Subject: [xsl] getting all nodes from a certain level in the xml hierarchy From: "Peter Menzel" <mai00bfy@xxxxxxxxxxxxxxxxxxxxxxx> Date: Fri, 27 Sep 2002 11:14:12 +0200 |
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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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