Subject: Re: generate-id() problem
From: "Nima Kaviani" <nkaviani@xxxxxx>
Date: Sat, 25 Nov 2006 13:59:58 -0800
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Hi Florent,
thanks for the msg. However your used syntax ( <rule:foo
rdf:ID="rule_{ generate-id(.) }">) is not working for me. My xalan
XSLT Processor complains about the use of curly braces in the string.
So I have to use it this way: <rule:foo rdf:ID="concat('rule_',
generate-id(.))">
ok I guess at this point it doesn't really matter. I just mentioned it
to check whether this is a problem at the level of the processor or
something else.
thanks for the idea you gave to me. It helps to generate unique ids.
-Nima
On 11/24/06, Florent Georges <darkman_spam@xxxxxxxx> wrote:
Nima Kaviani wrote:
Hi
> <rule:foo rdf:ID="rule_N2320">
> <rule:bar rdf:resource="constraint_N3456"/>
> </rule:foo>
> <constraint:test rdf:ID="constraint_N3456">
> <t1/>
> <t2/>
> <constraint:tes>
> ok so as it is obvious in the code above, I need to
> generate two unique values, here "N2320" and "N3456",
> using a funciton and attach them to my terms, "rule" and
> "constraint".
So you can use generate-id() iff you can make a link
between these generated elements and nodes in your input
tree. For example, say that those elements are generated in
the template for the element "myelem", that have a mandatory
attribute "myattr". Then you can write:
<xsl:template match="myelem">
<!-- Will always generate the same string for the same
myelem node in the same transformation. -->
<rule:foo rdf:ID="rule_{ generate-id(.) }">
...
</rule:foo>
<!-- Will always generate the same string for the same
@myattr node in the same transformation. -->
<constraint:test
rdf:ID="constraint_{ generate-id(@myattr) }">
..
<constraint:tes>
</xsl:template>
Regards,
--drkm
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