Subject: Re: [xsl] Feedback on grouping solution From: "Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Sat, 26 Oct 2019 19:06:18 -0000 |
I need to process the <step> child elements so that the <figure> elements are always on the "right" (even-numbered position) in the output. Immediate children of the <procedure> do not factor into the odd/even sequence.
The first child of each group of adjacent <step> elements starts a new odd/even series. To ensure that the each <figure> is in an even-numbered position, I want to insert a <spacer> element where it is required.
Here is my stylesheet. My basic question is: is there a better or more efficient way to do this?
I think with XSLT 3 it is possible to use xsl:iterate on the child elements of the adjacent steps found by grouping:
<xsl:template match="procedure"> <xsl:copy> <xsl:for-each-group select="*" group-adjacent="boolean(self::step)"> <xsl:choose> <xsl:when test="current-grouping-key()"> <xsl:iterate select="current-group()/*"> <xsl:param name="position-in-output" select="1"/> <xsl:apply-templates select="."> <xsl:with-param name="position-in-output" select="$position-in-output"/> </xsl:apply-templates> <xsl:next-iteration> <xsl:with-param name="position-in-output" select="if (self::figure and $position-in-output mod 2 = 1) then $position-in-output + 2 else $position-in-output + 1"/> </xsl:next-iteration> </xsl:iterate> </xsl:when> <xsl:otherwise> <xsl:apply-templates select="current-group()"/> </xsl:otherwise> </xsl:choose> </xsl:for-each-group> </xsl:copy> </xsl:template>
<xsl:template match="step/figure"> <xsl:param name="position-in-output"/> <xsl:if test="$position-in-output mod 2 = 1"> <spacer/> </xsl:if> <xsl:next-match/> </xsl:template>
Now waiting for Dimitre posting a less "fancy" but equally compact XSLT 1 solution :)
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