Re: [xsl] creating list from structured paragraphs

[Antonio Fiol Bonnín <fiol@xxxxxxxxxx>]

This should be quite straightforward, though I have not tested it:

<xsl:template match="para-1">
<list-item>
<xsl:value-of select=".">
</list-item>
</xsl:template>

<xsl:template match="para">
<para>
<xsl:value-of select=".">
<list>
<xsl:apply-templates />
</list>
</para>
</xsl:template>

The former should work for the first case. For the second case, it may 
be a little bit more complicated. I'd suggest using a xsl:for-each, 
using position() and/or following-sibling and/or something else. As you 
see, I am not an expert ;-)

Antonio Fiol


David Santamauro wrote:

> I'm hoping this is simple. Here is the XML:
>
> <doc>
> <para>The following text is a list
>  <para-1>List item No. 1</para-1>
>  <para-1>List item No. 2</para-1>
>  <para-1>List item No. 3</para-1>
> </para>
> </doc>
>
> I need:
>
> <doc>
> <para>The following text is a list
>  <list>
>   <list-item>List item No. 1</list-item>
>   <list-item>List item No. 2</list-item>
>   <list-item>List item No. 3</list-item>
>  </list>
> </para>
> </doc>
>
> I'd cut and paste my XSL but it doesn't work so I felt it not worth the
> bandwidth.
>
> It should also be noted that I do have influence over the XML to an extent.
> I can remove the parent-child relationship so that the XML looks like this:
>
> <doc>
> <para>The following text is a list</para>
> <para-1>List item No. 1</para-1>
> <para-1>List item No. 2</para-1>
> <para-1>List item No. 3</para-1>
> </doc>
>
> but it didn't bring me closer to a solution. Any help would be much
> appreciated.
>
> thanks
>
> David
>
>
>
>
> XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>
> .
>
>

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