How to get ContextPath in init() method of Servlet version 2.4

Asked 14/12, 2012 at 13:45 Answered 14/12, 2012 at 13:58

I'm using version 2.4 of Servlet and I need to get the ContextPath through the init() method which is called on server start, so I don't have any Request object that could call getContextPath() and because the Servlet version I do not have getContextPath() method in the ServletContext either.

Is there a way to get this ContextPath() somehow ?

Omnipotent answered 14/12, 2012 at 13:45 Comment(2)

ServletContext.getServletContextName() - ServletConfig.getServletContext().getServletContextName() – Heel 14/12, 2012 at 13:53

this is null in my case :/ – Omnipotent 14/12, 2012 at 13:56

One web application can be published at several different context paths, so the context path (singular) is only meaningful in the context of a particular request. Servlet 2.5 added getContextPath() to ServletContext, specified to return the "primary" context path for this web application, but there's no container-independent way to access this information in earlier spec versions.

There may be tricks that work for certain containers, for example on Tomcat the ServletContext.getResource() method returns URLs with a custom scheme, of the form jndi://hostname/context/.... Thus you may be able to use

ctx.getResource("/").getPath()

to get the context path on Tomcat (or possibly getResource("/WEB-INF/web.xml") and trim off the tail, as getResource() is specified to return null if you ask it for a file that does not exist). You will have to experiment with different containers to find similar tricks that work on those.

Brayton answered 14/12, 2012 at 13:58 Comment(0)

It seems to be only possible form servlet 2.5 as explained in this post: ServletContext getContextPath()

Neoimpressionism answered 14/12, 2012 at 13:51 Comment(0)

You are right in Servlet 2.4 the object ServeltContext does not have the method getContextPath.

I can suggest two options:

Set the context path as parameter of the servlet:

<servlet>

<servlet-name>initServlet</servlet-name>

<servlet-class>net.cirrus-it.InitServlet`</servlet-class>

<init-param>
        <param-name>contextPath</param-name>
        <param-value>/myApp</param-value>
</init-param>

</servlet>

Try to determine the context path from the method getRealPath()

http://tomcat.apache.org/tomcat-5.5-doc/servletapi/javax/servlet/ServletContext.html#getRealPath(java.lang.String)

According to the documentation:

Returns a String containing the real path for a given virtual path. For example, the path "/index.html" returns the absolute file path on the server's filesystem would be served by a request for "http://host/contextPath/index.html", where contextPath is the context path of this ServletContext.

Antoninaantonino answered 14/12, 2012 at 13:58 Comment(0)

-2

Try this code:

class demo extends HttpServlet {
       public void init(ServletConfig config) {
             String path = config.getServletContext().getRealPath("/");
       }
}

It should work

Injurious answered 14/12, 2012 at 13:50 Comment(1)

That will give the real filesystem path to the webapp directory (e.g. /var/tomcat/webapps/foo), not the context path at which the app is deployed (/foo). As krampstudio says, Servlet 2.5 added getContextPath() to ServletContext, returning the "primary" context path for the application, but in 2.4 or earlier you can only get the path for a particular request. – Brayton 14/12, 2012 at 13:55

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