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Re: [xsl] Continue seqlist numbering


Subject: Re: [xsl] Continue seqlist numbering
From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>
Date: Mon, 05 Jan 2004 11:43:22 -0500

At 2004-01-05 10:51 -0500, Norma Yeazell wrote:
I have two seqlist lists in a row in my xml like:

<seqlist>
<item>this is item 1</item>
<item>this is item 2</item>
<item>this is item 3</item>
</seqlist>
<seqlist>
<item>this is item 1</item>
<item>this is item 2</item>
<item>this is item 3</item>
</seqlist>

Again, you are trying to infer a hierarchy where none exists.


I can get it to number them but I need the second list to be numbered
starting with the last item from the first list.
I've tried <xsl:number count="item" level="multiple" format="1.1.1"/>

That relies on hierarchy in your input and one isn't there.


The outcome should look like.

1. this is item 1
2. this is item 2
3. this is item 3

    3.1. this is item 1
    3.2. this is item 2
    3.3. this is item 3

When doing the second <seqlist>, count the members in the first. To generalize it, you will probably need to do something along the lines of the following from within an item:


  <xsl:for-each select="../preceding-sibling::seqlist">
    <xsl:value-of select="count(item)"/>.<xsl:text/>
  </xsl:for-each>
  <xsl:number/>. <xsl:text/>

The above should give you the following in the third <seqlist> list:

3.3.1 this is item 1

I hope this helps.

................... Ken

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