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Making balanced two-column tables from one-column data
Subject: Making balanced two-column tables from one-column data From: Philip Borenstein <pb@xxxxxxxxxx> Date: Wed, 19 Jan 2000 13:35:12 -0500 |
I have a solution to my problem, but it seems so un-XSLT-like, and ugly that I'm sure there must be a better solution. i think that part of the problem is that I just don't "get" XPaths. Pointers to good XPath learning resources much appreciated. I have the following XML: <TASKS> <TASK> <COMPONENTS> <COMPONENT>A</COMPONENT> <COMPONENT>B</COMPONENT> <COMPONENT>C</COMPONENT> <COMPONENT>D</COMPONENT> <COMPONENT>E</COMPONENT> <COMPONENT>F</COMPONENT> <COMPONENT>G</COMPONENT> <COMPONENT>H</COMPONENT> <COMPONENT>I</COMPONENT> <COMPONENT>J</COMPONENT> <COMPONENT>K</COMPONENT> </COMPONENTS> </TASK> </TASKS> I want to produce a two-column table like this: A G B H C I D J E K F (The number of COMPONENTS may or may not be even.) This is my XSL template for the COMPONENT element: <xsl:template match="COMPONENTS"> <P>There are <xsl:value-of select="count(COMPONENT)"/> components.</P> <TABLE> <xsl:variable name="h" select="round(count(COMPONENT) div 2)" /> <xsl:for-each select="COMPONENT"> <xsl:variable name="p" select="position()" /> <xsl:if test="$h>=$p"> <TR> <TD><xsl:value-of select="../COMPONENT[position()=$p]" /></TD> <TD><xsl:value-of select="../COMPONENT[position()=$p+$h]" /></TD> </TR> </xsl:if> </xsl:for-each> </TABLE> </xsl:template> This just strikes me as the wrong way to go about it. --philip borenstein. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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