call StreamingServlet via Hyperlink

[NikiG]

Hello!

I tried to open a Servlet with Hyperlink.openLocation("/pdfstream?filename=Test.pdf") but it doesn't work. The Servlet definition looks like this:
    <servlet>
        <servlet-name>PdfStream</servlet-name>
        <display-name>Pdf-Streaming Servlet</display-name>
        <servlet-class>servlets.PdfStreamingServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>PdfStream</servlet-name>
        <url-pattern>/pdfstream/*</url-pattern>
    </servlet-mapping>

What do I have to do to call a Servlet within my Thinwire application? I just want to stream the file from the Server to the browser/client.

Thanks for help

[al0]

location is open location is calculated as if it was typed in client browser, so it shall be "http://yourhost:8080/yourapplication/pdfstream?filename=Test.pdf"

[NikiG]

Ok, I thought it cannot be called relative to the webapps root dir. I put the path in a configfile, so it can get changed dynamically.
Thanks for the fast answer.

[al0]

Really, you even need not it in the config - it can be calculated dynamically - host may be obtained from the parameters of main method of the main class of your TW application and then combined with Application.current().getBaseFolder()

[NikiG]

Thank you, I will check it.