stupidly basic xslt question

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Dan 1288 posts 3921 karma points c-trib

Jan 29, 2010 @ 14:51
0

Stupidly basic xslt question
Just as I think I'm getting the hang of xslt, being able to traverse nodes etc, I'm stuck with something mega mega basic.

How do I output the current node name (I know I could do this with an umbraco field in the template but it's more complicated than that)? I've done the following but it returns nothing:
```
<xsl:variable name="pageNode" select="$currentPage/node"/>
<h2>
    <xsl:value-of select="@nodeName"/>
</h2>
```
Please somebody put me out of my misery!
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Thomas Höhler 1237 posts 1709 karma points MVP

Jan 29, 2010 @ 15:01
0
You are at the current node so it has to be:
```
<xsl:value-of select="$currentPage/@nodeName" />
```
hth, Thomas
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Dan 1288 posts 3921 karma points c-trib

Jan 29, 2010 @ 15:02

0

Bingo, thanks Thomas.

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Nik Wahlberg 639 posts 1237 karma points MVP

Jan 29, 2010 @ 15:03
0
You should be able to do
```
<xsl:value-of select="$currentPage/@nodeName"/>
```
HTH

-- Nik
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Nik Wahlberg 639 posts 1237 karma points MVP

Jan 29, 2010 @ 15:03

0

way too late...:)

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Thomas Höhler 1237 posts 1709 karma points MVP

Jan 29, 2010 @ 15:04

0

For documentation, take a look at the wiki:

http://our.umbraco.org/wiki/reference/xslt

or Dougs great posts:

http://blog.percipientstudios.com/2009/4/11/anatomy-of-an-umbraco-xslt-file.aspx

http://blog.percipientstudios.com/2009/5/5/your-first-umbraco-xslt-macro.aspx

Cheers, Thomas

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