Re: [xsl] Dynamic numbering of lists in xslt

[xslt.new <xslt.new@xxxxxxxxx>]

ok I think i spoke too fast. I forogot another condition. What if
there is a <travel1> tag without any condition in it, but I still want
to count it in sequence with the ones which satisfy the condition?

That is:
<travel1>
<location
<place>NY<place>
<time>EST</time>
</location>
  <travel2>
<location
<place>CaL<place>
<time>PST</time>
</location>
</travel2>
</travel1>

<travel1>
<location
<place>Chi<place>
<time>CST</time>

</location>
</travel1>
<travel1>
<location
<place>NY<place>
<time>EST</time>

</location>
</travel1>
<!-- This is the travel1 without any condition that needs to be counted-->
<travel1>
<text>Some text</text>
</travel1>

I used <xsl:number from="/" count="travel1[location/time=$time and
location/place=$place]"/>

This counts only those travel1 tags which satisfy the conditions and
gives an output like this:

1. NY, EST
2. NY, EST

Now since I have another <travel1> tag without any conditions around
it, I want that to be counted as #3 and not as #4. But now, i get

1. NY, EST
2. NY, EST
4.Some text

Instead I want to count this sequentially with the <travel1> tags
which have the place and time conditions satisfied.

So I would want:
1. NY, EST
2. NY, EST
3.Some text

I tried:

<xsl:template match="travel1">

<xsl:choose>

<xsl:when test=".//location">
<xsl:if test=".//location/place=$place and .//location/time=$time">
<fo:list-block space-before="6pt" space-before.conditionality="retain">
      <fo:list-item>
              <fo:list-item-label end-indent="label-end()">
                      <fo:block>
                              <xsl:number format="1"/>
                      </fo:block>
              </fo:list-item-label>
              <fo:list-item-body start-indent="body-start()" end-indent="0pt">
                      <fo:block>
                              <xsl:apply-templates/>
                      </fo:block>
              </fo:list-item-body>
      </fo:list-item>
</fo:list-block>
</xsl:if>
</xsl:when>
<!-- This is for the <travel tags without any cnditions-->
without any condition in it, but I still want to count it in sequence
with the ones which satisfy the condition?

That is:
<travel1>
<location
<place>NY<place>
<time>EST</time>
</location>
  <travel2>
<location
<place>CaL<place>
<time>PST</time>
</location>
</travel2>
</travel1>

<travel1>
<location
<place>Chi<place>
<time>CST</time>

</location>
</travel1>
<travel1>
<location
<place>NY<place>
<time>EST</time>

</location>
</travel1>
<!-- This is the travel1 without any condition that needs to be counted-->
<travel1>
<text>Some text</text>
</travel1>

I used <xsl:number from="/" count="travel1[location/time=$time and
location/place=$place]"/>

This counts only those travel1 tags which satisfy the conditions and
gives an output like this:

1. NY, EST
2. NY, EST

Now since I have another <travel1> tag without any conditions around
it, I want that to be counted as #3 and not as #4. But now, i get

1. NY, EST
2. NY, EST
4.Some text

Instead I want to count this sequentially with the <travel1> tags
which have the place and time conditions satisfied.

So I would want:
1. NY, EST
2. NY, EST
3.Some text

I tried:

<xsl:template match="travel1">

<xsl:choose>

<xsl:when test=".//location">
<xsl:if test=".//location/place=$place and .//location/time=$time">
<fo:list-block space-before="6pt" space-before.conditionality="retain">
      <fo:list-item>
              <fo:list-item-label end-indent="label-end()">
                      <fo:block>
                <xsl:number from="/"
count="travel1[location/time=$time and location/place=$place]"
format="1"/>
                      </fo:block>
              </fo:list-item-label>
              <fo:list-item-body start-indent="body-start()" end-indent="0pt">
                      <fo:block>
                              <xsl:apply-templates/>
                      </fo:block>
              </fo:list-item-body>
      </fo:list-item>
</fo:list-block>
</xsl:if>
</xsl:when>
<xsl:otherwise>
<fo:list-block space-before="6pt" space-before.conditionality="retain">
      <fo:list-item>
              <fo:list-item-label end-indent="label-end()">
                      <fo:block>
                <xsl:number from="/" count="travel1" format="1"/>
                      </fo:block>
              </fo:list-item-label>
              <fo:list-item-body start-indent="body-start()" end-indent="0pt">
                      <fo:block>
                              <xsl:apply-templates/>
                      </fo:block>
              </fo:list-item-body>
      </fo:list-item>
</fo:list-block>

</xsl:otherwise>
</xsl:choose>
</xsl:template>




On 1/10/07, David Carlisle <davidc@xxxxxxxxx> wrote:
> > I need to use a numberd list but this time using XSLFO and not XSLT.
>
> what do you mean? XSLFO allows the specification of typesetting
> properties but has no capability to do any programming of any sort.
> The rest of tyour posting consists of some example XSLT stylesheets
> (the XSLT happens to use XSLFO literal result elements, but that is
> irrelevant to both the XSLt engine and your problem)
>
> as far as i can tell you want to count travel nodes which have a
> location child that matches your params, so use something like
>
> <xsl:number from="/" count="travel1[location/time=$time and location/place=$place]"/>
>
>
> Do you really want to make a list with one item for each travel element,
> why not make a single list outside the code that you have shown and then
> just generate   <fo:list-item> in the travel1 template.
>
>
> > Please let me know if any of this is achievable in XSLFO using
> > renderx.
>
> Your question has nothing to do with xslfo, it's about how to generate
> the right number using xslt.
>
> David