tomcat - How to send servlet response to web.xml configured error page -

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i using tomcat. defined <error-page> in web.xml , mapped 404 error page /error/error.jsp. need detect if resource exist , set response status 404 if resource not available.

response.setstatus(404);

but tomcat not redirect 404 page defined, question is, there api page location defined in web.xml? don't want parse web.xml myself.

just use httpservletresponse#senderror() status code. e.g.

file resource = new file(path, name); if (!resource.exists()) {     response.senderror(httpservletresponse.sc_not_found); // sends 404.     return; }

the servletcontainer display suitable errorpage.

note: return statement isn't there decoration. avoid remant of code in same method block continue run , might produce illegalstateexceptions in appserver logs! starters namely think methods sendredirect(), forward(), senderror(), etc somehow automagically exits method block when invoked. not true ;)


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