grouping nodes in list

Press Ctrl / CMD + C to copy this to your clipboard.

Copied to clipboard

Flag this post as spam?

This post will be reported to the moderators as potential spam to be looked at

Claushingebjerg 886 posts 2415 karma points

Jan 19, 2012 @ 10:38

Grouping nodes in list

Hi im trying to wrap subpages from an id in pairs of 2, but im kinda stuck.
So any pointer would be very nice.
Right now i have the following code, which i read up on somewhere, but i think either my scpoping or matching is off.

<xsl:variable name="source" select="1056"/>
<xsl:variable name="n" select="number(2)"/>

<xsl:template match="/">
  <div id="testimonials">
    <xsl:apply-templates select="umbraco.library:GetXmlNodeById($source)/node [position() mod $n = 1]"/>
</div>
</xsl:template>  
  
<xsl:template match="node">
<div class="clients-testimonials">
  <xsl:for-each select=". | following-sibling::item[position() &lt; $n]">
    <blockquote>
      Output node values here
    </blockquote > 
  </xsl:for-each>
</div>    
</xsl:template>        
</xsl:stylesheet>

Copy Link

Rodion Novoselov 694 posts 859 karma points

Jan 19, 2012 @ 11:01

Hi. It can be done e.g. the following way:

  <xsl:for-each select="$node/*[@isDoc][string(umbracoNaviHide) != '1']">
    <xsl:if test="position() mod 2 = 1">
      <xsl:value-of select="'&lt;div&gt;'" disable-output-escaping="yes"/>
    </xsl:if>
    <p>
    <xsl:value-of select="@nodeName"/>
    </p>
    <xsl:if test="position() mod 2 = 0 or position() = last()">
      <xsl:value-of select="'&lt;/div&gt;'" disable-output-escaping="yes"/>
    </xsl:if>
  </xsl:for-each>

Copy Link

Chriztian Steinmeier 2726 posts 8320 karma points MVP 4x admin c-trib

Jan 19, 2012 @ 11:07

Hi Claus,

In your node template you're selecting <item> siblings - probably a typo, right? Try this instead:

<xsl:variable name="source" select="1056" />
<xsl:variable name="n" select="2" />

<xsl:template match="/">
    <div id="testimonials">
        <xsl:apply-templates select="umbraco.library:GetXmlNodeById($source)/node[position() mod $n = 1]" />
    </div>
</xsl:template>  

<xsl:template match="node">
    <div class="clients-testimonials">
        <xsl:for-each select=". | following-sibling::node[position() &lt; $n]">
            <blockquote>
                Output node values here
            </blockquote> 
        </xsl:for-each>
    </div>
</xsl:template>

/Chriztian

Copy Link

Claushingebjerg 886 posts 2415 karma points

Jan 19, 2012 @ 11:36

0

Hmmm Chriztian.
It still doesnt work.It renders the main div "Testemonials" but doesnt render "client-testemonials" at all.

Copy Link
Chriztian Steinmeier 2726 posts 8320 karma points MVP 4x admin c-trib

Jan 19, 2012 @ 11:42

0

Hi Claus - works on my machine :-)

Ok, you're using the old schema, right? (Hence the "node" element)

/Chriztian

Copy Link
Claushingebjerg 886 posts 2415 karma points

Jan 19, 2012 @ 11:43

0

no, im on 4.7.1

Copy Link

Chriztian Steinmeier 2726 posts 8320 karma points MVP 4x admin c-trib

Jan 19, 2012 @ 11:48

OK - that explains it - you need to use *[@isDoc] instead of node (or better: If they're all the same Document Type - use the alias, e.g.: Testimonial)

<xsl:templatematch="/">
               <divid="testimonials">
                <xsl:apply-templatesselect="umbraco.library:GetXmlNodeById($source)/*[@isDoc][position() mod $n = 1]"/>
                </div>
        </xsl:template>  

        <xsl:templatematch="*[@isDoc]">
                <divclass="clients-testimonials">
                        <xsl:for-eachselect=". | following-sibling::*[@isDoc][position() &lt; $n]">
                                <blockquote>
                                        <xsl:value-ofselect="@nodeName"/>
                                </blockquote> 
                        </xsl:for-each>
                </div>
        </xsl:template>

/Chriztian

EDIT: Updated to use GetXmlNodeById()

Copy Link

Claushingebjerg 886 posts 2415 karma points

Jan 19, 2012 @ 12:52
0
im going crazy!

Ive copy/pasted your code, but it still doesnt render the inner div... Ive been staring at it for ages, but cant seem to find any errors!

Doing a normal
```
<xsl:for-each select="umbraco.library:GetXmlNodeById(1056)/* [@isDoc and string(umbracoNaviHide) != '1']">
      <xsl:value-of select="@nodeName"/>
</xsl:for-each>
```
Works fine, so the nodes are published and accessible...
Copy Link
Rodion Novoselov 694 posts 859 karma points

Jan 19, 2012 @ 12:57

0

It seems there's a misspelling - "node" means any node with the name "node".

To select all nodes you need to spell it as "node()".

Copy Link
Chriztian Steinmeier 2726 posts 8320 karma points MVP 4x admin c-trib

Jan 19, 2012 @ 13:00

0

Hi Claus,

My bad!! Really - I forgot to use GetXmlNodeById() - (on my local machine I use something else that doesn't work in Umbraco context).

I've updated the code above - please try again.

@Rodion: "node" in this context is actually from the old XML Schema in Umbraco; Bit confusing, yes, if you're used to XSLT :-)

Copy Link
Claushingebjerg 886 posts 2415 karma points

Jan 19, 2012 @ 13:05
0
SUCCESS!!!! i thought
```
//*
```
was som kind of fancy shorthand :)
Copy Link
Chriztian Steinmeier 2726 posts 8320 karma points MVP 4x admin c-trib

Jan 19, 2012 @ 13:12

0

It's a very bad shorthand - but works fine when you've only got a small set of nodes for testing :-)

/Chriztian

Copy Link
abhijeetwaychal 5 posts 25 karma points

Dec 02, 2012 @ 14:40

0

Thanks Chriztian and Claus

Updatd XSLT worked at it is with modifcation //*

Copy Link
is working on a reply...

This forum is in read-only mode while we transition to the new forum.

You can continue this topic on the new forum by tapping the "Continue discussion" link below.

Please Sign in or register to post replies