[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] problem using param passed
Hi, Kindly have a look at the code below , here I am not able to pass $whichNode parameter , the code doesn't work for some reason , I hope this is the right way to use parameters passed to a template. In the XML file "Article/ArticleTitle" contains several "Author" nodes containing author names. Also surprisingly the code works if "Author" is put in $whichNode's place, but that's not the way I want the code to work. Any help would be greatly appreciated. <xsl:template match="*" mode="fieldInNewLine"> <xsl:param name="fieldName"/> <xsl:param name="whichNode"/> <xsl:choose><!-- dont print Authors if none is available --> <xsl:when test="'' != string($whichNode[position()=1])"> <td align="right" valign="top"><b>Author(s): </b></td> <td colspan="7" class="left"> <xsl:for-each select="$whichNode"> <xsl:apply-templates select="."/> <xsl:if test="position() != last()"> ,<xsl:text> </xsl:text> </xsl:if> </xsl:for-each>. </td> </xsl:when> </xsl:choose> </xsl:template> <xsl:apply-templates select="Article/ArticleTitle" mode="fieldInNewLine"> <xsl:with-param name="fieldName">Article Title:</xsl:with-param> <xsl:with-param name="whichNode">Author</xsl:with-param> </xsl:apply-templates> Regards, ~Manish Jamadagni ______________________________________________________ 123India.com - India's Premier Portal Get your Free Email Account at http://www.123india.com XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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