How to ignore nested tags in an element while applying XSLT

kaps77 · August 4th, 2011, 05:04 AM

I am converting following html to xml

Code:

<html>
	<body>
		something in body
		<table>
			<thead>
				<tr>
					<th>Heading 1</th>
				</tr>
				<tr>
					<th>Heading 2</th>
				</tr>
			</thead>
			<tbody>
				<tr>
					<td>content starts<b>bold <i>start italic in bold ends here </i>bold continue</b>plain text here</td>
				</tr>
				<tr>
					<td>content 2<p>para here</p>continue</td>
				</tr>
			</tbody>
		</table>
	</body>
</htm

> I would like it to convert to following xml

Code:

<topic>
 <table>
     <row>
<cell>cel<b>bold <i>start italic in bold ends</i>here</b>1 is here</cell></row>
     <row>
<cell>content 2<p>para here</p>continue</cell>
</row>
</topic>

Basically I want to copy everything that comes in TD to the cell element i XML. However when I apply XSLT, the <b> and <i> tags are supressed. How can I copy all the contents of TD to cell element of XML?

thanks

Kapil

samjudson · August 4th, 2011, 05:10 AM

You are probably using xsl:value-of. Use xsl:copy-of instead.

This will copy the node, rather than creating a text node with the string value of the select statement (which is what xsl:value-of does).

kaps77 · August 4th, 2011, 05:14 AM

Thanks Samjudson,
Copy-of is working to perfection..

Thanks
Kapil

kaps77 · August 4th, 2011, 05:37 AM

There is one issue with copyof. It is also emitting the tag name for TD.
Whereas I just need the string value in the TD.

samjudson · August 4th, 2011, 05:41 AM

Assuming your current context is the TD element and you are doing <xsl:copy-of select="."/> then change it to <xsl:copy-of select="node()"/> instead.