node's first child node - XSLT

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Sony 20 posts 40 karma points

Apr 12, 2011 @ 12:32

0

node's first child node

XSLT

hello friends,

i have subnodes(say 10 subnodes) .each subnodes have 5 to 10 childnodes..

how to find each subnode's first child alone..

thank you

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gilad 185 posts 425 karma points

Apr 12, 2011 @ 12:58

I'm not sure but I think it can work


 <xsl:for-each select="$currentPage/* [@isDoc and string(umbracoNaviHide) != '1']">
       <xsl:for-each select="current()/* [@isDoc and string(umbracoNaviHide) != '1']">
           <xsl:if test="position() = 1">
              { i am the first }
           </xsl:if>      
       </xsl:for-each>
</xsl:for-each>

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elspiko 133 posts 302 karma points

Apr 12, 2011 @ 13:04

Alternativly you could use:

<xsl:for-each select="$currentPage/* [@isDoc and string(umbracoNaviHide) != '1']">

<xsl:value-of select="current()/*[@isDoc and string(umbracoNaviHide) != '1'][1]"/>

</xsl:for-each>

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Sony 20 posts 40 karma points

Apr 12, 2011 @ 13:52

0

here are my nodes

-home(home page)

      - gallery(subpage)

         -men wear(subpage)

                  -image1(picture documentype)

-imag2(picture documentype)

            -women wear(subpage)

-image3(picture documentype)

-imag4(picture documentype)

i want to get the image1, image3 as such.. i am using v4.0.3

thank you

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elspiko 133 posts 302 karma points

Apr 12, 2011 @ 13:53
0
Whats the doctype alias for gallery?

EDIT

Either way it'll be something like...
```
<xsl:for-each select="$gallery/node">
Name: <xsl:value-of select="./node[1]/@nodeName" />
</xsl:for-each>
```
Where $gallery is the gallery node. You might wish to apply further filtering on doc types
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Sony 20 posts 40 karma points

May 23, 2011 @ 16:27

0

if this could be done using razor?

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