Tree xml logic

[ghenaweb]

I need to get this results with this xml tree.

Ciccio-Salvatore
lollo-sasa

How can I do using xsl with a for-each statement ??



<root>
<rossi>
<padre>
<nome>Ciccio</nome>
</padre>
</rossi>
<rossi>
<padre>
<nome>lollo</nome>
</padre>
</rossi>
<verdi>
<padre>
<nome>Salvatore</nome>
</padre>
</verdi>
<verdi>
<padre>
<nome>Sasa</nome>
</padre>
</verdi>

</root>

 

[Joseph Kesselman]

Describe the *logic* of what you want to do, not just the results.
Otherwise people are likely to give you solutions that don't work for
your next data file.

 

[ghenaweb]

Hi Joseph...


in primis I wriong the group.

in secundis...

If you have not the right solutions...
please dont write anything.

Thanks a lot.


Joseph Kesselman ha scritto:

 

[Joseph Kesselman]

If you have not the right solutions...
> please dont write anything.

There are multiple solutions to the question you asked. Which one do you
want?

http://www.catb.org/~esr/faqs/smart-questions.html

 

[Ghena]

<xsl:for-each
select="/CommandList/CheckRouting/RouterList/Router[Complete='true']//GroupList/Group/OutwardList/Outward"
<xsl:variable name="root"
select="/CommandList/CheckRouting/RouterList/Router[Complete='true']/GroupList/Group/ReturnList/Return"
/>

<xsl:variable name="totale">
<xsl:choose>
<xsl:when test="Price/Amount">
<xsl:value-of select="number(Price/Amount) +
number($root/Price/Amount)"/>
</xsl:when>
<xsltherwise>
<xsl:value-of select="//Group/Price/Amount/text()"/>
</xsltherwise>
</xsl:choose>
</xsl:variable>

</xsl:for-each>


the problem is :

number($root/Price/Amount)
display always the same (first) value

I dont know how get the relative value of OutwardList/Outward with the
ReturnList/Return ones.

Perhaps using ancestor or some axes... ?
you could find the document here...

www.lastminutesud.it/test/901968736.xml
Thanks a lot.

 

[Joseph Kesselman]

number($root/Price/Amount)
display always the same (first) value

Because $root is always the same set of nodes, and number() returns the
numeric value if the first node in that set.

I dont know how get the relative value of OutwardList/Outward with the
ReturnList/Return ones.

Yes, you need relative paths. The simplest solution is

<xsl:value-of select="number(Price/Amount) +
number(../../OutwardList/Outward/Price/Amount)"/>

 

[Ghena]

Perhaps is not clear....

Do you mean ?

<xsl:value-of select="number(Price/Amount) +
number(../../ReturnList/Return/Price/Amount)"/>

with your method I get two similar data...

number(Price/Amount) is equal to
.../../OutwardList/Outward/Price/Amount


Joseph Kesselman ha scritto:

 

[Joe Kesselman]

number(Price/Amount) is equal to
../../OutwardList/Outward/Price/Amount

In that case, yes, you're in the context of the Outward element, and you
need to navigate up and back down to the Return to find the other value.

 

[Andy Dingley]

please dont write anything.

Granted. Plonked and killfiled too.