Selecting the first node of a Sorted Group

IcedDante · Dec 9, 2006

Working with a sorted group, the inability to use following-sibling
(which uses Document Order) and convert an RTF (not avaible with the
Parser that we are using) hampered our ability to solve the following
problem. Consider the following set of Data:

<example>
<Credentials time="3">
<UserId>kiss</UserId>
</Credentials>
<Credentials time="1" break="true">
<UserId>bob</UserId>
</Credentials>
<Credentials time="6" break="true">
<UserId>my</UserId>
</Credentials>
<Credentials time="0">
<UserId>Look,</UserId>
</Credentials>
<Credentials time="2">
<UserId>can</UserId>
</Credentials>
<Credentials time="9">
<UserId>tookish</UserId>
</Credentials>
</example>

When sorted in ascending order, the data reads: "Look, bob can kiss my
tookish". For our requirements I am processing it as a descending set:
"tookish my kiss can bob Look,"

That is the first requirement. However, in the even that a node with
the property of "break" equal to "true" is found, processing should
halt (multiple nodes can have the break property, but we really only
care about the first one).
So the output should read: "tookish my"

I devised a solution by using the substring-before operator to select
the first break node's sorted position.

<xsl

aram name="breakPos">
<xsl:for-each select="example/Credentials">
<xsl:sort select="@time" order="descending" />
<xsl:call-template name="countSequence">
</xsl:call-template>
</xsl:for-each>
</xsl

aram>
<xsl

aram name="endElem">
<xsl:value-of select="substring-before($breakPos,'|')" />
</xsl

aram>

<xsl:template match="/">
<html>
<body>
<xsl:for-each select="example/Credentials">
<xsl:sort select="@time" order="descending" />
<xsl:if test="($endElem = '') or (position() <= $endElem)">
<xsl:value-of select="UserId" /> 
</xsl:if>
</xsl:for-each>
</body>
</html>
</xsl:template>

Yeah, this renders the html output:
tookish 
my 

but I felt like the "endElem" parameter derivation was kind of a hack.
Would there be a better solution- possibly using Meunchian grouping to
perform this fix?

IcedDante · Dec 9, 2006

Whoops... I forgot the countSequence template. It basically builds the
list of nodes that have break="true" based on their position in the
sorted tree, delimited by the '|' symbol:

<xsl:template name="countSequence">
<xsl:choose>
<xsl:when test="(@break = 'true')">
<xsl:value-of select="position()" /><xsl:text>|</xsl:text>
</xsl:when>
<xsl

therwise></xsl

therwise>
</xsl:choose>
</xsl:template>

p.lepin · Dec 11, 2006

IcedDante said:
Working with a sorted group, the inability to use
following-sibling (which uses Document Order) and convert
an RTF (not avaible with the Parser that we are using)

Mentioning what processor you're using would've been a good
idea, instead of mentioning just some of its limitations.

hampered our ability to solve the following
problem. Consider the following set of Data:
[XML]

When sorted in ascending order, the data reads: "Look,
bob can kiss my tookish". For our requirements I am
processing it as a descending set: "tookish my kiss can
bob Look,"

That is the first requirement. However, in the even that
a node with the property of "break" equal to "true" is
found, processing should halt (multiple nodes can have
the break property, but we really only care about the
first one). So the output should read: "tookish my"

[partial solution reeking of imperative programming]

It would've been a better idea to post the entire
transformation instead of just parts of it.

Yeah, this renders the html output:
tookish 
my

but I felt like the "endElem" parameter derivation was
kind of a hack. Would there be a better solution-
possibly using Meunchian grouping to perform this fix?

I've grown accustomed to XSLT 2.0, so I can't think of any
elegant 1.0 solution off the top of my head. The following
works:

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<result>
<xsl:apply-templates select="example/Credentials">
<xsl:sort select="@time" order="descending"/>
</xsl:apply-templates>
</result>
</xsl:template>
<xsl:template match="Credentials"/>
<xsl:template
match=
"
Credentials
[
not
(
../Credentials
[@time>current()/@time][@break='true']
)
]
">
<xsl:value-of select="UserId"/> 
</xsl:template>
</xsl:stylesheet>

....but specifying the sorting order in two separate places
in two different formats is a bit ugly, too, of course.

With XSLT 2.0, a much more elegant solution is possible:

<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="@*|node()" mode="copy">
<xsl:copy-of select="."/>
</xsl:template>
<xsl:template match="/">
<xsl:variable name="sorted">
<xsl:apply-templates
select="example/Credentials" mode="copy">
<xsl:sort select="@time" order="descending"/>
</xsl:apply-templates>
</xsl:variable>
<xsl:apply-templates select="$sorted/Credentials"/>
</xsl:template>
<xsl:template match="Credentials"/>
<xsl:template
match=
"
Credentials
[
not
(
preceding-sibling::Credentials[@break='true']
)
]
">
<xsl:value-of select="UserId"/> 
</xsl:template>
</xsl:stylesheet>

[Tested with Saxon-8B]

XSL node as string	1	Feb 26, 2009
Extracting the first child node of a parent node	2	Mar 16, 2008
xsl variable $node/text() but $node can non-node-set help!	2	Aug 23, 2006
selecting the value of the node based on the value of parameter	3	Feb 7, 2006
Selecting subelements from a node-set	2	Jan 25, 2004
[XSLT] How to establish a context node without <xsl:for-each>	4	Nov 17, 2008
distinct grouping of sorted results	4	Feb 8, 2005
Compare node set against a string	6	Mar 13, 2008

Selecting the first node of a Sorted Group

IcedDante

IcedDante

p.lepin

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