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Re: How to sort attribute?

2005-08-13 08:57:00
I thought this code should have sorted attribute nodes by names(seems
logical). I tested with Saxon 8.4, and it sorted the attributes by
name.

<xsl:for-each select="@*">
   <xsl:sort select="name()"/>

I remember David Carlisle telling some time back on XSL-List, that XML
Spec is notoriously famous for such anomalies.

Regards,
Mukul

http://gandhimukul.tripod.com

On 8/13/05, Ross, Douglas <DRoss(_at_)kronos(_dot_)com> wrote:
John,

Since Michael Kay points out that attribute order is implementation
dependent and not quaranteed, you might consider transforming your
attributes into elements, where element order is quaranteed to be
maintained in document order. You could change Mukul's stylesheet
slightly:

<xsl:template match="x">
  <x>
    <xsl:for-each select="@*">
     <xsl:sort select="name()"/>
       <!-- change this to element tag:
     <xsl:attribute name="{name()}"><xsl:value-of
select="."/></xsl:attribute> -->
      <xsl:element name="{name()}"><xsl:value-of
select="string(.)"/></xsl:element>
    </xsl:for-each>
  </x>
</xsl:template>


Douglas Ross
Senior Software Engineer, Advanced Development
Kronos
www.kronos.com
Improving the Performance of People and Business(tm) by making software
Smaller, Faster, Sharper, Easier


-----Original Message-----
From: Mukul Gandhi [mailto:gandhi(_dot_)mukul(_at_)gmail(_dot_)com]
Sent: Saturday, August 13, 2005 5:33 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] How to sort attribute?

Hi John,
 This could be the use-case for the problem you are trying to solve.

XML file -
<root>
 <x a="1" d="2" c="4" />
</root>

XSLT file -
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">

<xsl:output method="xml" indent="yes" />

<xsl:template match="/root">
  <xsl:apply-templates select="x" />
</xsl:template>

<xsl:template match="x">
  <x>
    <xsl:for-each select="@*">
     <xsl:sort select="name()"/>
     <xsl:attribute name="{name()}"><xsl:value-of select="."
/></xsl:attribute>
    </xsl:for-each>
  </x>
</xsl:template>

</xsl:stylesheet>

Regards,
Mukul

On 8/13/05, John Li <johnli121(_at_)163(_dot_)com> wrote:
Hi,

When exporting one node and its attribute, I want to sort its
attributes
lexicographic. I try it as below but always fail. Anyone could help?

<xsl:for-each select="@*">
   <xsl:sort select="name()"/>
   <xsl:copy/>
</xsl:for-each>

Thanks,
John



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