select="node1", select="node2"...

Baldo · March 12th, 2004, 08:30 AM

I've got (XML):

Code:

<root>
   <node1> bla bla </node1>
   <node2> bla bla </node1>
   <node3> bla bla </node1>
   <node4> bla bla </node1>
   <node5> bla bla </node1>
</root>

I've got (XSLT):

Code:

<xsl:call-template name="FOO">
   <xsl:with-param name="number" select="1"/>
</xsl:call-template>

and the template:

Code:

<xsl:template name="FOO">
   <xsl:param name="number"/>
   <xsl:copy-of select="concat(node,$number)"/>
</xsl:template>

I wanna copy-of node1, node2, node3
but the concat(node,$number) is not the right way
(Can not convert #STRING to a NodeList!)

is there a solution?

xslt learner :)

pgtips · March 12th, 2004, 08:48 AM

I don't know of any way to dynamically create selection paths. If you had written your xml like this instead:

Code:

<root>
   <node num="1"> bla bla </node>
   <node num="2"> bla bla </node>
   <node num="3"> bla bla </node>
   <node num="4"> bla bla </node>
   <node num="5"> bla bla </node>
</root>

then you could just use

Code:

<xsl:copy-of select="node[@num=$number]"/>

joefawcett · March 12th, 2004, 08:54 AM

Try:

Code:

<xsl:template name="FOO">
   <xsl:param name="number"/>
   <xsl:copy-of select="/root/*[name() = concat('node', $number)]"/>
</xsl:template>

or

Code:

<xsl:template name="FOO">
   <xsl:param name="number"/>
   <xsl:copy-of select="//*[name() = concat('node', $number)]"/>
</xsl:template>

if you don't the exact position of the nodes to find. If yur source document has namespaces you will need to change name() to local-name() or alter the XPath to include the namespace.

Joe (MVP - xml)

Baldo · March 12th, 2004, 09:39 AM

Quote:

quote:Originally posted by pgtips
I don't know of any way to dynamically create selection paths. If you had written your xml like this instead:

tnx, but source XML was like I wrote (and it's a problem since it's not the classic <node id="3"/>

xslt learner :)

Baldo · March 12th, 2004, 09:49 AM

Quote:

quote:Originally posted by joefawcett
Try:

Code:

<xsl:template name="FOO">
   <xsl:param name="number"/>
   <xsl:copy-of select="/root/*[name() = concat('node', $number)]"/>
</xsl:template>

GREAT. it worked!
what is the "*" ?

(try to find it in my wrox manual)

xslt learner :)

Baldo · March 12th, 2004, 09:50 AM

page 394 :)
NameTest

xslt learner :)

joefawcett · March 12th, 2004, 10:33 AM

Quote:

quote:GREAT. it worked!
what is the "*" ?

It is short for all nodes of the current type on the child axis, in this case the type is element so it's short for all child elements.

Joe (MVP - xml)

Baldo · March 12th, 2004, 10:38 AM

Quote:

quote:Originally posted by joefawcett

Quote:

quote:GREAT. it worked!

Quote:

what is the "*" ?

It is short for all nodes of the current type on the child axis, in this case the type is element so it's short for all child elements.

I'm so sorry. I already knew "*". It's an important "element" of xslt. :)

xslt learner :)