Subject: [xsl] Using Saxon 8.5 and collection() to process a directory of XML fi les From: "Welch Andrew (ELS)" <A.Welch@xxxxxxxxxxxx> Date: Thu, 4 Aug 2005 13:08:25 +0100 |
I've just had a requirement to process a directory of XML files, generating 10 output files per source XML file and placing them in a folder named the same as the input file. Mike mentioned in his annoucement email for Saxon 8.5 something along these lines, so I want to try and achieve it all in a single stylesheet, rather then needing the host language. So, given a folder of source XML files: 001.xml 002.xml ... 999.xml The result of tranform should be a folder per XML file, named the same as the XML file, with the 10 ouput files within it: [001] <- folder first.html second.html ... [002] first.html second.html .... ... [999] first.html etc I believe it's possible now to use the collection() function to get a sequence of the input XML files. Is it possible to get the filenames so that I can create the corresponding output folder? This is the code I think I need once I have the collection bit sorted (which will wrap this code): <xsl:for-each select="for $i in $outputFileNames return $i"> <xsl:result-document href="{$dir}/{$folderName}/{.}.html"> .... </xsl:result-document> </xsl:for-each> $outputFileNames is the sequence of the 10 filenames eg 'first', 'second' and so on. $dir is the output folder. $folderName needs to come from the filename of the input XML - not sure if this is possible yet. Once this is ready, I think the end user should be able to run this stylesheet (with the correct path parameters but without any input XML) to generate all of the necessary files. It would be great to do all this in the stylesheet, rather than requiring the host language to perform each transform separately. My question is - is this possible? cheers andrew
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