Omit unneeded namespaces from the output

Question

My XSLT is outputiung some tags with xmlns:x="http://something" attribute... How to avoid this redundant attribute? The output XML never use, neither in a the x:tag, nor in an x:attribute.

EXAMPLE OF XML:

<root><p>Hello</p><p>world</p></root>

EXAMPLE OF XSL:

<xsl:transform version="1.0" 
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
   xmlns:xlink="http://www.w3.org/1999/xlink">
<xsl:output encoding="UTF-8" method="xml" version="1.0" indent="no"/>

<xsl:template match="root"><foo>
   <xsl:for-each select="p">
    <p><xsl:value-of select="." /></p>
   </xsl:for-each></foo>
   <xsl:for-each select="x">
    <link xlink:href="{x}" />
   </xsl:for-each></foo>
</xsl:template>

EXAMPLE OF XML OUTPUT:

<foo>
   <p xmlns:xlink="http://www.w3.org/1999/xlink">Hello</p>
   <p xmlns:xlink="http://www.w3.org/1999/xlink">world</p>
</foo>

The xmlns:xlink is a overhead, it is not used!

A typical case where XSLT must use namespace but the output not:

 <xsl:value-of select="php:function('regFunction', . )" />

Please, edit the question and provide the source XML document and the transformation code. Without these one has to guess. — Dimitre Novatchev, Mar 23 '13 at 02:52

score 9 · Accepted Answer · answered Mar 23 '13 at 06:06

9

As Dimitre has already said, if you are not using the xlink namespace anywhere in your XSLT, you should just remove its namespace declaration. If, however, your XSLT is actually using it somewhere that you haven't shown us, you can prevent it from being output by using the exclude-result-prefixes attribute:

<xsl:transform version="1.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
   xmlns:xlink="http://www.w3.org/1999/xlink"
   exclude-result-prefixes="xlink">

answered Mar 23 '13 at 06:06

JLRishe

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19
131
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Thanks a lot, exclude-result-prefixes is the tool! About the "... just remove its namespace declaration..." solution, I edited the question, for avoid this "trivial case". – Peter Krauss Mar 23 '13 at 16:27

score 0 · Answer 2 · answered Mar 23 '13 at 04:24

0

Just remove this namespace declaration from the xsl:stylesheet instruction -- it isn't used (and thus necessary) at all:

xmlns:xlink="http://www.w3.org/1999/xlink"

The whole transformation now becomes:

<xsl:transform version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output encoding="UTF-8" method="xml" version="1.0" indent="no"/>

 <xsl:template match="root"><foo>
   <xsl:for-each select="p">
   <p class="a"><xsl:value-of select="." /></p>
   </xsl:for-each></foo>
 </xsl:template>
</xsl:transform>

and when applied on the provided XML document:

<root><p>Hello</p><p>world</p></root>

produces result that is free of namespaces:

<foo>
    <p class="a">Hello</p>
    <p class="a">world</p>
</foo>

answered Mar 23 '13 at 04:24

Dimitre Novatchev

240,661
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Thanks, I edited with a typical case. The other one is a `xsl:if` that is not used. – Peter Krauss Mar 23 '13 at 10:25
@PeterKrauss, exclude-result-prefixes="{}space-separated list of prefixes" can help in many cases. In other cases still one needs to remove the namespaces by code. BTW, I don't see any `xsl:if` in your posted code. – Dimitre Novatchev Mar 23 '13 at 15:34
1

@PeterKrauss, Yes, in this specific case using the `exclude-result-prefixes` attribute solves the problem. There are more tricky cases -- not on literal result elements, where one needs to actually prevent the copying of namespace nodes -- by code. – Dimitre Novatchev Mar 23 '13 at 15:49

Omit unneeded namespaces from the output

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