Re: [xsl] How to read an XML file that is in a subfolder or sub-subfolder or sub-sub-subfolder or ...?

["John Lumley john@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

On 07/04/2020 14:06, Costello, Roger L. costello@xxxxxxxxx wrote:
> Hi Folks,
>
> My XSLT program has a variable $filename which contains the name of an XML file. The XML file is in a folder named "files" or one of its subfolders or one of its sub-subfolders or one of its sub-sub-subfolders or ...
>
> In other words, it's somewhere in the tree of folders rooted at "files."
If you're using Saxon PE or EE the EXPath file:list() 
(http://www.saxonica.com/documentation/index.html#!functions/expath-file/list) 
function will help you find such files.
> I want to open the XML file (wherever it's located within "files") and read it into an XSLT variable. Is there a way to do this?
>
> /Roger


--
*John Lumley* MA PhD CEng FIEE
john@xxxxxxxxxxxx <mailto:john@xxxxxxxxxxxx>
on behalf of Saxonica Ltd