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Rank: Member
Joined: 11/17/2016
Posts: 12
Location: DE
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Hello,
is it possible to use wildcards (*) in Windows RaptorXML command line parameter?
Example:
RaptorXML.exe xslt --xslt-version=2 --input=E:\Input\Input*.xml --output=E:\Output\Output*.xml MappingMapToCarLoAuftragIn.xslt
Thank you.
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Rank: Advanced Member
Joined: 12/13/2005
Posts: 2,856
Location: Mauritius
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No, definitely not. This would require RaptorXML guessing how to build output file name depending on the input file name and placement of the wildcard.
As far as I know, it should be possible to write a small XSLT file instead, which reads input directory and outputs one file for each input file. There you can have whole control about output file name generation
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Rank: Member
Joined: 11/17/2016
Posts: 12
Location: DE
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Hello,
Thank you for your feedback. Do you know about an example for this anywhere? Here in the forum search I was not able to find anything.
Kind regards
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Rank: Member
Joined: 11/17/2016
Posts: 12
Location: DE
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Hello,
I found another solution: I'm running a Windows batch command before doing the RaptorXML command which always picks the oldest file out of my directory, moves it to another folder and renames it to a fixed name. RaptorXML is then pointing on this file with the fixed file name.
Here is the Windows batch command to achieve it:
Code:for /f "delims=" %%a in (
'dir /b /a-d /tw /od "E:\Directory1\*.xml"'
) do move "E:\Directory1\%%a" "E:\Directory2\FixedName.xml" & goto fileMoved
:fileMoved
Kind regards
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Rank: Advanced Member
Joined: 12/13/2005
Posts: 2,856
Location: Mauritius
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This has nothing to do with this forum, but with XSLT knowledge.
With fn:collection you should be able to retrieve file contents of your input files.
With xsl:result-document you can produce different result documents for each of them
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