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XSLT: <xsl:strip-space> does not work

Tags:

xml

xslt

I have a servlet filter in my application that intercepts all the incoming requests and tries to strip the whitespaces from the incoming XML and write the resulting 'clean' XML to the response. I am using XSLT to achieve this. Please see the XSLT below:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

<xsl:output method="xml" omit-xml-declaration="no" indent="yes"/>
<xsl:strip-space elements="*"/>

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

However, this is not working as expected. The resulting XML is the same as the original XML (despite of using the <xsl:strip-space elements="*"/> in the stylesheet.

Please help me getting this right.

Regards,
- Ashish

like image 283
Vini Avatar asked Jul 15 '09 21:07

Vini


1 Answers

It is not clear what you intend to get as an output, and what you expect from xsl:strip-whitespace in the first place. But one thing to note is that it doesn't strip all whitespace, but only that which is deemed insignificant under the "usual" rules. In particular, from XSLT 1.0 spec:

A text node is never stripped unless it contains only whitespace characters.

So, for example, this:

<foo>
  <bar>   </bar>
</foo>

will be stripped down to:

<foo><bar/></foo>

because it had 3 whitespace-only text nodes (after <foo> and before <bar>, between <bar> and </bar>, and after </bar> and before </foo>).

Note also that because you have <xsl:output indent="yes"> in your stylesheet, it will end up being transformed to:

<foo>
   <bar/>
<foo>

in the output.

On the other hand, this:

<foo>
   text1
   <bar>  text2  </bar>
   text3
</foo>

Will not be stripped at all, because all text nodes it contains are not purely whitespace nodes.

like image 83
Pavel Minaev Avatar answered Sep 25 '22 10:09

Pavel Minaev