[Zoe]

> I want to display the specific service id="01"'s name.
> how can I approach this? The result will be 'My News'.
> does it have any condition that I can use? Thanks.
>
> <xsl:for-each select="dynamictree/category/service"....? >
> <!-- serviceid =01, name="My News" -->
> </xsl:for-each>
>

If you literally mean what you ask, then you only want to see services
having a id value of "01".  You can get them, using your selector in the
xsl:for-each element, like this:

<xsl:if test='@id="01"><xsl:value-of  select='@name'/></xsl:if>

But if you only are interested in those services, you could include them in
the specification of the selected nodes:

select='dynamictree/category/service[@id="01"]'

If so, you could omit the xsl:if test for an "01" value.

If, on the other hand, you actually meant that you want to simply extract
the id and name of each service regardless of its id, you could use
something like this:

id: <xsl:value-of select='@id'/>, name:<xsl:value-of select='@name'/>

It all depends on what you really want to accomplish.

Cheers,

Tom P

> XML file:
> - <dynamictree version="1" statuscode="0" statusdesc="OK">
> - <category id="1" name="Favorite">
> - <service id="01" name="My News">
>   <querykeyword id="7003" name="News USA" />
>   <querykeyword id="7004" name="News Africa" />
>   </service>
> - <service id="02" name="My Stocks">
>   <querykeyword id="7001" name="Stock Quotes" />
>   </service>
>   </category>
>   </dynamictree>
>
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list