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Subject: Re: [xsl] What is the simplest method for using xsl:sort without losing attribute names and values? From: "Mark Wilson" <mark@xxxxxxxxxxxx> Date: Sun, 20 Jul 2008 10:58:23 -0700 |
Thanks, Mark
You suggested:
<xsl:template match="*">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates>
<xsl:sort select="Article/Year" />
<xsl:sort select="Article/IssueNumber"/>
<xsl:sort select="Article/Page" />
</xsl:apply-templates>
</xsl:copy>
</xsl:template>My input:
<List>
<Item>
<Cat pofis="4" pofis-number="116"/>
<Subject>Errors</Subject>
<Subject>Color Errors</Subject>
<Subject>Olympic Games. Atlanta</Subject>
<Article>
<Person>Fritz, Zdenek</Person>
<Person is-author="false">Kleskovic, Peter Z.</Person>
<Title>Imperfect color se-tenant</Title>
<Year>2000</Year>
<IssueNumber>1</IssueNumber>
<IssueName>Jan/Feb</IssueName>
<Page>10</Page>
</Article>
</Item>
</List>The output from your template:
<List>
<Item>
<Cat>4116</Cat>
<Subject>Errors</Subject>
<Subject>Color Errors</Subject>
<Subject>Olympic Games. Atlanta</Subject>
<Article>
<Person>Fritz, Zdenek</Person>
<Person>falseKleskovic, Peter Z.</Person>
<Title>Imperfect color se-tenant</Title>
<Year>2000</Year>
<IssueNumber>1</IssueNumber>
<IssueName>Jan/Feb</IssueName>
<Page>10</Page>
</Article>
</Item>
</List>| Current Thread |
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