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[xsl] Transformation of hierarchy, changing levels, and... back
Subject: [xsl] Transformation of hierarchy, changing levels, and... back
From: Marcin Miłkowski <milek_pl@xxxxx>
Date: Tue, 02 Aug 2005 17:13:20 +0200
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hi,
I've been trying to make some transformation from a structure:
<Tu>
<Tuv>
@Lang
some text1
<df>some text2</df>
<ut>some optional text</ut>
some text
<ut>some more text</ut>
and some text
</Tuv>
<Tuv>
@Lang
some text
<df>
some text
</df>
<ut>some optional text</ut>
some text
<ut>some more text</ut>
and some text
</Tuv>
</Tu>
to a structure
<trans-unit> (=Tu)
@id
<source> (=Tu/Tuv@Lang=$blabla)
some text1
<g@id@barfoo/> (<df>)
some text2
<g@id@foobar/> (</df>)
<ph@id>some optional text</ph>(=ut)
some text
<ph@id>some more text</ut>
and some text
</source>
<target> (=Tu/Tuv@Lang=$foobar)
some text1
<g@id@barfoo/> (<df>)
some text2
<g@id@foobar/> (</df>)
<ph@id>some optional text</ph>(=ut)
some text
<ph@id>some more text</ut>
and some text
</target>
</trans-unit>
I tried to match //Tu, and then use for-each to match all Tuv[1]/ut and
Tuv[2]/ut. This approach turned quite difficult because I had to insert
ut at exactly the same position in source as it was in Tuv[1]. Probably
I should use more templates and match //Tu/Tuv or something like
that, but I wouldn't know how to close the trans-unit tag (is there a
way to check if this Tuv is Tu/Tuv[1] or Tu/Tuv[2]?). Additionally, I
would need to produce a second file with ids replacing all text between
tags in Tu/Tuv[1]@Lang=foobar and Tu/Tuv[2]@lLang=barfoo. (So the text
must be uniquely numbered, or indexed in the other file). The third step
would be to backconvert the transformed files into a source one.
Any ideas what method would be the best in this case? All help very
appreciated.
Best regards,
Marcin Milkowski
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