Two templates: an identity template that copies everything by default
<xsl:template match="*">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:apply-templates/>
</
</
and a template for the nodes you want to change:
<xsl:template match="b[a][count(child::node()=1]">
<a>
<b>
<xsl:apply-templates select="a/child::node()"/>
</b>
</a>
</xsl:template>
Regards,
Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay
-----Original Message-----
From: Kenneth Reid Beesley [mailto:krbeesley(_at_)gmail(_dot_)com]
Sent: 18 February 2010 17:55
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Noobie: normalize <b><a>...</a></b> to
<a><b>...</b></a>
I'm converting a non-XML data-dump into XML, and the document
contains examples of both
<a><b>...</b></a>
and
<b><a>...</a></b>
which (in this document) are equivalent. I'd like to use
XSLT to convert all examples of the latter to the former,
with the following
caveats:
1. <b> can contain mixed text, in which case nothing should
be changed.
2. <b><a>...</a></b> should be changed to <a><b>...</b></a>
only if the <a>...</a> element is the unique child node of <b>...</b>
How can I do this?
Thanks,
Ken
******************************
Kenneth R. Beesley, D.Phil.
P.O. Box 540475
North Salt Lake, UT
84054 USA
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