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Resolve XML references

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  • 10-02-2011 5:30pm
    #1
    tetsujin1979
    Registered Users Posts: 495 ✭✭


    hi
    I have a large XML file, filled with elements like this 
    <myItems>
	<items>
		<item href="#id0"></item>
		<item href="#id1"></item>
	</items>
	<multiRef id="id0">
		<description>id0 description</description>
		<title>id0 title</title>
	</multiRef>
	<multiRef id="id1">
		<description>id1 description</description>
		<title>id1 title</title>
	</multiRef>
</myItems>

    The XML itself is valid, but I'm looking for a tool that will read in the source XML and resolve the references to produce an output like this.
    <myItems>
	<items>
		<item>
			<description>id0 description</description>
			<title>id0 title</title>
		</item>
		<item>
			<description>id1 description</description>
			<title>id1 title</title>
		</item>
	</items>
</myItems>

    I could do it manually, but there are literally hundreds of item references, and some child items have references in them.

    I was thinking possible an XSLT stylesheet might do it?


Comments

  • Options
    #2
    BlackWizard
    Registered Users Posts: 3,282 ✭✭✭BlackWizard


    I would just write some C# or other language to just parse and recreate the XML.

    Just read every multiref and get the description and title, then write it to a a file in the format you want.


  • Options
    #3
    tetsujin1979
    Registered Users Posts: 495 ✭✭tetsujin1979


    yeah, that's probably what I'll end up doing

    just wondering if there's a simpler way of doing it


  • Options
    #4
    ron_darrell
    Users Awaiting Email Confirmation Posts: 351 ✭✭ron_darrell 


    <?xml version="1.0" ?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/myItems">
    &lt;myItems&gt;
    	<xsl:for-each select="multiRef">
			<ul style="list-style:none;">
			<li>&lt;item&gt;
				<ul style="list-style:none;">
					<li>&lt;description&gt;
						<ul style="list-style:none;">
							<li><xsl:value-of select="description" /></li>
						</ul>&lt;/description&gt;
					</li>
					<li>&lt;title&gt;
						<ul style="list-style:none;">
							<li><xsl:value-of select="title" /></li>
						</ul>&lt;/title&gt;
					</li>
				</ul>&lt;/item&gt;
			</li>
			</ul>
		</xsl:for-each>
    &lt;/myItems&gt;
</xsl:template>
</xsl:stylesheet>

    Save the above XSLT file and add it as the stylesheet for your XML document. Looking at your requirements, you don't seem to need/use the item tag values at all so I've ignored them. The above will generate a html file with the necessary tags. Just copy and paste.

    -RD


  • Options
    #5
    ron_darrell
    Users Awaiting Email Confirmation Posts: 351 ✭✭ron_darrell 


    <?xml version="1.0" ?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/myItems">
    &lt;myItems&gt;&lt;items&gt;
    	<xsl:for-each select="multiRef">
			<ul style="list-style:none;">
			<li>&lt;item&gt;
				<ul style="list-style:none;">
					<li>&lt;description&gt;
						<ul style="list-style:none;">
							<li><xsl:value-of select="description" /></li>
						</ul>&lt;/description&gt;
					</li>
					<li>&lt;title&gt;
						<ul style="list-style:none;">
							<li><xsl:value-of select="title" /></li>
						</ul>&lt;/title&gt;
					</li>
				</ul>&lt;/item&gt;
			</li>
			</ul>
		</xsl:for-each>
    &lt;/items&gt;&lt;/myItems&gt;
</xsl:template>
</xsl:stylesheet>

    Slight change to code to allow for extra tag


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