问题
With the following file as input of xsltprocessor :
mylibrary.xml :
<library>
<book isbn="1"/>
<book isbn="3"/>
<book isbn="5"/>
</library>
and this one that can be used : bookreference.xml :
<reference>
<book isbn="1">
<category>SF</category>
</book>
<book isbn="2">
<category>SF</category>
</book>
<book isbn="3">
<category>SF</category>
</book>
<book isbn="4">
<category>Comedy</category>
</book>
<book isbn="5">
<category>Comedy</category>
</book>
</reference>
i want to get the numbers of book i got in mylibrary, groupby category, using xslt 1-0.
output wanted:
SF : 2 book(s)
Comedy : 1 book(s)
here is the xsl i write using Martin Honnen method explains in 'Grouping when using 2 different XML files as sources?' , i think that solve the problem but i do not validate yet and perhaps someone have a better solution.
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:exsl="http://exslt.org/common"
exclude-result-prefixes="exsl"
version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:key name="book-by-category" match="book" use="category"/>
<xsl:param name="bookref" select="'bookreference.xml'"/>
<xsl:variable name="doc" select="document($bookref)/reference"/>
<xsl:variable name="rtf">
<xsl:apply-templates select="//library" mode="merge"/>
</xsl:variable>
<xsl:template match="library" mode="merge">
<xsl:element name="mybookcat">
<xsl:for-each select="book">
<xsl:variable name="isbn" select="@isbn"/>
<xsl:element name="book">
<xsl:attribute name="isbn"><xsl:value-of select="$isbn"/></xsl:attribute>
<xsl:for-each select="$doc/book[@isbn=$isbn]">
<xsl:element name="category">
<xsl:value-of select="./category"/>
</xsl:element>
</xsl:for-each>
</xsl:element>
</xsl:for-each>
</xsl:element>
</xsl:template>
<xsl:template match="/">
<xsl:apply-templates select="exsl:node-set($rtf)/mybookcat"/>
</xsl:template>
<xsl:template match="mybookcat">
<xsl:for-each select="book[count(. | key('book-by-category',category)[1]) = 1]">
<xsl:sort select="category"/>
<xsl:value-of select="category" /><xsl:text> : </xsl:text>
<xsl:variable name="current-cat" select="key('book-by-category', category)"/>
<xsl:value-of select="count($current-cat)"/><xsl:text> book(s)
</xsl:text>
<xsl:for-each select="key('book-by-category', category)">
<xsl:sort select="@isbn" data-type="number" />
<xsl:value-of select="@isbn" /><xsl:text>
</xsl:text>
</xsl:for-each>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
回答1:
here is the xsl i write using Martin Honnen method explains in 'Grouping when using 2 different XML files as sources?' , i think that solve the problem but i do not validate yet and perhaps someone have a better solution
.
Here is a simpler XSLT 1.0 solution that doesn't use any extension functions or xsl:for-each:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="kBookByCat" match="book"
use="category"/>
<xsl:variable name="vRef" select=
"document('file:///c:/temp/delete/reference.xml')"/>
<xsl:variable name="vMyIsbns" select="/*/*/@isbn"/>
<xsl:variable name="vResult">
<xsl:apply-templates select="$vRef/*"/>
</xsl:variable>
<xsl:template match="/">
<xsl:copy-of select="$vResult"/>
</xsl:template>
<xsl:template match=
"book[generate-id()
=
generate-id(key('kBookByCat', category)[1])
]
">
<xsl:variable name="vBooksinCat" select=
"key('kBookByCat', category)"/>
<xsl:value-of select="category"/> : <xsl:text/>
<xsl:value-of select="count($vBooksinCat[@isbn=$vMyIsbns])"/>
<xsl:text> book(s)
</xsl:text>
</xsl:template>
<xsl:template match="text()"/>
</xsl:stylesheet>
When applied on the provided XML document contained in MyLibrary.xml:
<library>
<book isbn="1"/>
<book isbn="3"/>
<book isbn="5"/>
</library>
and having this provided XML document contained in the file C:\temp\delete\reference.xml:
<reference>
<book isbn="1">
<category>SF</category>
</book>
<book isbn="2">
<category>SF</category>
</book>
<book isbn="3">
<category>SF</category>
</book>
<book isbn="4">
<category>Comedy</category>
</book>
<book isbn="5">
<category>Comedy</category>
</book>
</reference>
the wanted, correct result is produced:
SF : 2 book(s)
Comedy : 1 book(s)
II. An XSLT 2.0 solution:
This is slightly simpler, as we can define the key to be dependent on the second document.
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="kBookByCat" match="book[@isbn = $vMyIsbns]"
use="category"/>
<xsl:variable name="vRef" select=
"document('file:///c:/temp/delete/reference.xml')"/>
<xsl:variable name="vMyIsbns" select="/*/*/@isbn"/>
<xsl:variable name="vResult">
<xsl:apply-templates select="$vRef/*"/>
</xsl:variable>
<xsl:template match="/">
<xsl:copy-of select="$vResult"/>
</xsl:template>
<xsl:template match=
"book[generate-id()
=
generate-id(key('kBookByCat', category)[1])
]
">
<xsl:variable name="vBooksinCat" select=
"key('kBookByCat', category)"/>
<xsl:value-of select="category"/> : <xsl:text/>
<xsl:value-of select="count($vBooksinCat)"/>
<xsl:text> book(s)
</xsl:text>
</xsl:template>
<xsl:template match="text()"/>
</xsl:stylesheet>
来源：https://stackoverflow.com/questions/8334375/groups-two-xml-files-like-a-sql-group-by