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RE: XML to XML

2003-03-26 15:54:07
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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