Spring controller exception handling for 404 statuses and url pattern
Question
Is it possible to handle 404 status for particular url pattern? I would like display login page if someone go to 'localhost:8080/login/other' but there is no request mapping for 'login/other'. Controller class is as follow:
@RequestMapping("/login")
@Controller
public class UserManagerController {
@RequestMapping({"/", ""})
public String getLoginPage() {
return "login.html";
}
}
I cannot add '/login/**' because this match my static content and any request for js or css match that endpoint. Example html:
<!DOCTYPE html>
<html lang="en">
<head>
...
</head>
...
<script src="/login/some.js"</script>
</html>
2 answers
solution1
0 2020-06-03 13:29:17
You can use something like this:
import java.util.Objects;
import javax.servlet.http.HttpServletRequest;
import org.springframework.boot.web.servlet.error.ErrorController;
import org.springframework.http.HttpStatus;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;
@Controller
public class GlobalErrorController implements ErrorController {
@RequestMapping("/error")
public String handleError(HttpServletRequest request, Model model) {
Integer statusCode = (Integer) request.getAttribute("javax.servlet.error.status_code");
Exception exception = (Exception) request.getAttribute("javax.servlet.error.exception");
model.addAttribute("errorMessage", exception);
model.addAttribute("statusCode", statusCode);
if (Objects.nonNull(statusCode)
&& HttpStatus.NOT_FOUND.value() == statusCode) {
return "resource-not-found";// in your case you should use: return "redirect:/login";
}
return "exception";
}
@Override
public String getErrorPath() {
return "/error";
}
}
That code opens the 'resource-not-found' page in case that user will open an invalid url, for instance localhost:8080/invalid-page
For your case you could use the method 'handleError()' to return "redirect:/login"; if statusCode is 404
In my project I thought that it will be more logical to open a 'resource-not-found' page and a button that will redirect the user to home page, you can look how it works here: https://macari-home-finance.herokuapp.com/login/invalid-url and here is my project https://github.com/ruslanMacari/HomeFinance
hope it will help you
solution2
0 2020-06-05 14:23:04
I found solutions to redirect http 404 responses for defined url pattern using Filters and HttpServletResponseWrapper.
In spring boot define filter for pattern.
@Bean
public FilterRegistrationBean<LoginFilter> loginFilter(){
FilterRegistrationBean<LoginFilter> registrationBean = new FilterRegistrationBean<>();
registrationBean.setFilter(new LoginFilter());
registrationBean.addUrlPatterns("/login/*");
return registrationBean;
}
In filter instance create HttpServletResponseWrapper and override sendError's methods to redirect all 404 responses. Filters look as follow:
public class LoginFilter implements Filter {
@Override
public void doFilter(ServletRequest servletRequest,ServletResponse servletResponse,
FilterChain chain) throws IOException, ServletException {
HttpServletResponseWrapper servletResponseWrapper = new HttpServletResponseWrapper((HttpServletResponse) servletResponse) {
@Override
public void sendError(int sc, String msg) throws IOException {
if (sc == 404) {
sendRedirect();
} else {
super.sendError(sc, msg);
}
}
@Override
public void sendError(int sc) throws IOException {
if (sc == 404) {
sendRedirect();
} else {
super.sendError(sc);
}
}
private void sendRedirect() throws IOException {
((HttpServletResponse)getResponse()).sendRedirect("/login");
}
};
chain.doFilter(servletRequest, servletResponseWrapper);
}
}
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