I wanted to try sql2 because XPATH and SQL are marked as deprecated in jcr 2.0
After trying SQL2, I still will use XPATH for a while for this kind of query... BR, Michael Coldewey -----Ursprüngliche Nachricht----- Von: Marcel Reutegger [mailto:[email protected]] Gesendet: Freitag, 30. Juli 2010 14:25 An: dev Betreff: Re: One Question about SQL2 hi, On Thu, Jul 29, 2010 at 13:22, Michael Coldewey <[email protected]> wrote: > Thank you a lot... > > I will try it, but it seems not so simple as with xpath-query. why don't you use xpath then? regards marcel > BR > Michael Coldewey > > -----Ursprüngliche Nachricht----- > Von: Alexander Klimetschek [mailto:[email protected]] > Gesendet: Donnerstag, 29. Juli 2010 11:05 > An: [email protected] > Betreff: Re: One Question about SQL2 > > I think something like this could work (beware, untested): > > SELECT * FROM [my:a] AS p INNER JOIN child AS c ON ISDESCENDANTNODE(p, > c) WHERE c.[jcr:primaryType] = 'my:b' AND c.Prop1 = 'test' > > Regards, > Alex > > On Wed, Jul 28, 2010 at 21:45, Michael Coldewey <[email protected]> wrote: >> Hello, >> >> >> >> sorry, I know, ist the wrong mailinglist, but i am actually not member of >> the users list. >> >> >> >> Perhaps someone would and could answer me. >> >> >> >> Till now I use XPATH for querying the repository, but now I want to change >> to SQL2. The samples in the test case explains the basics, but I didnt >> found, how to make a query to find nodes, which have childs with specific >> properties. >> >> >> >> [my:a] >> >> [my:b] >> >> Prop1=test >> >> >> >> >> >> I want every node [my:a] which have any child with the property Prop1 > and >> value test. >> >> >> >> In xpath, this is really simple to solve, but i dont know to make it in >> SQL2. >> >> >> >> It would be great, if someone give me a hint on this. >> >> >> >> Thanks a lot >> >> BR, >> >> Michael Coldewey >> >> >> >> > > > > -- > Alexander Klimetschek > [email protected] > > >
