getting nodes on the same level including the nodes above - XSLT

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Dennis Aaen 4499 posts 18254 karma points admin hq c-trib

Dec 30, 2010 @ 17:05

Getting nodes on the same level including the nodes above $currentPage

XSLT

Hi,

I am working on a solution where the sub menu should be shown if $currentPage has no subpages.

So if $currentPage has no childs the sub menu must consist of pages on the same level as you stand on.

I got it to list all following-sibling, but my problem is that I also need the siblings above $CurrentPage.

I'm using umbraco umbraco v 4.5.1.

My XSLT code looks like:

<xsl:variable name="count" select="count($currentPage/child::* [@isDoc and string(umbracoNaviHide) != '1'])"/>

<!-- The fun starts here -->

<div id="submenu"> 
<xsl:choose>
<xsl:when test="$count != 0">
<ul>
<xsl:for-each select="$currentPage/child::* [@isDoc and string(umbracoNaviHide) != '1']">
  <li>
    <a href="{umbraco.library:NiceUrl(@id)}">
      <xsl:value-of select="pcd_master_menuTitle"/>
    </a>
  </li>
</xsl:for-each>
</ul>
</xsl:when>
<xsl:otherwise>
  <ul>
<xsl:for-each select="$currentPage/following-sibling::* [@isDoc and string(umbracoNaviHide) != '1']">
  <li>
    <a href="{umbraco.library:NiceUrl(@id)}">
      <xsl:value-of select="pcd_master_menuTitle"/>
    </a>
  </li>
</xsl:for-each>
</ul>
</xsl:otherwise>
</xsl:choose>

 </div>

Hope someone can help me:)

/ Dennis Aaen

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Kim Andersen 1447 posts 2196 karma points MVP

Dec 30, 2010 @ 19:23
2
Hi Dennis

Could you try changing the for-each from this:
```
<xsl:for-each select="$currentPage/following-sibling::* [@isDoc and string(umbracoNaviHide) != '1']">
```
To this:
```
<xsl:for-each select="$currentPage/parent::*[@isDoc]/* [@isDoc and string(umbracoNaviHide) != '1']">
```
Then we'll go up to the parent node, and list all the childnodes from there. Haven't tested it though, but I think it will work.

/Kim A
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Dennis Aaen 4499 posts 18254 karma points admin hq c-trib

Dec 30, 2010 @ 19:27

0

Thanks Kim,

It worked just as it should:)

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Kim Andersen 1447 posts 2196 karma points MVP

Dec 30, 2010 @ 19:33

0

Great to hear Dennis.

As far as I know there's not a way of getting all of the siblings with XPath like you can get the parent, child, following-sibling etc. It would be nice though, to have an axis just called "sibling" or "siblings" that would get all of the siblings to the current node.

/Kim A

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Dennis Aaen 4499 posts 18254 karma points admin hq c-trib

Dec 30, 2010 @ 19:42

0

Yes I also looked after an XPath expression as sibling or siblings, which would then list all Siblings, but unfortunately there was not such an XPath expression:)

So glad that you could help me.

/Dennis Aaen

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Chriztian Steinmeier 2798 posts 8788 karma points MVP 7x admin c-trib

Dec 30, 2010 @ 23:22
0
Just chippin' in with a couple of cents...

Since an XML node can only ever have one single parent node, I actually think that it's pretty efficient to go up to that one and then select all children (which itself is a fast XPath selection), instead of trying to combine preceding-sibling::, self:: and following-sibling:: in another axis...

One tiny optimization to the selection would be to skip the [@isDoc] predicate in the selection - currentPage is always a document; and a document can only be the child of another document, so...

In fact, you can just use the ../ shortcut to select the parent with:
```
<xsl:for-each select="$currentPage/../*[@isDoc][not(umbracoNaviHide = 1)]">
```
Anyways - long reply to an already answered, closed & archived question - must be the Christmas Spirit in me :-)

...and if you really want that siblings axis - you could just create it yourself - here's a hint:
```
<!ENTITY siblings "$currentPage/../*">
```
:-)

/Chriztian
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Dennis Aaen 4499 posts 18254 karma points admin hq c-trib

Dec 30, 2010 @ 23:45

0

Hi Chriztian,

Thanks for your suggestions.

It's always good to see several different ways to solve the same problems on. Especially for me who has worked with Umbraco professionally for about 4 months.

So I'm glad to see several ways to solve the same problem.:-)

/Dennis

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