XSLT problem

[Gendag]

Hi all,
I have a simple xml file:

<imgs>
<img name="1"/>
<img name="2"/>
<img name="3"/>
<img name="4"/>
</imgs>

And i have the following (simplified) XSLT:

<?xml version="1.0"?>
<xsl:styleshe et xmlns:xsl="http ://www.w3.org/1999/XSL/Transform">

<xsl:template match="/imgs>
<for-each select="img">
<for-each select="../img">
<xsl:choose>
<xsl:when test="???">
<span style="color: red;">
<xsl:value-of select="@name"/>
</span>
</xsl:when>

<xsl:otherwis e>
<xsl:value-of select="@name"/>
</xsl:otherwise>
</xsl:choose>
</for-each>
</for-each>
</xsl:template>

</xsl:stylesheet>

What I want to do is: for each IMG, print _all_ the IMG and if the IMG
in the second for-each is equal to the IMG in the first for-each, color
it of red. But how can I say with XPath "if the IMG name in equal to the
name of the IMG in the outer for-each"? I remember that there is the
possibility to express it with XPath, but i don't remember how.

Thx!

Gendag

[Joris Gillis]

Re: XSLT problem

Hi,
Tempore 21:28:13, die Sunday 25 September 2005 AD, hinc in foro {comp.text.xml} scripsit Gendag <thenews-no_spam_please@ libero.it>:
[color=blue]
> What I want to do is: for each IMG, print _all_ the IMG and if the IMG
> in the second for-each is equal to the IMG in the first for-each, color
> it of red. But how can I say with XPath "if the IMG name in equal to the
> name of the IMG in the outer for-each"? I remember that there is the
> possibility to express it with XPath[/color]

There is no such Xpath solution.
You'll have to use at least one variable.
You can use 'current()' to select to context node within Xpath predicates, but there's no way to recall the context node of an outer loop.

regards,
--
Joris Gillis (http://users.telenet.be/root-jg/me.html)
«Numquam mores quos extuli refero» - Seneca , Ad Lucilium I, 7

[toudidel]

Re: XSLT problem

don't use <xsl:for-each> - it's terrible construction (instead of externl
xml document's "looping" with document() function). Instead of it you should
use <xsl:apply-templates match="img"> and <xsl:template match="img">... to
do it

[Gendag]

Re: XSLT problem

Joris Gillis wrote:
[color=blue]
> There is no such Xpath solution.
> You'll have to use at least one variable.
> You can use 'current()' to select to context node within Xpath
> predicates, but there's no way to recall the context node of an outer loop.[/color]

Thx for the advice, I think I use a variable to solve the problem.
The function of which i didn't remember the name was 'current()', but as
you say, it woks only insiede an Xpath. Thx to rember me the nome of it :-)

Gendag

[Gendag]

Re: XSLT problem

toudidel wrote:[color=blue]
> don't use <xsl:for-each> - it's terrible construction (instead of externl
> xml document's "looping" with document() function). Instead of it you should
> use <xsl:apply-templates match="img"> and <xsl:template match="img">... to
> do it[/color]

Thx of the advice.
I know that a recursive style is better of the iterative one, but I used
the 'for-each' beacuse the stylesheet is very simple and i wrote it
quickly :-)

Gendag