How to get rid of not required namespace from result

mmmathur06 · September 5th, 2011, 03:25 AM

Hi,
Can anyone suggest me the work around to get rid of redundant xmlns:xlink from result xml.
I declared xmlns:xlink as root stylesheet attribute but it is posing the problem for me in result xml, as the element which are not process specifically or newly created in stylesheet itself adds default namespace xmlns:xlink.

stylesheet:

Code:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:aid="http://ns.adobe.com/AdobeInDesign/4.0/" xmlns:xlink="http://www.w3.org/1999/xlink">

input xml:

Code:

<volume>0</volume><issue>0</issue><fpage>506</fpage><lpage>513</lpage>

result xml:

Code:

<!--Below tags are not proessed specificly -->

<volume xmlns:xlink="http://www.w3.org/1999/xlink">0</volume>
         <issue xmlns:xlink="http://www.w3.org/1999/xlink">0</issue>
         <fpage xmlns:xlink="http://www.w3.org/1999/xlink">506</fpage>
         <lpage xmlns:xlink="http://www.w3.org/1999/xlink">513</lpage>

One way to solve this issue to process one by one all elements like below but it is pretty uncomfortable.

Code:

<xsl:template match="volume">
    <xsl:element name="volume">
      <xsl:apply-templates/>
    </xsl:element>
  </xsl:template>
  <xsl:template match="issue">
    <xsl:element name="issue">
      <xsl:apply-templates/>
    </xsl:element>
  </xsl:template>
  <xsl:template match="fpage">
    <xsl:element name="fpage">
      <xsl:apply-templates/>
    </xsl:element>
  </xsl:template>
  <xsl:template match="lpage">
    <xsl:element name="lpage">
      <xsl:apply-templates/>
    </xsl:element>
  </xsl:template>

So can any one let me know what I am doing wrong and a decent workaround?

Mohan

mhkay · September 5th, 2011, 03:36 AM

Add exclude-result-prefixes="xlink" to the xsl:stylesheet element.

mmmathur06 · September 5th, 2011, 04:49 AM

as per defination also your answer is absoultly correct but I don't know what is wrong with my result, still showing the same result(with redundant xlink attrib)

Code:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:aid="http://ns.adobe.com/AdobeInDesign/4.0/" xmlns:xlink="http://www.w3.org/1999/xlink" exclude-result-prefixes="xlink">

I am using saxon 9.0 processor with windows xp

Mohan

mhkay · September 5th, 2011, 05:57 AM

You'll have to show more of your code. I can't see what you've done wrong unless you show me what you've done.

mmmathur06 · September 5th, 2011, 08:01 AM

Okay... caught it... since namespace is also given in input xml, stylesheet was not able to remove it.

Code:

<article article-type="research-article" dtd-version="2.2" xml:lang="en" xmlns:xlink="http://www.w3.org/1999/xlink">

now if i remove it from input xml am able to overcome from this issue.

but now another question since I need this namespace in output xml as an attribute of <article>, if I try to generate it something like below but it produce the error [Invalid attribute name: {xmlns:xlink}
Failed to compile stylesheet. 1 error detected. ]

Code:

<xsl:template match="article">
		<xsl:element name="article">
      <xsl:attribute name="article-type">research-article</xsl:attribute>
      <xsl:attribute name="dtd-version">2.2</xsl:attribute>
      <xsl:attribute name="xml:lang">en</xsl:attribute>
 <xsl:attribute name="xmlns:xlink">http://www.w3.org/1999/xlink</xsl:attribute>
<xsl:apply-templates/>
</xsl:element >
</xsl:template>

where I am sliping here?

Thanks for taking this issue
Mohan

Martin Honnen · September 5th, 2011, 08:19 AM

with XSLT a namespace declaration is not considered an attribute. If you want to output a namespace declaration in XSLT 2.0 use xsl: namespace http://www.w3.org/TR/xslt20/#creating-namespace-nodes, not xsl: attribute.
But usually putting the namespace declaration on your xsl: stylesheet element is enough to have it in scope:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:xlink="http://www.w3.org/1999/xlink"
  version="2.0">

...

</xsl:stylesheet>

mhkay · September 5th, 2011, 08:49 AM

You've got me very confused because you started by saying that you didn't want to declare the xlink namespace because it was unused, and now you're asking how to generate an attribute that's in the xlink namespace, so it clearly isn't unused.

Go back to first principles: show us your input document, your desired output, your attempt at an XSLT stylesheet, and explain how the output differs from what you want.

mmmathur06 · September 6th, 2011, 01:09 AM

Sorry for getting you confuse here is my sample input and output fragment:

input xml:

Code:

<?xml version="1.0" encoding="UTF-16" standalone="yes"?>
<article article-type="research-article" dtd-version="2.2" xml:lang="en"<!-- xmlns:xlink="http://www.w3.org/1999/xlink"-->>
<journal-id>JO</journal-id>
<journal-title>The JO</journal-title>
<journal-subtitle>Journal of Ornithology</journal-subtitle>
<abbrev-journal-title>JO</abbrev-journal-title>
<issn>0234-8038</issn>
<issn>2938-4254</issn>
<email>[email protected]</email>
<alt-title>Covino and Holberton</alt-title>
<alt-title>Energy and Migratory Decisions</alt-title>
<volume>56</volume>
<issue>2</issue>
<fpage>506</fpage>
<lpage>513</lpage>
</article>

desierd output:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<article article-type="research-article" dtd-version="2.2" xml:lang="en" xmlns:xlink="http://www.w3.org/1999/xlink">
   <front>
      <journal-meta>
         <journal-id journal-id-type="publisher-id">JO</journal-id>
         <journal-title>The JO</journal-title>
         <issn pub-type="ppub">0234-8038</issn>
         <issn pub-type="epub">2938-4254</issn>
	 <email xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="[email protected]">[email protected]</email>
      </journal-meta>
      <article-meta> 
         <volume>56</volume>
         <issue>2</issue>
         <fpage>506</fpage>
         <lpage>513</lpage></article-meta></front>
	</article>

As could see I need xmlns:xlink in some of elements as attrib but not in all the elements as it was in preceding post. Defining xmlns:xlink at stylesheet level add this namespace to all unprocess tags which is what I asked about and you suggested in one of your post resolve my issue by using exclude-result-prefixes="xlink", but now as this namespace required in some specific element how could I add this in those specific element?

Thanks
Mohan

mhkay · September 6th, 2011, 04:11 AM

exclude-result-prefixes should work here. It will exclude the namespace from elements where it is not needed, and cause it to be declared at the level where it is actually used.

(Your example has a declaration of xmlns:xlink at the top level of your "desired" output - I'm assuming that was a mistake.)

mmmathur06 · September 6th, 2011, 06:02 AM

No, It was not a mistake my whole theread is all about this since namespace is required in only and only <article> and <ext-link> elements. so I have to declare it at top level but not in others so I need to prevent it to add in other elements also.

Using exclude-result-prefix solve one issue but also adds another issue: [how to add namespace in <article> or <ext-link> ?]

Thanks
Mohan