I love to play with Kernow's XSLT Playground feature and your case way
well prepared for that.
With the help of xsl:for-each-group and regular expressions your case
can be solved. I did not try to exclude the possibly last P from the
OL and I did not check for the alphabetical order. You would have to
specify more clearly what should happen, but look at this:
<xsl:template match="root">
<xsl:copy>
<xsl:for-each-group select="*"
group-adjacent="if (self::P and
(self::P[matches(., '^[A-Z][.]')] or
preceding-sibling::P[matches(., '^[A-Z][.]')]))
then 0 else position()">
<xsl:choose>
<xsl:when test="current-grouping-key() = 0">
<OL>
<xsl:for-each-group select="current-group()"
group-starting-with="P[matches(., '^[A-Z][.]')]">
<xsl:choose>
<xsl:when test="./self::P[matches(., '^[A-Z][.]')]">
<LI>
<xsl:apply-templates select="current-group()"
mode="join"/>
</LI>
</xsl:when>
<xsl:otherwise>
<xsl:apply-templates select="current-group()"/>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each-group>
</OL>
</xsl:when>
<xsl:otherwise>
<xsl:apply-templates select="."/>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
<xsl:template match="node()" mode="join">
<xsl:apply-templates select="@*|node()"/>
<xsl:value-of select="' '"/>
</xsl:template>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
It creates
<root>
<H1>Heading</H1>
<P>List Heading</P>
<OL>
<LI>A. One big sentence incorrectly placed in P tags </LI>
<LI>B. Another long sentence spanning P tags. </LI>
<LI>C. This should really be one long list item that spans
randomly. Hopefully I am some unrelated text. </LI>
</OL>
</root>
Good luck!
- Michael
Am 13.01.2009 um 23:45 schrieb Graeme Kidd:
Hi everyone,
I am using XSLT 2.0 and I have three questions about an ordered list
in this format:
<root>
<H1>Heading</H1>
<P>List Heading</P>
<P>A. One big</P>
<P>sentence incorrectly</P>
<P>placed in P tags</P>
<P>B. Another long</P>
<P>sentence spanning P tags.</P>
<P>C. This should really</P>
<P>be one</P>
<P>long</P>
<P>list item</P>
<P>that spans randomly.</P>
<P>Hopefully I am some unrelated text.</P>
</root>
Which I want converted to this format:
<root>
<H1>Heading</H1>
<P>List Heading</P>
<OL>
<LI>One big sentence incorrectly placed in P tags.</LI>
<LI>Another long sentence spanning P tags.</LI>
<LI>This should really be one long list item that spans
randomly.</LI>
</OL>
<P>Hopefully I am some unrelated text.</P>
</root>
Due to the original XML file being rather bad the list may not start
at A. Before I have been able to catch a numbered list just by
checking if the P tag starts with a number and its preceding sibling
does not, then when I am inside the list check if the next P tag
starts with a number a well. This list is different though.
1) I imagine you can check if the P tags starts with a letter by
doing something like this:
P[starts-with(translate(., 'vUppercaseChars_CONST',
'vUppercaseAChar_CONST'), 'A')]
How would you then check it begins with letter followed by a dot?
2) Is it possible to find a letter followed by a dot then check if
the next P node starts with the next letter of the alphabet followed
by a dot?
3) Is it possible to check if the next 10 P tags contain the next
letter of the alphabet plus a dot. Previously I have been able to
pick up lists no problem when they had a predictable pattern but
this one doesn't. I can only assume that the list ends after about
10 P tags or it finds a character in a previous position in the
alphabet or it hits some other tag that is not a P tag. I would end
the list item at the first full stop it found after the last P tag
that started with character plus a dot. Is something like this
possible in XSLT and if so how?
Thanks for your time,
Graeme
--
_______________________________________________________________
Michael Müller-Hillebrand: Dokumentation Technology
Adobe Certified Expert, FrameMaker
Consulting and Training, FrameScript, XML/XSL, Unicode
Blog [de]: http://cap-studio.de/
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