Subject: Re: Find distinct nodes from a sequence
From: "Imsieke, Gerrit, le-tex" <gerrit.imsieke@xxxxxxxxx>
Date: Thu, 07 Jul 2011 10:11:33 +0200
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a) upgrade to a more recent version of Saxon
b) Something like
<xsl:for-each-group select="person" group-by="@name">
<xsl:sequence select="." />
</xsl:for-each-group>
should do the trick. You could also group-by="generate-id(.)" if the
nodes are really identical.
<xsl:sequence select="." /> selects the context item of each group which
is, per spec, the first item in the group.
Gerrit
On 2011-07-07 10:05, Pushkar Khadilkar wrote:
Hi List,
I have a function getNodes() that returns a sequence of nodes(>= 1) from source tree by xsl:sequence statement.
I want to remove duplicate nodes from this sequence.
The duplicate nodes in the sequence are not different nodes from source tree with same values for some attributes but actually refer to same node in source tree.
For example , Consider following tree
<person name=a/>
<person name=b/>
The function getNodes() might return a sequence containing 2 nodes both of which refer to same node (<person name=a/>) of the source tree.
I cannot use some logic like select node whose generate-id(.) is not equal to generate-id(preceding-node) as the duplicate nodes actually refer to same node of source tree and hence preceding wont work.
Im using Saxon HE 8.8 Java.
Is there a function for selecting unique nodes from a sequence ?
There was a saxon:distinct function in older versions of saxon but is not present in Saxon 8.
Any help in this regard is highly appreciated.
Thanks ,
Pushkar Khadilkar
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