Hi, I'm observing a behavior (Tomcat 9.0.10) with respect to getServletPath() that I don't understand when reading the servlet spec.
For a request http://localhost:8080/nuxeo/ui/ I get the following: httpRequest.getRequestURL() http://localhost:8080/nuxeo/ui/ httpRequest.getRequestURI() /nuxeo/ui/ httpRequest.getContextPath() /nuxeo httpRequest.getServletPath() /ui/index.jsp httpRequest.getPathInfo() null httpRequest.getQueryString() null But my understanding from the Servlet spec is that we should always have (Servlet 4.0 spec, ยง3.5): requestURI = contextPath + servletPath + pathInfo which you can see is not the case here. I don't think getServletPath() should return the index.jsp part, which is a welcome file. FYI the mapping that matches the above is: httpRequest.getHttpServletMapping().getServletName() (java.lang.String) jsp httpRequest.getHttpServletMapping().getMappingMatch() (javax.servlet.http.MappingMatch) EXTENSION httpRequest.getHttpServletMapping().getPattern() (java.lang.String) *.jsp httpRequest.getHttpServletMapping().getMatchValue() (java.lang.String) ui/index Is this a bug in Tomcat or did I miss something from the spec? Thanks, Florent -- [image: Nuxeo Logo] <https://www.nuxeo.com/> Florent Guillaume Head of R&D [image: LinkedIn] <https://www.linkedin.com/in/fguillaume/> [image: Twitter] <https://twitter.com/efge> [image: Github] <https://github.com/efge> Nuxeo Content Services Platform. Stay ahead.
