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Asked: 2026-05-26T00:55:23+00:00

I’m trying to create a menu template in JSF where the link for the

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I’m trying to create a menu template in JSF where the link for the current directory has a different “current” or “active” class. The code currently looks like:

<ul>
    <li><h:outputLink value="#{request.contextPath}/a/">A</h:outputLink></li>
    <li><h:outputLink value="#{request.contextPath}/b/">B</h:outputLink></li>
    <li><h:outputLink value="#{request.contextPath}/c/">C</h:outputLink></li>
</ul>

I’m thinking of using something like styleClass="#{(thisDir == currentDir) ? currentLinkClass : normalLinkClass}". But how do I get the current path? Is this even correct, or is there a better way to do this?

Also, I want the links to base on the current path, not just the page. For example, myapp/a/1.jsf and myapp/a/2.jsf (that is, myapp/a/*.jsf) should trigger the active class for the A link. (I hope my explanation is clear.) Is this possible? How should this be done?

Thank you very much!

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1 Answer

  1. Editorial Team

    You can use #{request.requestURI} to get the current request URI. You can if necessary use several EL functions from JSTL fn taglib to do some string comparisons/manipulations in EL.

    Your proposed EL styleClass suggestion is perfectly fine. There is no other easy way anyway. Best optimization which you could do so far is to render those links in a loop by an <ui:repeat> so that code duplication is at least eliminated.

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