RE: [xsl] can restructure xml??

[Jarno.Elovirta@xxxxxxxxx]

Hi,

Yes, but hard coding the structure like that is evil. I don't know what's the best practice on these kinds of grouping problems nowadays (i.e. how to use xsl:key etc.), but I would recommend something like this:

<xsl:key name="title" match="list/*[not(self::title)]" use="generate-id(preceding-sibling::title[1])" />

<xsl:template match="list">
  <xsl:copy>
    <xsl:apply-templates select="@*"/>
    <xsl:apply-templates select="title" mode="group"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="title" mode="group">
  <item>
    <xsl:attribute name="num">
      <!-- you could use position() here, too -->
      <xsl:number count="title"/>
    </xsl:attribute>
    <xsl:apply-templates select=".|key('title', generate-id())"/>
  </item>  
</xsl:template>

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

Cheers,

Jarno - Feindflug: Vollstreckung

> -----Original Message-----
> From: ext Laura [mailto:xsl_list@xxxxxxxxxxx]
> Sent: 15 January, 2003 14:28
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] can restructure xml??
> 
> 
> You could do simple things like
> <?xml version="1.0"?>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
>     <xsl:output omit-xml-declaration="yes" indent="yes" method="xml"/>
>     <xsl:template match="/">
>         <root>
>             <list>
>                    <xsl:for-each select="/root/list/title">
>                     <item num = "{position()}">
>                         <title>
>                             <xsl:value-of select="."/>
>                         </title>
>                         <writer>
>                             <xsl:value-of
> select="following-sibling::writer"/>
>                         </writer>
>                         <type>
>                             <xsl:value-of 
> select="following-sibling::type"/>
>                         </type>
>                         <remark>
>                             <xsl:value-of
> select="following-sibling::remark"/>
>                         </remark>
>                     </item>
>                 </xsl:for-each>
>             </list>
>         </root>
>     </xsl:template>
> </xsl:stylesheet>
> this gets first writer,type and remark elements after each 
> title element and
> puts them under the item element. The number attribute takes 
> the value of
> the position() function. so the number of <item> elements you 
> would have
> will be the number of  <title> element you have.
> But as Jarno suggested, you could as well modify the code 
> that generates the
> XML to produce the right kind of structure
> Hope this helps
> Vasu
> -- Original Message -----
> From: "a847356549/mail.h7.dion.ne.jp" <motom@xxxxxxxxxxxxx>
> To: <XSL-List@xxxxxxxxxxxxxxxxxxxxxx>
> Sent: Wednesday, January 15, 2003 11:32 AM
> Subject: [xsl] can restructure xml??
> 
> 
> > hello.
> > I just started putting my data into xml.
> > now I only have a xml like ---
> >
> > <root>
> >   <list>
> >     <title>aaa</title>
> >     <writer>bbb</writer>
> >     <type>ccc</type>
> >     <remark>ddd</remark>
> >     <title>eee</title>
> >     <writer>fff</writer>
> >     <type>ggg</type>
> >     <remark>hhh</remark>
> >     ...
> >   </cd>
> > </list>
> >
> > my ideal structure is like ---
> >
> > <root>
> >   <list>
> >    <item num="1">
> >     <title>aaa</title>
> >     <writer>bbb</writer>
> >     <type>ccc</type>
> >     <remark>ddd</remark>
> >    </item>
> >    <item num="2">
> >     <title>eee</title>
> >     <writer>fff</writer>
> >     <type>ggg</type>
> >     <remark>hhh</remark>
> >    </item>
> >     ...
> >   </list>
> > </root>
> >
> > is there a way to do this throgh xsl?  any advice and hint will
> >  be grateful.
> >
> > ttkaya
> >
> >  XSL-List info and archive:  
http://www.mulberrytech.com/xsl/xsl-list
>

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list