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>
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