Michael,
A problem, reversing (descending) works, but selecting the top of the
list does not work with position(). I have been testing by outputting
some numbers from the following 2 functions.
<a><xsl:value-of select="count(ancestor::node())"/></a>
<b><xsl:value-of select="position()"/></b>
This the output I am getting for the first item in the list:
<a>5</a>
<b>74</b>
So the position() function does not work in my case. That is why I think
I need to match in an if statement with count(ancestor::node()), e.g.
<xsl:if test="count(ancestor::node())=5">
Then I am getting the top of the list, a problem is however that I am
reading multiple documents and match with multiple keywords. So I cannot
set it adhoc this way.
Could you help me? This is my stylesheet, I hope it is readable. I am
loading 2 files (myfiles.xml, mykeywords.xml) in my stylesheet, which
each consist of a list of files and words respectively.
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:saxon="http://saxon.sf.net/";
extension-element-prefixes="saxon"
xmlns:f="http://fxsl.sf.net/";
exclude-result-prefixes="f">
<xsl:output method="xml" encoding="iso-8859-1" indent="no"/>
<xsl:template match="/">
<xsl:for-each select="document('myfiles.xml')//element">
<xsl:variable name="filename" select="."/>
<xsl:for-each select="document('mykeywords.xml')//element">
<xsl:variable name="term" select="."/>
<xsl:for-each select="document($filename)">
<xsl:for-each select="//*">
<xsl:sort select="count(ancestor::node())" order="descending"/>
<xsl:if test="matches(string(self::node()), concat('\W', $term, '\W'),
'im')">
<a><xsl:value-of select="count(ancestor::node())"/></a>
<b><xsl:value-of select="position()"/></b>
<c><xsl:number /></c>
<xsl:variable name="content" select="string(self::node())" />
<xsl:text>
</xsl:text>
<topic><xsl:value-of select="$term"/></topic>
<xsl:variable name="before"
select="string-length(substring-before($content, $term))" />
<xsl:variable name="after"
select="string-length(substring-after($content, $term))" />
<path>/<xsl:value-of select="concat(saxon:path(), ' ', '0', ' ',
$after)"/></path>
<content><xsl:value-of select="$content"/></content>
<xsl:text>
</xsl:text>
</xsl:if>
</xsl:for-each>
</xsl:for-each>
</xsl:for-each>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Why won't it work in my case, and is there a solution?
Michael Kay wrote:
The simplest change to your code is to add an xsl:sort to your for-each that
sorts the matching nodes by descending depth. That's <xsl:sort
select="count(ancestor::node)" order="descending"/>. Then you can return a
value only when position()=1.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: J. Zhang [mailto:j(_dot_)zhang(_at_)uva(_dot_)nl]
Sent: 12 July 2007 15:47
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Select Deepest Node only in Iteration
I have already checked the XSL FAQ at
http://www.dpawson.co.uk/ and explored the history of this
mailinglist. I could not find the answer, that is why I am
posting here.
I am matching the content of XML elements this way in a
for-each iteration:
xsl:if test="matches(string(self::node()), concat('\W',
$term, '\W'), 'im')">
I return the absolute path where $term occurs in the
transformation. The result after the transformation is for
example this for Word 1:
//article
//article/body[1]
//article/body[1]/section[6]
//article/body[1]/section[6]/normallist[6]
//article/body[1]/section[6]/normallist[6]/item[1]
The problem is that I only want the longest path, e.g. the
deepest node of this leave. I need to set a condition that
only returns something when the for-each iteration has come
to the longest path of this leave, but I could not find a way
to do this.
Any help would be greatly appreciated! Thanks!
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