Re: Conversion from xsl to xsl:fo

You wrote:

> There are (at least) two ways to get from 
> "(1): a stylesheet producing X" and "(2): a styleheet converting X to Y"
> to "(3): a stylesheet producing Y"
> 
> plan a)
> write a stylesheet that takes (1) as input and modifies all the
> templates so that where they use constructions from the language X
> the result is a template using constructions from language Y.

IIRC what stopped me doing this was this particular case: Suppose the
input DTD of (1) has elements <foo> and <bar> that are like xhtml's ul
and li (a <foo> contains <bar>s and a <bar> can contain <foo>s)

(1) has

<xsl:template match="foo"><ul><xsl:apply-templates/></ul></xsl:template>
<xsl:template match="bar"><li><xsl:apply-templates/></li></xsl:template>

(2) converts XHTML to FO

<template match="ul">
  <fo:list-block><apply-templates/></fo:list-block>
</template>

<template match="li">
  <fo:list-item>
    <fo:list-item-label>
      <choose>
        <when test="count(ancestor::ul)=1">-</when>
        <when test="count(ancestor::ul)=2">*</when>
      </choose>
    </fo:list-item-label>
    <fo:list-item-body><apply-templates/></fo:list-item-body>
  </fo:list-item>
</template>

How do I rewrite template 2 of (1) if I don't know that <bar>s can
descend from <foo>s. When I rewrite <li> I guess I can go look for all
the templates that produce <ul> in (1), find out that it is produced
by <foo> and write a <choose> appropriately, but I don't think you can
infer it on more complex cases (where <li> are produced in more than
one template).

> plan b)
> is to use a xx:node-set extension and just write stylesheet (3) like so:

Ah yes. I was young at the time and thought I could do without
extensions.  This is equivalent to running (1) and (2) sequentially,
which is what I did.

Max.


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