Subject: Re: [xsl] Passing attributes in a node-set From: David Carlisle <davidc@xxxxxxxxx> Date: Tue, 17 Jan 2006 11:07:45 GMT |
> ..I want to have all values, then you want a node set of the attribute nodes. You can't do that by copying the attribute nodes to a newly constructed root element as an element can only hold one attribute with a given name, you want to just select the attributes, so <xsl:with-param name="value" select="//element/@avalue"/> now if you use select= rather than using xsl:with-param with content you generate a node set, which you can iterate over. I doubt you need to use an explict integer parameter: the system will do the counting for you, something like ... <xsl:for-each select="$value"> <xsl:variable name="name" select="."/> <xsl:for-each select="$value"> <xsl:element name="{$name}_{position()}"><xsl:value-of select="."/> </xsl:for-each> </xsl:for-each> which will make <foo_1>foo</foo_1> <foo_2>bar</foo_2> <bar_1>foo</bar_1> <bar_2>bar</bar_2> David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Passing attributes in a n, Frank Klasens | Thread | Re: [xsl] Passing attributes in a n, Frank Klasens |
Re: [xsl] Passing attributes in a n, Frank Klasens | Date | [xsl] integer comparision, T Uma Shankari |
Month |