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position() of the parent
Subject: position() of the parent From: "Alexandru-Ionut Albu" <albu@xxxxxxxxxxxx> Date: Wed, 19 Jan 2000 16:59:47 -0500 |
Hello, I am new to XML and I am trying to write an XSL sheet that "flattens" the source XML tree, but preserves information about child-parent relationships. For example: <mynode name="xxx"> <mynode name="yyy"> <mynode name="zzz"> </mynode> </mynode> </mynode> should become <mynode name="xxx" id="1" parentid=""/> <mynode name="yyy" id="2" parentid="1"/> <mynode name="zzz" id="3" parentid="2"/> I don't really care what the data type of the id's is. I've tried something like: <xsl:template match="mynode"> <mynode name="{@name}"> <xsl:attribute name="id"> <xsl:value-of select="position()"/> </xsl:attribute> </mynode> </xsl:template> but I don't know how to apply the position() function to the parent node. Thank you, Alex LYCOShop. Thousands of products! One location! http://shop.lycos.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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