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getting all nodes from a certain level in the xml hierarchy
- From: "Peter Menzel" <mai00bfy at studserv dot uni-leipzig dot de>
- To: <XSL-List at lists dot mulberrytech dot com>
- Date: Fri, 27 Sep 2002 11:14:12 +0200
- Subject: [xsl] getting all nodes from a certain level in the xml hierarchy
- Reply-to: xsl-list at lists dot mulberrytech dot com
Hallo
my problem is the following:
my XML document maps the structure of a folder tree (just like the unix file
hierarchy) but without files, e.g.:
<Folder NAME="/">
<Folder NAME="a"/>
<Folder NAME="b">
<Folder NAME="ba"/>
<Folder NAME="bb"/>
</Folder>
...
<Folder NAME="z">
<Folder NAME="za">
...
<Folder NAME="very deep folder"/>
...
</Folder>
</Folder
</Folder>
The real folder NAMEs are like real folder names, and have no specific
length or content.
The depth of the deepest node is unknown.
I need to access all nodes that have the same depth.
So first i need root, then all direct childs of root, then all nodes that
are two levels under root, because i want tu put them in a table like:
level | 0 | 1 | 2 ... x
------+---+-----+------ ---------------
| / | a | ba very deep folder
| | b | bb
...
| | z | za
first I started to try <xsl:for-each select="//Folder"> and then
<xsl:for-each select="//Folder/Folder"> but because I do not know the depth
of the tree, this won't work..
Can anybody help me, what is the direction i should go?
I don't think there is an easy way to use XPath for adressing nodes on the
same level?
Nice greetings, Peter
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