为什么HTML中ajax函数不生效
来源：2-4 实现动态图表
weixin_慕少0536252
2020-03-11 22:55:39
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Insert title here</title>
<script type="text/javascript" src="js/echarts.min.js"></script>
<script type="text/javascript" src="js/jquery-3.4.1.min.js"></script>
</head>
<body>
<div id="main" style="width: 600px;height:400px;"></div>
<script type="text/javascript">
$.ajax({
"url ": "/listener/rc",
"type" : "get",
"dataType" : "json",
"success" : function(json){
var myChart = echarts.init(document.getElementById('main'));
// 指定图表的配置项和数据
var option = {
title: {
text: '请求流量数据统计'
},
tooltip: {},
legend: {
data:['访问量']
},
xAxis: {
data: json.timeList
},
yAxis: {},
series: [{
name: '访问量',
type: 'line',
data: json.valueList
}]
};
// 使用刚指定的配置项和数据显示图表。
myChart.setOption(option);
}
});
// 基于准备好的dom，初始化echarts实例
</script>
</body>
</html>
package com.bbwan.servlet;
import java.io.IOException;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
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 ReuquestCountServlet
*/
@WebServlet("/rc")
public class ReuquestCountServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public ReuquestCountServlet() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
List<String> timeList=(List)request.getServletContext().getAttribute("timeList");
List<String> valueList=(List)request.getServletContext().getAttribute("valueList");
response.setContentType("text/html;charset=utf-8");
// response.getWriter().println(timeList.toString()+"<br/>");
// response.getWriter().println(valueList.toString());
Map map=new HashMap();
map.put("timeList", timeList);
map.put("valueList", valueList);
String json=JSON.toJSONString(map);
response.getWriter().println(json);
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
doGet(request, response);
}
}
1回答
-
好帮手慕小脸
2020-03-12
同学你好，这里ajax里面的内容无需加上双引号。具体修改如下：
如果我的回答解决了你的疑惑，请采纳，祝学习愉快~
1
相似问题
回答 2
回答 1