Remove unwanted attribute

thava · March 11th, 2009, 05:25 AM

Dear all,

I am very new to xslt kidly help me to remove unwanted attribute from xml using xsl

here my example i dont want attribute for <job> element.

here my sample input xml
===============
<persons>
<person>
<name id="001">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job type="NIT">Singer</job>
<gender>Male</gender>
</person>
<person>
<name id="002">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job type="IT">SoftwareDeveloper</job>
<gender>Male</gender>
</person>
</persons>

output xml should be
==============

<persons>
<person>
<name id="001">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job>Singer</job>
<gender>Male</gender>
</person>
<person>
<name id="002">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job>SoftwareDeveloper</job>
<gender>Male</gender>
</person>
</persons>

joefawcett · March 11th, 2009, 05:35 AM

Search online for xslt identity template. Then add a template that matches the attribute and does nothing:

Code:

<xsl:template match="job/@type" />

mrame · March 11th, 2009, 05:40 AM

As Joe has told use the below stylesheet.

Code:

 
<xsl:stylesheet version="2.0" xmlns:xsl=http://www.w3.org/1999/XSL/Transform>
<xsl:output method="xml" indent="yes"/>

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

<xsl:template match="job/@type"/>
<xsl:stylesheet>

thava · March 11th, 2009, 09:19 AM

Dear Joe and rummy

Thanks for your solution.

Its working fine, what i was expected. Now when i using my xsl by the tag of <xsl:copy-of.. for the node person that would inherit the root element namespace but i don't want to inherit the root element namespace and when i implement to ignore the attribute that does not work.

Here is my sample input xml
===========================
<persons article-type="research-article" dtd-version="1.1" xml:lang="EN" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:mml="http://www.w3.org/1998/Math/MathML">
<person>
<name id="001">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job type="NIT">Singer</job>
<gender>Male</gender>
</person>
<person>
<name id="002">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job type="IT">SoftwareDeveloper</job>
<gender>Male</gender>
</person>
</persons>

output should be
===============
<persons article-type="research-article" dtd-version="1.1" xml:lang="EN" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:mml="http://www.w3.org/1998/Math/MathML">
<person>
<name id="001">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job>Singer</job>
<gender>Male</gender>
</person>
<person>
<name id="002">
<firstname>Paul</firstname>
<lastname>McCartney</lastname>
</name>
<job>SoftwareDeveloper</job>
<gender>Male</gender>
</person>
</persons>

Here my xsl
=======================================
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="persons">
<xsl:element name="persons">
<xsl:attribute name="article-type">research-article</xsl:attribute>
<xsl:attribute name="dtd-version">1.1</xsl:attribute>
<xsl:copy-of select="descendant::person"/>
</xsl:element>
</xsl:template>
<xsl:template match="persons/person/job/@type"/>
</xsl:stylesheet>

Since i am new to the xsl, i don't know where i am getting fail. Please do the needful.

Thanks,
Thava

joefawcett · March 11th, 2009, 10:36 AM

As you didn't use the suggested code it's hard to know what to suggest.
As far as I can see the root element isn't namespaced.

Martin Honnen · March 11th, 2009, 10:41 AM

Joe's suggestion is to use the following stylesheet:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">
  
  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>
  
  <xsl:template match="job/@type"/>

</xsl:stylesheet>

That produces exactly the output result you posted.

If that is not what you want to achieve then you need to show us the output you want. You mention some 'root element namespace' but in your sample input the root element is a 'persons' element in no namespace.

Please also mention whether you want to use XSLT 1.0 or 2.0.

thava · March 12th, 2009, 01:19 AM

Dear all,

As told that the code which you given is working fine in normal browser, but when i import into Adobe IndesingCS3 it shows error message "DOM transformation error:Invalid namespace". Pls. advise.

Thanks,

mrame · March 12th, 2009, 02:38 AM

When you dont use the prefixes, why do you use
xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:mml=http://www.w3.org/1998/Math/MathML. If possible try without those namespaces.

mhkay · March 12th, 2009, 05:45 AM

>When you dont use the prefixes, why do you use
xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:mml=http://www.w3.org/1998/Math/MathML. If possible try without those namespaces.

Rummy, I know you are trying to be helpful. But this is like suggesting to someone who's car won't start that it would help to remove the unnecessary newspapers sitting on the back seat.

Yes, simplifying the problem to its bare essentials can sometimes be useful, by eliminating all distractions and helping you to focus on the essence of the problem. But don't imagine that these redundant namespaces are anything to do with the problem.

mrame · March 12th, 2009, 05:51 AM

Thanks Mike. I thought that could help him in anyway. Is there any other solution?