servlet错误,请帮忙
html文件:
<html>
<title>
<head> me </head>
</title>
<body>
<form action= "http://localhost:8080/test/login " method= "post ">
<input type= "text " name= "name ">
<input type= "password " name= "pass ">
<input type= "submit ">
</body>
</html>
servlet文件:
import java.io.*;
import java.sql.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class LoginServlet extends HttpServlet
{
public void doPost(HttpServletRequest req,HttpServletResponse resp)throws
ServletException,
IOException {
PrintWriter out=resp.getWriter();
String n=req.getParameter( "name ");
String p=req.getParameter( "pass ");
out.print(n);
out.print(p);
}
}
web.xml配置文件:
<?xml version= "1.0 " encoding= "UTF-8 "?>
<web-app version= "2.4 "
xmlns= "http://java.sun.com/xml/ns/j2ee "
xmlns:xsi= "http://www.w3.org/2001/XMLSchema-instance "
xsi:schemaLocation= "http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd ">
<servlet>
<servlet-name> login </servlet-name>
<class-name> LoginServlet </class-name>
</servlet>
<servlet-mapping>
<servlet-name> login </servlet-name>
<url-pattern> /login/* </url-pattern>
</servlet-mapping>
</web-app>
错误:
Servlet login is not available
The requested resource (Servlet login is not available) is not available.
以上简单的servlet试验不知道错误发生在哪里,请求帮助。
------解决方案--------------------1.把servlet.jar包加入classpath中;
2.jsp页面中的action= "http://localhost:8080/test/login "应该与web.xml中 <servlet-name> login </servlet-name> 相对应,即改成action= "login "
3.web.xml中 <url-pattern> /login/* </url-pattern> ,你为什么要加个*号,这个路径的定义指明的是你在浏览器中访问时的地址,所以定义简单一些要好。