无法显示名字请检查
来源：3-6 自由编程
warren_au
2020-06-11 10:20:23
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Insert title here</title>
</head>
<body>
<input type="button" id="btn1" value="员工列表">
<input type="button" id="btn2" value="职位列表">
<input type="button" id="btn3" value="部门列表">
<div id="divContext"></div>
<script type="text/javascript" src="js/jquery-3.5.1.js"></script>
<script type="text/javascript">
$("#btn1").click(function(){
$.ajax({
"url":"/ajax/prictice1",
"type":"get",
"data":"t",
"dataType":"json",
"success":function(json){
var html="";
for(var i=0;i<json.length;i++){
var emp=json[i];
html=html+emp.name;
}
$("divContext").append(html);
}
})
})
</script>
</body>
</html>
Servlet
package wjb.prictice1;
import java.io.IOException;
import java.util.HashSet;
import java.util.Set;
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 Prictice1Servlet
*/
@WebServlet("/prictice1")
public class Prictice1Servlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public Prictice1Servlet() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
Set set=new HashSet();
set.add(new Emploee("小红","职员","外勤"));
set.add(new Emploee("小明","经理","整理"));
set.add(new Emploee("小白","CEO","批文"));
String json=JSON.toJSONString(set);
response.setContentType("text/html;charset=utf-8");
response.getWriter().println(json);
}
}
1回答
-
同学你好，并没有divContext标签，则建议同学使用id选择器，获取到id为divContext的标签。修改后代码如下所示：
如果我的回答解决了你的疑惑，请采纳，祝学习愉快~
0
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