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Re: [xsl] Selectively "delete" elements/attributes from the output

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yes
If you have this template:

<xsl:template match="@*|node()" priority="-1">
   <xsl:copy>
      <xsl:apply-templates select"@*|node()"/>
   </xsl:copy>
<xsl:template>

it should pick up all templates that are not catered for by another template and copy them over.

Cheers.

Kamal.

avo@xxxxxxxxxxxxx wrote:

I'm doing XML to XML transformation (newbie when it comes to XML and XSLT)
Say I have an XML document like this

<element1 att1="a" att2="b">
<element11>text1</element11>
</element1>
<element2>text2</element2>
</root>

I want to delete att1 and att2 from element1 so that output will look like this:
<root>
<element1>
<element11>text1</element11>
</element1>
<element2>text2</element2>
</root>

I could write a template that call <xsl:copy-of select=3D"..." to basically copy everything except the two attributes there. But this is just small examples, with more complex XML files, where I want to delete things here and there, this becomes unscalable (the problem applies to both attributes elements, and text, not just attributes)
Basically what I want to do is to have a template for each element/attribute that I want to "not copy" (delete), for ex:

<xsl:template match=3D"root/element1/@att1"/>
(doing nothing, essentially not copying the attribute over)

Since the built-in templates for everything else do not copy the element nodes over (only text nodes are output), I have to provide templates for the rest to copy them over (not sure how to do this effectively, either),
which is not very efficient.

So my question comes down to:
Can I provide rules for each of the thing I want to "delete", and then have one template to copy over everything else
~Anh

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