Question

NewServlet.java：这是我的servlet

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class NewServlet extends HttpServlet 
{
    protected void processRequest(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException 
    {
        response.setContentType("text/html;charset=UTF-8");

        String url = request.getParameter("url");
        System.out.println("my url"+url);
   }

this is my javascript file in which i am getting the url from document.URL and want to pass it on servlet NewServlet.java
//demo.js: this is my javascript

var url=document.URL;
alert(url);
$.ajax({
        url:"NewServlet.java",
        type:"POST",
        dataType:'url',
        data: {url:url}
        });

这是我的代码，我收到了错误

"Uncaught Reference Error: $ is not defined".

我想将当前网页网址从我的JavaScript值传递到我的servlet。

如果我正在使用以下流程： -

var url=document.URL;
alert(url);
var url1="TMServlet?val1="+url;
if(window.XMLHttpRequest)
{
    request=new XMLHttpRequest();
}
else if(window.ActiveXObject)
{
    request=new ActiveXObject("Microsoft.XMLHTTP");
}
try
{ 
   request.open("GET",url1,true);
   request.send();
}
catch(e)
{
    alert("unable to connect server");
}

然后我收到以下错误： -

1. Failed to load resource: the server responded with a status of 404 (Not Found)
Http://www.mca.gov.in/TMServlet?val1=http://www.mca.gov.in/ 

2. Uncaught TypeError: Failed to execute 'animate' on 'Element': 1 argument required, but only 0 present.                                    
jquery-1.7.2.min.js:3

在错误1中，我只想要val1的值，但为什么要追加http://www.mca.gov.in/TMServlet?val1=这个？

在第二个错误中，我想知道在哪里需要添加jquery-1.7.2.min.js：3文件。

如果有人尽快知道答案，请提供帮助。

使用ajax从javascript传递url值到servlet并获取错误＆＃34; Uncaught Reference Error：$未定义＆＃34;

0 个答案: