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Re: creating list from structured paragraphs
Subject: Re: creating list from structured paragraphs
From: Antonio Fiol Bonnín <fiol@xxxxxxxxxx>
Date: Mon, 20 May 2002 07:55:39 +0200
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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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