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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: for-each order
Nope, That doesn't do it . ----- Original Message ----- From: "Long Zhao" <LZhao@xxxxxxxxxxxxxxxx> To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Sent: Friday, December 14, 2001 9:34 AM Subject: RE: for-each order > how about this > > <xsl:variable name="position"> > <xsl:for-each select="$path"> > <xsl:value-of select="concat(@value, position)"/>, > </xsl:for-each> > </xsl:variable> > > Long > > -----Original Message----- > From: Charly [mailto:cohana@xxxxxxxxxxxxxxx] > Sent: Friday, December 14, 2001 2:35 PM > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: for-each order > > > Hello, > Is there a way to do <xsl:for-each ..... and looping backwards. > > My XML looks like > <histo> > <bar value="60" legend="Jan" /> > <bar value="77" legend="Feb" /> > <bar value="53" legend="Mar" /> > <bar value="14" legend="Apr" /> > </histo> > > and I want to store 14,53,77,60 in a variable . > > > <xsl:variable name="position"> > <xsl:for-each select="$path"> > <xsl:value-of select="@value"/>, > </xsl:for-each> > </xsl:variable> > > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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