所以我想学习 AJAX,并且我想制作与 in here 相同的应用程序 我几乎复制了它,但它不起作用。我不知道为什么,我试图自己解决它,但找不到任何解决方案。
我的 .js 文件是:
function ajaxAsyncRequest(reqURL) {
var xmlhttp;
xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", reqURL, true);
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4) {
if (xmlhttp.status == 200) {
// How to get message
alert('It\'s K');
document.getElementById("message").innerHTML = xmlhttp.responseText;
alert(xmlhttp.responseText);
} else {
alert('Something is wrong !');
}
}
};
xmlhttp.send(null);
}
索引.jsp是:
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<script type="text/javascript" src="javascript.js"></script>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<input type="button" value="Show Server Time" onclick='ajaxAsyncRequest("getTime")' />
</body>
</html>
我的 servlet 代码是:
import java.io.IOException;
import java.io.PrintWriter;
import java.time.LocalDate;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
@WebServlet("/getTime")
public class GetTimeServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public GetTimeServlet() {
super();
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
public void doGet (HttpServletRequest request,HttpServletResponse response)
throws ServletException, IOException
{
response.setHeader("Cache-Control", "no-cache");
response.setHeader("Pragma", "no-cache");
PrintWriter out = response.getWriter();
LocalDate currentTime= LocalDate.now();
String message = "Currently time is "+currentTime.toString();
out.write(message);
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
}
}
但是当我单击按钮时,我收到了我在行 document.getElementById("message").innerHTML = xmlhttp.responseText;
的标题中指出的消息
在 .js 文件中。
我通过 http://localhost:8080/HelloAjax/
运行它,所以没有本地,它加载晚于页面,所以我不知道它会是什么。
最佳答案
document.getElementById("message") 为 null,因为 DOM 中没有 id 为“message”的元素。尝试更改您的 HTML:
<body>
<div id="message"></div>
<input type="button" value="Show Server Time" onclick='ajaxAsyncRequest("getTime")' />
</body>
关于java - 无法将属性 'innerHTML' 设置为 null。 AJAX 问题 (java),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24654734/