Hi again,
I will try if this example explains it better...
With the example XML:
<X>
<Y>
<Z>lv 1 (one lv. descendants)</Z>
<X>
<Y>
<Z>lv 2 (zero lv. descendants)</</Z>
</Y>
</X>
</Y>
<Y>
<Z>lv 1 (two lv. descendants)</</Z>
<X>
<Y>
<Z>lv 2 (one lv. descendants)</</Z>
<X>
<Y>
<Z>lv 3 (zero lv. descendants)</Z>
</Y>
</X>
</Y>
<Y>
<Z>lv 2 (two lv. descendant)</Z>
<X>
<Y>
<Z>lv 3 (zero lv. descendant)</Z>
</Y>
</X>
<X>
<Y>
<Z>lv 3 (one lv. descendants)</Z>
<X>
<Y>
<Z>lv 4 (zero lv. descendants)</Z>
</Y>
<Y>
<Z>lv 4 (zero lv. descendants)</Z>
</Y>
</X>
</Y>
</X>
</Y>
</X>
</Y>
</X>
With a parameter $lv="2", and an id-transformation, I should get the xml
without (zero lv. descendants) and (one lv. descendants), that is:
<X>
<Y>
<Z>lv 1 (two lv. descendants)</</Z>
<X>
<Y>
<Z>lv 2 (two lv. descendant)</Z>
</X>
</Y>
</X>
With $lv="1" I should get all except (zero lv. descendatnts), that is,
without leaf Y-nodes (and without the X-node containing the leaf Y-nodes):
<X>
<Y>
<Z>lv 1 (one lv. descendants)</Z>
</Y>
<Y>
<Z>lv 1 (two lv. descendants)</</Z>
<X>
<Y>
<Z>lv 2 (one lv. descendants)</</Z>
</Y>
<Y>
<Z>lv 2 (two lv. descendant)</Z>
<X>
<Y>
<Z>lv 3 (one lv. descendants)</Z>
</Y>
</X>
</Y>
</X>
</Y>
</X>
So in all, instead of wanting, say the first two levels of Y-nodes, instead
I want all the nodes top-down, except for ie. buttom 2 levels.
I would be happy if I, from a given node, could count how many levels of
Y-nodes where below the current. I do not know if that was the answer that
David Carlisle gave, I did not understand the answer that well (which is
fair enough given that my question was perhaps not understood :-) )
I hope that with this, my question is clear, and that an XPath solution is
possible.
Regards,
Ragulf Pickaxe :-)
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