## [xsl] Finding sequences of same element

 Subject: [xsl] Finding sequences of same element From: Simon Kissane Date: Wed, 9 Feb 2005 18:58:35 +1100
```Hi

Suppose I have an input document:
<A><B X="1"/><B X="2"/><B X="3"/><C X="4"/><B X="5"/><B X="6"/><B X="7"/></A>

Now, suppose I wish to group together consecutive B elements, giving a
result document like this:
<A><D><B X="1"/><B X="2"/><B X="3"/></D><C X="4"/><D><B X="5"/><B
X="6"/><B X="7"/></D></A>

How can I do this? (I would prefer not to use recursive templates, but
rather for-each, if that is at all possible...)

I think I can find the initial element of these sequences, like so:
B[not(preceding-sibling::*) or preceding-sibling::*[not(self::B)]]
This, I think, should select all B for which there are either no
preceeding sibling elements, or for which the immediately preceeding
sibling element is not a B element. Thus, in the above example, it
would pick B[@X=1] and B[@X=5].

But, given each initial sequence element, how can I find the remaining
nodes in the sequence?  With the initial sequence element as the
context node, I could do:
.|following-sibling::B
But that will also pick up B[@X=5] and B[@X=6] when the context node is B[@X=1].

Is there a predicate test I could use on following-sibling::B to
restrict it only to the current sequence of B elements?

Thanks

Simon Kissane

```