如何在xmlhttprequest加入参数以便servlet筛选
来源:2-8 自由编程
慕设计3517557
2019-06-19 16:59:00
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Insert title here</title>
</head>
<body>
<button id="btn1">员工列表</button>
<button id="btn2">职位列表</button>
<button id="btn3">部门列表</button>
<div id="container" style="text-align:center"></div>
<script type="text/javascript" src="js/jquery-3.4.1.js"></script>
<script type="text/javascript">
$("#btn1").click(function(){
//1.创建xmlhttp请求
var xmlhttp ;
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
//2.发送请求
xmlhttp.open("GET","/ajax/twoeight",true);
xmlhttp.send();
//3.处理服务器响应
xmlhttp.onreadystatechange = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200 ){
var text = xmlhttp.responseText ;
var json = JSON.parse(text) ;
var temp = "" ;
for (var i = 0 ; i< json.length ; i ++ ){
temp = temp + "<h1>" + json[i] + "</h1>" ;
}
$("#container").html(temp);
}
}
});
</script>
</body>
</html>
package com.imooc.ajax;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import com.alibaba.fastjson.JSON;
/**
* Servlet implementation class twoeight
*/
@WebServlet("/twoeight")
public class twoeight extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public twoeight() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
List listone = new ArrayList();
listone.add("小张");
listone.add("小明");
listone.add("小李");
String lone = JSON.toJSONString(listone);
List listtwo = new ArrayList();
listtwo.add("职员");
listtwo.add("经理");
String ltwo = JSON.toJSONString(listtwo);
List listthree = new ArrayList();
listthree.add("人事部");
listthree.add("技术部");
listthree.add("无线事业部");
String lthree = JSON.toJSONString(listthree) ;
response.setContentType("text/html;charset=UTF-8");
response.getWriter().println(lone);
}
}
1回答
好帮手慕柯南
2019-06-19
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