使用ajax将url值从javascript传递到servlet并获取错误;未捕获引用错误:$未定义“;
java:这是我的servlet使用ajax将url值从javascript传递到servlet并获取错误;未捕获引用错误:$未定义“;,javascript,java,jquery,ajax,servlets,Javascript,Java,Jquery,Ajax,Servlets,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
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,我得到了错误
"Uncaught Reference Error: $ is not defined".
我想将当前网页URL从JavaScript值传递到servlet
如果我使用以下过程:-
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");
}
然后我得到以下错误:-
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,我只需要val1的值,但为什么它要追加http://www.mca.gov.in/TMServlet?val1=
这是什么
在第二个错误中,我想知道我需要在哪里添加jquery-1.7.2.min.js:3文件
如果有人知道答案,请尽快帮助