LoginServlet页面404
来源：6-1 登录案例
laoLangLoveProgram
2017-03-02 01:24
@WebServlet(name="LoginServlet", urlPatterns={"/LoginServlet"})
public class LoginServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public LoginServlet() {
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
System.out.println("doPost");
String username = request.getParameter("username");
String password = request.getParameter("password");
System.out.println(request.getContextPath());
if("admin".equals(username) && "admin".equals(password)) {
//校验通过
HttpSession session = request.getSession();
session.setAttribute("username", username);
System.out.println(request.getContextPath());
response.sendRedirect(request.getContextPath()+ "/success.jsp");
}else {
//校验失败
response.sendRedirect(request.getContextPath()+ "/failure.jsp");
}
}
}这是LoginServlet.java的代码，输入用户名密码后跳转页面出错，404
<body> <form action="<%=request.getContextPath() %>/servlet/LoginServlet" method="post"> 用户名：<input type="text" name="username" /> 密码：<input type="password" name="password" /> <input type="submit" value="提交" /> </form> </body>
上面这是login.jsp的代码
输入用户名密码登录以后，显示页面为：
HTTP Status 404 - /LoginFilter/servlet/LoginServlet type Status report message /LoginFilter/servlet/LoginServlet description The requested resource is not available.
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- laoLangLoveProgram
- 2017-03-02 17:02:02
我算弄明白了，注解的路径不对urlPatterns={"/LoginServlet"}改成urlPatterns={"/servlet/LoginServlet"}
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