Thanks for the answer, but my question was : is there another way only using XPath? I should
have been clearer, sorry.
Frédéric Esnault
-----Message d'origine-----
De : Marcel Reutegger [mailto:marcel.reutegger@gmx.net]
Envoyé : vendredi 22 juin 2007 12:28
À : dev@jackrabbit.apache.org
Objet : Re: jcr:deref() in predicate
Frédéric Esnault wrote:
> Thansk I will because if we don't have, how can we do this query.
>
> I want all contracts having as an internal contractor the company Lycos.
> Knowing that contractors are references in the contract type.
>
> I did /jcr:root/lgw:root/lgw:contracts/element(*, lgw:contractType)[jcr:deref(@lgw:internalContractor,
'lgw:contractorType')/@lgw:companyName = 'Lycos']
>
> Is there antoher way to make this query?
yes, there is. only search for the company node:
lgw:root//element(*, lgw:company)[@lgw:companyName = 'Lycos']
then use the API to get the references to the lycos company node:
Node lycosCompany = ...;
for (PropertyIterator it = lycosCompany.getReferences(); it.hasNext(); ) {
Property p = it.nextProperty();
if (p.getName().equals("lgw:internalContractor")) {
Node contractType = p.getParent();
// do something with this node
}
}
or even better, if there were no same name siblings you could directly address
the lycos company by path:
Node rootNode = ...;
rootNode.getNode("lgw:root/companies/Lycos");
that way you don't even have to execute a query.
in contrast to databases you don't necessarily have to use joins to get to your
content along references. the JCR API provides methods to do that much quicker
than a query.
regards
marcel
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