java.lang.IllegalArgumentException: Invalid <url-pattern> in servlet mapping

MM picture MM · Aug 25, 2008 · Viewed 81.2k times · Source
<servlet>
    <servlet-name>myservlet</servlet-name>
    <servlet-class>workflow.WDispatcher</servlet-class>
    <load-on-startup>2</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>myservlet</servlet-name>
    <url-pattern>*NEXTEVENT*</url-pattern>
</servlet-mapping>

Above is the snippet from Tomcat's web.xml. The URL pattern *NEXTEVENT* on start up throws

java.lang.IllegalArgumentException: Invalid <url-pattern> in servlet mapping

It will be greatly appreciated if someone can hint at the error. ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­

Answer

McDowell picture McDowell · Aug 25, 2008
<url-pattern>*NEXTEVENT*</url-pattern>

The URL pattern is not valid. It can either end in an asterisk or start with one (to denote a file extension mapping).

The url-pattern specification:

  • A string beginning with a ‘/’ character and ending with a ‘/*’ suffix is used for path mapping.
  • A string beginning with a ‘*.’ prefix is used as an extension mapping.
  • A string containing only the ’/’ character indicates the "default" servlet of the application. In this case the servlet path is the request URI minus the context path and the path info is null.
  • All other strings are used for exact matches only.

See section 12.2 of the Java Servlet Specification Version 3.1 for more details.