Subject: RE: [xsl] Transforming multiple XML files into one file From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Mon, 3 Jul 2006 23:57:55 +0100 |
If the argument to the document() function is a string containing a relative URI, then it is interpreted relative to the base URI of the stylesheet element in which the document() call appears. So in layman's terms, if you say document('thing.xml') then you are referring to a file in the same directory as the stylesheet. Michael Kay http://www.saxonica.com/ > -----Original Message----- > From: Chad Chelius [mailto:cchelius@xxxxxxxxxxxxxxx] > Sent: 03 July 2006 20:32 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: Re: [xsl] Transforming multiple XML files into one file > > David, > How does the XSLT know the exact location of the files that > you are listing. Do they need to be located in the same directory? > > > <chad/> > > Chad Chelius > AGI Training > cchelius@xxxxxxxxxxxxxxx > > > On Jul 3, 2006, at 9:12 AM, David Carlisle wrote: > > > <xsl:copy-of select="(document('file1.xml')|document('file2.xml')| > > document('file3.xml'))/story"/> > > > > or if it's more convenient have a list.xml that looks like <files> > > <file>file1.xml</file> <file>file2.xml</file> > <file>file3.xml</file> > > </files> > > > > then you can co > > > > <xsl:copy-of select="document(document('list.xml')/files/file)/ > > story"/> > > > > in xslt2 you can use the collection() function which essentially > > automates (and hides) the generation of the list.xml file, but in > > XSLT1 > > you need to have this file list to hand either by making it > manually > > or 9for example) modifying a directory listing. Some web service > > setups have an option of supplying a listing in XMl format so > > sometimes you don't need to make an actual file list xml file, just > > use an http URI pointing at the directory. > > > > David
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