[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: to extract the longest string (fwd)
I tested the following code in SAXON and I could not get the expected longest string, which is "cxxyycc". Please check Thanks <?xml version="1.0"?> <table> <row> <col align="l"/> <col align="l"/> <col align="l"/> </row> <row> <col align="cxxyy"/> <col align="c"/> <col align="c"/> </row> <row> <col align="cxx"/> <col align="c"/> <col align="c"/> </row> <row> <col align="cx"/> <col align="c"/> <col align="c"/> </row> </table> ********************************************************************** <?xml version="1.0"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xsl:version="1.0"> <xsl:template match="table"> <xsl:variable name="longest"> <xsl:apply-templates select="row[1]" mode="get-longest" /> </xsl:variable> </xsl:template> <xsl:template match="row" mode="get-longest"> <xsl:variable name="current"> <xsl:for-each select="col"> <xsl:value-of select="@align" /> </xsl:for-each> </xsl:variable> <xsl:variable name="longest"> <xsl:apply-templates select="following-sibling::row[1]" mode="get-longest" /> </xsl:variable> <xsl:choose> <xsl:when test="string-length($longest) > string-length($current)"> <xsl:value-of select="$longest"/> </xsl:when> <xsl:otherwise><xsl:value-of select="$current" /></xsl:otherwise> </xsl:choose> </xsl:template> </xsl:stylesheet> Rajagopal XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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