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Subject: RE: XML to XML
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>
Date: Wed, 26 Mar 2003 17:54:07 -0500
Jim,

Ken's solution is also a good one (assuming he wrote what he intended to write in his untested code ;-). Note that he's optimized for performance on the assumption the nodes are correctly ordered in the input. If they're not, his solution will be slow on large datasets compared to mine, but it has the advantage of not caring how many levels deep you have to go. (Except for the output element naming bit: tricky, that. If you could name them 'Level2Category' instead of 'LevelTwoCategory', etc., this problem would go away).

Cheers,
Wendell

At 04:49 PM 3/26/2003, he wrote:
Untested but try the following approach

<xsl:template match="Categories">
<Categories>
<xsl:apply-templates select="Category[length(string(Code))=1]" />
</Categories>
</xsl:template>


<xsl:template match="Category">
<xsl:variable name="code" select="string(Code)"/>
<xsl:variable name="code-length" select="length(string(Code))"/>
<xsl:variable name="name">
<xsl:choose>
<xsl:when test="length(string(Code))=1">LevelOneCategory</xsl:when>
...etc
</xsl:choose>
</xsl:variable>
<xsl:element name="{$name}" Code="{$code}" Description="{Description}">
<xsl:apply-templates select="following-sibling::Category[substring(string(Code),1,$code-length)=string($code)]" />
</xsl:element>
</xsl:template>


following-sibling:: could be replaced with ../ if the order is not guaranteed.


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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Suite 207                                          Phone: 301/315-9631
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Current Thread
  • RE: XML to XML, (continued)
    • Michael Kay - Thu, 27 Mar 2003 04:01:35 -0500 (EST)
    • Jeni Tennison - Thu, 27 Mar 2003 04:35:16 -0500 (EST)
    • Ross Ken - Wed, 26 Mar 2003 16:46:16 -0500 (EST)
      • Wendell Piez - Wed, 26 Mar 2003 17:51:34 -0500 (EST) <=
    • Jim Han - Wed, 26 Mar 2003 17:57:48 -0500 (EST)
    • Jim Han - Wed, 26 Mar 2003 18:01:37 -0500 (EST)
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