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Editorial Team
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Asked: 2026-05-20T10:31:28+00:00

I created a JSF application which also offers some Webservices. The webservices are created

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I created a JSF application which also offers some Webservices. The webservices are created via annotations.

Now I want to create a webserviceInfo.xhtml Page , where I get all the needed webservice Information.

When I go to the address http://our.server.com/application/OurWebserviceName, I get all the information needed to access the webservice (this info page is generated automatically by Glassfish ).

To include this page, I did the following in the webserviceInfo.xhtml:

 <iframe scrolling="automatic"  width="971" height="1000" src="#{myBean.generateUrlToWebservice()}/OurWebserviceName"/>

Where:

public String generateUrlToWebservice(){
    FacesContext fc = FacesContext.getCurrentInstance();
    String servername = fc.getExternalContext().getRequestServerName();
    String port = String.valueOf(fc.getExternalContext().getRequestServerPort());
    String appname = fc.getExternalContext().getRequestContextPath();
    return "http://"+servername+":"+port+appname;
}

Is there a more elegant solution to this?

BR, Rene

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1 Answer

  1. Editorial Team

    Use a page-relative URL.

    <iframe src="OurWebserviceName"></iframe>

    Or make use of <base> tag with little help of JSTL functions taglib.

    <html xmlns:fn="http://java.sun.com/jsp/jstl/functions">
...
<base href="#{fn:replace(request.requestURL, request.requestURI, '')}#{request.contextPath}"></base>

    This way any URL which doesn’t start with scheme or / is always relative to this URL.

    Or if you really need to do it in JSF the following gives less headache with scheme and port.

    HttpServletRequest request = (HttpServletRequest) externalContext.getRequest();
return request.getRequestURL().toString().replace(request.getRequestURI, "") + request.getContextPath();
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