Subject: Re: [xsl] XPath expression that yields the same result as xsl:for-each-group? From: "Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Thu, 30 May 2019 13:20:40 -0000 |
What XPath expression will yield the desired value for $groups?
<xsl:variable name="groups" select="???" />
Is there an XPath 2.0 expression that can be used?
What kind of sequence type would you expect that XPath expression to return? You can't build sequences of sequences to somehow wrap the items belonging to a group.
So unless you could move to XPath 3.1 with maps and/or arrays I don't see an XPath type constructed with pure XPath to hold groups. And of course even there it would be easier and smarter to use XSLT to create the map or the nested sequence/array or array/sequence structure than to rely on pure XPath I think.
A pure XPath 3.1 expression to build the groups as a sequence of array(element(row)) would be
let $keys := distinct-values($rows/concat(ARPT__IDENT, '|', TRM__IDENT)) return for $key in $keys return array { $rows[$key = concat(ARPT__IDENT, '|', TRM__IDENT)] }
<xsl:template match="Test"> <xsl:variable name="rows" select="row" as="element(row)*"/> <xsl:variable name="groups" as="array(element(row))*" select="let $keys := distinct-values($rows/concat(ARPT__IDENT, '|', TRM__IDENT)) return for $key in $keys return array { $rows[$key = concat(ARPT__IDENT, '|', TRM__IDENT)] }"/> <results> <xsl:for-each select="$groups"> <group> <xsl:sequence select="?*" /> </group> </xsl:for-each> </results> </xsl:template>
But I think using XSLT 2/3 grouping to build a grouping structure, be it an XML one in XSLT 2 or such a nested sequence of arrays if needed in XSLT 3 is much easier and probably more efficient anyway.
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