Subject: Re: [xsl] How to read an XML file that is in a subfolder or sub-subfolder or sub-sub-subfolder or ...?|
From: "Wendell Piez wapiez@xxxxxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 7 Apr 2020 14:59:16 -0000
Roger, In Saxon, also, when consuming a path to a directory, the collection() function accepts a 'recurse' option, which could be used to return such a file. collection('file:///path/to/files?recurse=yes;select=filelist.xml') Of course you might get more than one filelist.xml this way. Cheers, Wendell On Tue, Apr 7, 2020 at 9:14 AM Michael Kay mike@xxxxxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: > > Assuming your processor can map a file system directory to a collection, you can use the fn:uri-collection() function to get all the URIs of files in the collection and then search them to find the one you are looking for. > > An alternative would be to use the functions in the EXPath file module, which gives explicit access to concepts like folders and files. > > Michael Kay > Saxonica > > On 7 Apr 2020, at 14:06, Costello, Roger L. costello@xxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> 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." > > 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 > > > XSL-List info and archive > EasyUnsubscribe (by email) -- ...Wendell Piez... ...wendell -at- nist -dot- gov... ...wendellpiez.com... ...pellucidliterature.org... ...pausepress.org... ...github.com/wendellpiez... ...gitlab.coko.foundation/wendell...