Subject: Re: [xsl] Show YEAR once if the YEAR same as preious element by xsl From: S Woodside <sbwoodside@xxxxxxxxx> Date: Sun, 27 Apr 2003 02:41:22 -0400 |
try this: <xsl:apply-template select="row[@year=2001]" mode="year_only"/> <xsl:apply-template select="row[@year=2001]" mode="month"/>
<xsl:template match="row" mode="year_only"> <xsl:value-of select="@year"/><xsl:text>/</xsl:text> </xsl:template> <xsl:template match="row" mode="month"> <xsl:value-of select="@month"/><br/> </xsl:template>
hi expert, sorry for the syntax error
<record> <row year="2001" month="1"/> <row year="2001" month="2"/> <row year="2001" month="3"/> <row year="2001" month="4"/> <row year="2002" month="1"/> <row year="2002" month="2"/> <row year="2002" month="3"/> <row year="2003" month="1"/> <row year="2003" month="2"/></record>
From: "Cheung Tin Po" <timothy298@xxxxxxxxxxx>
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CC: timothy298@xxxxxxxxxxx
Subject: [xsl] Show YEAR once if the YEAR same as preious element by xsl
Date: Sun, 27 Apr 2003 03:57:44 +0000
hi expert,
i get the following XML file
<record> <row year="2001" month="1"> <row year="2001" month="2"> <row year="2001" month="3"> <row year="2001" month="4"> <row year="2002" month="1"> <row year="2002" month="2"> <row year="2002" month="3"> <row year="2003" month="1"> <row year="2003" month="2"> </record>
i want to transform it to html with the following result
2001/1 2 3 4 2002/1 2 3 2003/1 2
as XSLT cannot store the variable so i cannot think any method to should the result.
:< From Timothy
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Re: [xsl] Show YEAR once if the YEA, Cheung Tin Po | Thread | [xsl] XSL FO - undeclared namespace, Mark Ivs |
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