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Asked: 2026-05-10T14:33:33+00:00

<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

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<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. ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­

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1 Answer

  1. <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.

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