Subject: [xsl] Count the node from zero instead of one. From: Rashi Bhardwaj <rashi.bhardwaj@xxxxxxxxx> Date: Fri, 26 Nov 2010 12:38:53 +0530 |
HI All, I m counting the node position from this logic in the below sample xml xml: <test> <a> <b name ='1'></b> <b name ='2'></b> <b name ='3'> <c>aaa</c> </b> <b name ='4'> <c>bbb</c> <c>ccc</c> </b> <b name ='4'> <c>dddd</c> <c>eeee</c> </b> </a> <a> <b name ='1'></b> <b name ='2'> <c>fffff</c> </b> <b name ='3'></b> <b name ='4'> <c>gggg</c> </b> </a> </test> <xsl:template name="CountNode"> <xsl:param name="node" select="//b[c[preceding-sibling::c]][not(@name =preceding::b[child::c]/@name)]/@name"/> <xsl:for-each select="$node"> <xsl:element name="position"> <xsl:number count="*"/> <xsl:text>,</xsl:text> </xsl:element> <xsl:if test="position()!=last()"/> </xsl:for-each> </xsl:template> it gives the result 3,4,2,....I want it should count from zero instead of one and the result shld be 2,3,1.or it can print the result by subracing 1 from it like 3-1=2. Please suggest some thing... Thanks.... Rashi
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