" + name + "\"," + "\"age"
+ "\":" + age + "}";
System.out.println(personJSON);
response.getWriter().write(personJSON);
}
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
doGet(request, response);
}
}
1.2.5 web.xml代码<servlet>
<servlet-name>JqueryAjaxServlet</servlet-name>
<servlet-class>com.jquery.ajax.com.JqueryAjaxServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>JqueryAjaxServlet</servlet-name>
<url-pattern>/JqueryAjaxServlet</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>原文：http://blog.csdn.net/zbw18297786698/article/details/51334103