问题
I am using Tomcat. I defined some <error-page> in web.xml and mapped 404 error to page /error/error.jsp. I need to detect if the resource exist and set the response status to 404 if the resource is not available.
response.setStatus(404);
But Tomcat does not redirect to the 404 page I defined, so my question is, is there any API to get the page location defined in web.xml? I don't want to parse the web.xml by myself.
回答1:
Just use HttpServletResponse#sendError() with a status code. E.g.
File resource = new File(path, name);
if (!resource.exists()) {
response.sendError(HttpServletResponse.SC_NOT_FOUND); // Sends 404.
return;
}
The servletcontainer will then display the suitable errorpage.
Note: the return statement isn't there for decoration. It will avoid that the remant of the code in the same method block will continue to run and might produce IllegalStateExceptions in the appserver logs! Starters namely often think that methods like sendRedirect(), forward(), sendError(), etc somehow automagically exits the method block when invoked. This is thus not true ;)
回答2:
No, there is no API. The setting is internal to Tomcat, it doesn't expose this to the outside world directly. (You can, as you said, parse the web.xml - since it's XML it would be simple to write an XQuery to pull this out).
Also I think you seem to be confused as to exactly how this is supposed to work. You supply a URI to Tomcat that it will use to serve the HTML body of a 404 response. It will need to be able to resolve this URI into an actual resource just as if it were supplied in a request. So in many cases, you don't need a servlet to detect if the resource exists - you need a servlet to contain the resource to be served. The exception being if you use Tomcat to serve static filesystem data for some URLs, which I believe is possible albeit seldom used.
If you have a Tomcat set up without any servlets, what are you expecting it to serve anyway? Where on earth are you expecting it to get your error.jsp page from? How do you expect it to know that? What you need to do is add at least one servlet to Tomcat (containing the error.jsp file), and then make sure that web.xml maps the /error/error.jsp URL to this servlet (and that the resource is positioned in the error subdirectory in the servlet).
Once this is done, you should be able to manually go to e.g. http://localhost:8080/<context>/error/error.jsp and have the response served (possibly with some weird content since there's no actual exception but the file should be found). Similarly, if you've set up the error-page directive correctly in web.xml, going to an umapped URL (e.g. http://localhost:8080/<context>/asdfghjasgh) should show you your error page as the 404 response.
来源：https://stackoverflow.com/questions/2047367/how-to-send-servlet-response-to-web-xml-configured-error-page