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나는 간단한 웹 응용 프로그램을 가지고 있습니다. 이 내 서블릿잘못된 요청 url 처리기 servlet
@WebServlet(urlPatterns = "/info_send", loadOnStartup = 1)
public class ApplicationController extends HttpServlet {
@Override
public void init() throws ServletException {
UsersService.addUser("admin", "admin");
}
@Override
protected void doPost(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse) throws ServletException, IOException {
PrintWriter writer = httpServletResponse.getWriter();
httpServletResponse.setContentType("text/html");
String login = httpServletRequest.getParameter("login");
String password = httpServletRequest.getParameter("pass");
if (!login.isEmpty() && !password.isEmpty() && UsersService.isLoginPresent(login)) {
if(UsersService.isUserExist(login, password)) {
getServletContext().getRequestDispatcher("/result.jsp").forward(httpServletRequest, httpServletResponse);
} else {
writer.println("User with such login is already registered.");
getServletContext().getRequestDispatcher("/").include(httpServletRequest, httpServletResponse);
}
} else {
writer.println("No such user in the system");
getServletContext().getRequestDispatcher("/").include(httpServletRequest, httpServletResponse);
}
writer.close();
}
@Override
protected void doGet(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse) throws ServletException, IOException {
httpServletResponse.getWriter().write("The request was wrong");
}
}
입니다 이건 내 web.xml을
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<error-page>
<location>/error.jsp</location>
</error-page>
<error-page>
<location>/AppExceptionHandler</location>e
</error-page>
</web-app>
나는 간단한 받는다는 웹 프로젝트를 사용하고 있습니다. 어떻게하면 잘못된 URL을 처리 할 수 있습니까? 예를 들어 localhost:8080/home/12
을 확인하면? 내 경우 오류 페이지 (error.jsp)와 오류 처리기 (다른 서블릿)가 작동하지 않았습니다.