Re: [xsl] Dynamic numbering of lists in xslt

Subject: Re: [xsl] Dynamic numbering of lists in xslt
From: cknell@xxxxxxxxxx
Date: Mon, 08 Jan 2007 14:50:13 -0500
That response reminded me of one of my favorite lawyer jokes.

Without seeing your XSLT, we can only give you a very general answer. Given what I see in your source code, and assuming that you have a template that matches the "section" element, I would use this:

<xsl:template match="section">
  <xslt:value-of select="position()" />.<xslt:value-of select="." />
</xslt:template>
-- 
Charles Knell
cknell@xxxxxxxxxx - email



-----Original Message-----
From:     xslt.new <xslt.new@xxxxxxxxx>
Sent:     Mon, 8 Jan 2007 13:20:41 -0600
To:       xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject:  Re: [xsl] Dynamic numbering of lists in xslt

Hi:

Thank you for your reply. Can you give me an idea of how this can be achieved?

On 1/8/07, Alexey Nickolaenkov <nikolaenkov@xxxxxxxxxxxx> wrote:
> yes.
>
> Monday, January 8, 2007, 10:04:47 PM, you wrote:
>
> xn> Hi:
>
> xn> I am a newbie to xslt and XSLFO. I have to do a numbering which offers the
> xn> flexibility to reorder elements if an element in between is being removed.
>
> xn> I will try to demonstrate with an example input.
>
> xn> <chapter1>
> xn>    <section>first section<section>
> xn> </chapter1>
> xn> <chapter1>
> xn>   <section>second section</section>
> xn> </chapter1>
> xn> <chapter1>
> xn>   <section>third section</section>
> xn> </chapter1>
> xn> <chapter1>
> xn>   <section>fourth section</section>
> xn> </chapter1>
>
> xn> So, I want my output to look like
>
> xn> 1. first section
> xn> 2. second section
> xn> 3. third section
> xn> 4. fourth section
>
> xn> and so on. Suppose one chapter1 element is removed in between, lets say
> xn> second section, then the numbering should reset itself so that now the
> xn> output looks like:
>
> xn> 1. first section
> xn> 2. third section
> xn> 3. fourth section
>
> xn> Is it possible to achieve this in xslt and XSLFO?
>
> xn> Please do let me know.
>
> xn> Thanks
>
>
> --
> Alexey                            mailto:alexey.nikolaenkov@xxxxxxxxxxxx

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