Getting the current working directory

• From: Patrick Gundlach <patrick@gundla.ch>
• Date: Wed, 25 Oct 2017 13:27:25 +0200
• To: XProc Dev <xproc-dev@w3.org>
• Message-Id: <EB6E9960-C7FD-451F-9B50-3BF9B5B1BF43@gundla.ch>

Hello all,

I try to use a simple pipeline which runs an XSLT stylesheet. Now I'd like to pass a parameter (it is a pointer to a file in the current working directory).

My pipeline so far is:

<p:declare-step xmlns:p="http://www.w3.org/ns/xproc"
    xmlns:c="http://www.w3.org/ns/xproc-step" version="1.0">
    <p:xslt name="Transformation">
        <p:input port="parameters">
            <p:empty/>
        </p:input>
        <p:input port="source">
            <p:document href="source.xml"/>
        </p:input>
        <p:input port="stylesheet">
            <p:document href="transformation.xsl"/>
        </p:input>
    </p:xslt>
    <p:store href="data.xml"/>
</p:declare-step>


I have found out how to pass a parameter to the stylesheet


<p:xslt name="Transformation">
    <p:with-param name="somefile" select="'somefile.xml'">
        <p:empty/>
    </p:with-param>
    ....


The transformation.xsl resides in a different directory than the current directory, but I'd like to tell the XSLT to use a file (somefile.xml) from the current directory. How is this possible?


Patrick


BTW: is there any manual for xproc?

Received on Wednesday, 25 October 2017 11:27:55 UTC