本人菜鸟，求高手帮帮忙。要做毕业设计了。。。。。
CheckAdmin:
package com.tyhg.action;
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 CheckAdmin extends HttpServlet{
protected void processRequest(HttpServletRequest request,HttpServletResponse response) throws IOException,ServletException{
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
String name = request.getParameter("username");
String pass = request.getParameter("password");
if(name.equalsIgnoreCase("admin") && pass.equals("12345")){
response.sendRedirect("http://www.baidu.com");
}
else{
response.sendRedirect("http://www.baidu.com");
}
out.close();
}
protected void doPost(HttpServletResponse response,HttpServletRequest request) throws IOException,ServletException{
processRequest(request,response);
}
protected void doGet(HttpServletResponse response,HttpServletRequest request) throws IOException,ServletException{
doPost(request,response);
}
}
web.xml:
<servlet>
<servlet-name>CheckAdmin</servlet-name>
<servlet-class>com.tyhg.action.CheckAdmin</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>CheckAdmin</servlet-name>
<url-pattern>/CheckAdmin</url-pattern>
</servlet-mapping>
jsp:
<form action="CheckAdmin" method="post" id="login" name="login" onsubmit="JavaScript: return check_Null();">
------解决方案--------------------
check_Null这里面写的什么啊，
还有你指示迁移画面而已，用get方式就可以了吧。
------解决方案--------------------
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
String name = request.getParameter("username");
String pass = request.getParameter("password");
if(name.equalsIgnoreCase("admin") && pass.equals("12345")){
response.sendRedirect("http://www.baidu.com");
}
else{
response.sendRedirect("http://www.baidu.com");
}
out.close();
把你的这段代码写在dopost方法里面
------解决方案--------------------
报的错，应该是没找到url