Yet another question about getting distinct values using XSLT 1.0. Here's a stupid, made-up example that should illustrate my problem.
<?xml version="1.0" encoding="UTF-8"?>
<moviesByYear>
<year1994>
<movie>
<genre>Action</genre>
<director>A</director>
</movie>
</year1994>
<year1994>
<movie>
<genre>Comedy</genre>
<director>A</director>
</movie>
</year1994>
<year1994>
<movie>
<genre>Drama</genre>
<director>B</director>
</movie>
</year1994>
<year1994>
<movie>
<genre>Thriller</genre>
<director>C</director>
</movie>
</year1994>
<year1995>
<movie>
<genre>Action</genre>
<director>A</director>
</movie>
</year1995>
<year1995>
<movie>
<genre>Comedy</genre>
<director>C</director>
</movie>
</year1995>
<year1996>
<movie>
<genre>Thriller</genre>
<director>A</director>
</movie>
</year1996>
</moviesByYear>
Now let's say that I'd like to list all years that produced movies that are either comedies or directed by director B. I use the following stylesheet:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:output method="text" encoding="UTF-8" indent="no"/>
<xsl:template match="/">
<xsl:for-each select="/moviesByYear/*[movie/genre='Comedy' or movie/director='B']">
<xsl:value-of select="name()"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
This gives me the following output:
year1994year1994year1995
I have not yet found any solution for getting distinct values that would work here. For example, using name(.) != name(following-sibling::*) causes year1994 to be excluded altogether.
In my real-world case I have a complex XML structure and an XPath with many criteria that picks out a number of nodes, from which I need to get an output of distinct node names.
Update: michael.hor257k gave an elegant solution to this, but using it I faced a problem with xsl:key. Allow me to alter the scenario a bit:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<genres>
<genre>Action</genre>
<genre>Comedy</genre>
<genre>Drama</genre>
<genre>Thriller</genre>
</genres>
<moviesByYear>
<year1994>
<movie>
<genre>Action</genre>
<director>A</director>
</movie>
</year1994>
<year1994>
<movie>
<genre>Comedy</genre>
<director>A</director>
</movie>
</year1994>
<year1994>
<movie>
<genre>Drama</genre>
<director>B</director>
</movie>
</year1994>
<year1994>
<movie>
<genre>Thriller</genre>
<director>C</director>
</movie>
</year1994>
<year1995>
<movie>
<genre>Action</genre>
<director>A</director>
</movie>
</year1995>
<year1995>
<movie>
<genre>Comedy</genre>
<director>C</director>
</movie>
</year1995>
<year1996>
<movie>
<genre>Thriller</genre>
<director>A</director>
</movie>
</year1996>
</moviesByYear>
</root>
Now let's say that I want a list of genres, each of which lists years that produced movies of that genre or movies directed by director B. Stylesheet:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:exsl="urn:schemas-microsoft-com:xslt"
extension-element-prefixes="exsl">
<xsl:output method="text" version="1.0" encoding="UTF-8" indent="no"/>
<xsl:template match="/">
<xsl:for-each select="/root/genres/genre">
<xsl:call-template name="output">
<xsl:with-param name="genre">
<xsl:value-of select="."/>
</xsl:with-param>
</xsl:call-template>
</xsl:for-each>
</xsl:template>
<xsl:param name="director" select="'B'"/>
<xsl:key name="year" match="year" use="." />
<xsl:template name="output">
<xsl:param name="genre"/>
<!-- first pass -->
<xsl:variable name="years">
<xsl:for-each select="/root/moviesByYear/*/movie[genre=$genre or director=$director]">
<year><xsl:value-of select="local-name(..)"/></year>
</xsl:for-each>
</xsl:variable>
<xsl:variable name="years-set" select="exsl:node-set($years)" />
<!-- final pass -->
<xsl:value-of select="concat($genre, ': ')"/>
<xsl:for-each select="$years-set/year[count(. | key('year', .)[1]) = 1]">
<xsl:value-of select="."/>
</xsl:for-each>
<xsl:text> </xsl:text>
</xsl:template>
</xsl:stylesheet>
This produces the following output:
Action: year1994year1995
Comedy:
Drama:
Thriller: year1996
As you can see, each year is listed only once. The desired output would have been:
Action: year1994year1995
Comedy: year1994year1995
Drama: year1994
Thriller: year1994year1996
Here's a different implementation of Muenchian grouping - one that allows you to parametrize the criteria by which the movies are selected.
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:exsl="http://exslt.org/common"
extension-element-prefixes="exsl">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:param name="genre" select="'Comedy'"/>
<xsl:param name="director" select="'B'"/>
<xsl:key name="year" match="year" use="." />
<xsl:template match="/">
<!-- first pass -->
<xsl:variable name="years">
<xsl:for-each select="moviesByYear/*/movie[genre=$genre or director=$director]">
<year><xsl:value-of select="local-name(..)"/></year>
</xsl:for-each>
</xsl:variable>
<xsl:variable name="years-set" select="exsl:node-set($years)" />
<!-- final pass -->
<output>
<xsl:for-each select="$years-set/year[count(. | key('year', .)[1]) = 1]">
<xsl:copy-of select="."/>
</xsl:for-each>
</output>
</xsl:template>
</xsl:stylesheet>
When the above is applied to your example input, the result is:
<?xml version="1.0" encoding="UTF-8"?>
<output>
<year>year1994</year>
<year>year1995</year>
</output>
With regard to your modified input, I believe I would do it this way:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:exsl="http://exslt.org/common"
extension-element-prefixes="exsl">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:param name="director" select="'B'"/>
<xsl:key name="movies-by-genre" match="movie" use="genre" />
<xsl:key name="movies-by-director" match="movie" use="director" />
<xsl:key name="year" match="year" use="." />
<xsl:template match="/">
<output>
<xsl:apply-templates select="root/genres/genre"/>
</output>
</xsl:template>
<xsl:template match="genre">
<!-- first pass -->
<xsl:variable name="years">
<xsl:for-each select="key('movies-by-genre', .) | key('movies-by-director', $director)">
<year><xsl:value-of select="local-name(..)"/></year>
</xsl:for-each>
</xsl:variable>
<xsl:variable name="years-set" select="exsl:node-set($years)" />
<!-- final pass -->
<genre name="{.}">
<xsl:for-each select="$years-set/year[count(. | key('year', .)[1]) = 1]">
<xsl:copy-of select="."/>
</xsl:for-each>
</genre>
</xsl:template>
</xsl:stylesheet>
The result here is:
<?xml version="1.0" encoding="UTF-8"?>
<output>
<genre name="Action">
<year>year1994</year>
<year>year1995</year>
</genre>
<genre name="Comedy">
<year>year1994</year>
<year>year1995</year>
</genre>
<genre name="Drama">
<year>year1994</year>
</genre>
<genre name="Thriller">
<year>year1994</year>
<year>year1996</year>
</genre>
</output>
Note: the two added keys are for efficiency only - they are not required for the main purpose here.
On second thought, we could do this all in a single pass, thus (hopefully) avoiding the issues Xalan and MSXSML have with processing a variable - but still using Muenchian grouping:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:param name="director" select="'B'"/>
<xsl:key name="year" match="moviesByYear/*" use="local-name()" />
<xsl:template match="/">
<output>
<xsl:apply-templates select="root/genres/genre"/>
</output>
</xsl:template>
<xsl:template match="genre">
<xsl:variable name="genre" select="." />
<genre name="{$genre}">
<xsl:for-each select="../../moviesByYear/*
[count(. | key('year', local-name())[1]) = 1]
[key('year', local-name())/movie[genre=$genre or director=$director]]">
<year>
<xsl:value-of select="local-name()"/>
</year>
</xsl:for-each>
</genre>
</xsl:template>
</xsl:stylesheet>