ajax提交到java后台之后处理数据的实现

Java  /  管理员 发布于 9年前   183

环境：eclipse+struts

要实现的效果：点击按钮提交数据到后台之后回到前台显示出来数据

index.jsp

<%@ page language="java" contentType="text/html; charset=UTF-8"   pageEncoding="UTF-8"%> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>Insert title here</title> </head> <body> <input type="text" id="userinput"> <input type="button" id="submit"> <div id="msg"></div> </body> <script type="text/javascript" src="https:/article/jquery-2.1.0.js"></script> <script type="text/javascript"> window.onload = function() {   document.getElementById("submit").onclick = test; } function test(){   var userinput = document.getElementById("userinput");   $.post("http://localhost:8080/TestSpring/TestAction",{username:userinput.value},   function(data, textStatus){     document.getElementById("msg").innerHTML = data;   });  } </script> </html>

 struts.xml

<action name="TestAction" class="com.action.Test">   <result>index.jsp</result> </action>

Test.java

package com.action;  import java.io.PrintWriter; import java.util.Map;  import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse;  import org.apache.struts2.ServletActionContext;  import com.opensymphony.xwork2.ActionContext; import com.opensymphony.xwork2.ActionSupport;  public class Test extends ActionSupport {   @Override   public String execute() throws Exception {     // TODO Auto-generated method stub     HttpServletRequest request = org.apache.struts2.ServletActionContext.getRequest();     HttpServletResponse response = ServletActionContext.getResponse();     request.setCharacterEncoding("utf-8");     response.setCharacterEncoding("utf-8");     PrintWriter out = response.getWriter();     out.write(request.getParameter("username"));     out.flush();     out.close();     return SUCCESS;   } }

以上这篇ajax提交到java后台之后处理数据的实现就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持。


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