Re: [xsl] Formatting issues

[ms <mina_hurray@xxxxxxxxx>]

Hi:

Thank you for your help so far. I forgot to mention
that I use xslt 1.0, I am not sure if we can use for
statements in 1.0

Here are my examples:

Input XML:

<root>
	<level1>
		<text>
			<para>This has values high and steep</para>
		</text>
		<locn>
			<cond1>high</cond1>
			<cond2>steep</cond2>
		</locn>
		<level2>
			<text>
				<para>This has values high and deep</para>
			</text>
			<locn>
				<cond1>high</cond1>
				<cond2>deep</cond2>
			</locn>
		</level2>
	</level1>
	<level1>
		<text>
			<para>This has values high and low</para>
		</text>
		<locn>
			<cond1>high</cond1>
			<cond2>low</cond2>
		</locn>
	</level1>
	<level1>
		<text>
			<para>This has values high and steep</para>
		</text>
		<locn>
			<cond1>			high</cond1>
			<cond2>steep</cond2>
		</locn>
		<level2>
			<text>
				<para>This has values high and steep</para>
			</text>
			<locn>
				<cond1>high</cond1>
				<cond2>steep</cond2>
			</locn>
		</level2>
	</level1>
</root>


XSLT for level1:

<xsl:template match="level1">
	<xsl:choose>
		<xsl:when test=".//locn">
			<xsl:if test=".//locn/cond1=$cond1 and
.//locn/cond2=$cond2">
				<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" from="level1"/>
							</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:choose>
</xsl:template>

This template repeats for level2...6

$cond1 and $cond2 are being passed thru the xslt for
testing purposes and have been set to high and steep
respectively for testing.

So given $cond1=high and $cond2=steep

The output when this XSLT is applied to the XML ,
should be:

1. This has values high and steep
2. This has values high and steep.
    A. This has values high and steep.

But what I am getting is this:

1. This has values high and steep
1. This has values high and steep.
    A. This has values high and steep.

Please let me know how this ca be resolved.

--- Abel Braaksma <abel.online@xxxxxxxxx> wrote:

> ms wrote:
> > Level1 - 1,2,3 etc
> > Level2 - A, B, C
> > Level3 - i,ii,..
> > Level4 - (1), (2)
> > Level5 - (a), (b)
> > Level6 - (i), (ii)
> >   
> 
> Aha. Well, just add more templates to my base
> template.
> 
> > The hierarchy for levels is that level 1 is the
> parent
> > for level2, level2 is he parent for level3 and so
> on.
> > i.e.
> >
> > [...]
> >
> > Now in an XML , there can be any number of these
> > levels, and I have a logic to filter these levels
> > based on certain conditions.
> >   
> 
> I see. So that part is working. Send your XSLT for
> where it is not 
> working. Make the smallest possible example that
> does not work, and add 
> the input, too. Also, add an example of what the
> output is that you expect.
> 
> > I tried <xsl:number from="level1" format="1"/>
> > But that only numbers all levels with 1 and does
> not
> > increment them sequentially. 
> >   
> 
> Indeed. In your <xsl:template> where you match for
> "level1", add a 
> variable with one <xsl:number /> instruction in it.
> That's your counter 
> (see my example).
> 
> > This is my problem. How do I write the XSLT to
> > basically check if previous levels have satisfied
> the
> > filtering condition
> 
> This is what my XSLT is about. You can use it for
> that. Though I assume 
> you will have to use the your own XPath expressions
> in the matches.
> 
> >  and if they have, include them for
> > counting, else remove them and count only those
> levels
> > that satisfy the condition and count sequentially.
> >   
> 
> Which is what it does, too ;)
> Your conditions will likely be different, though.
> 
> Please send what you've got (minimized to your
> problem domain).
> 
> Cheers,
> -- Abel
> 
> 


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