> -----Original Message-----
> From: sqlalchemy@googlegroups.com [mailto:sqlalchemy@googlegroups.com]
> On Behalf Of Jules Stevenson
> Sent: 09 June 2011 08:53
> To: sqlalchemy@googlegroups.com
> Subject: [sqlalchemy] Re: Trying to query a relationship of a
> relationship
>
> Sorry, for the spamming, code typo (was trying to simplify it),
> should read:
>
> invoices = query(ArkInvoice).\
> join(ArkInvoice.project).\
> join(ArkProject.client).\
>
> options(sa.orm.contains_eager(ArkInvoice.project.client)).\
> filter(ArkInvoice.project.client.id == id)
>
I think you probably want something like this (all untested):
invoices = (session.query(ArkInvoice)
.join(ArkInvoice.project)
.join(ArkProject.client)
.filter(ArkClient.id == id)).all()
If you need contains_eager (which is purely an optimisation allowing you
to access invoice.project without a subsequent query), I think it would
look like this:
invoices = (session.query(ArkInvoice)
.join(ArkInvoice.project)
.join(ArkProject.client)
.options(contains_eager(ArkInvoice.project),
contains_eager(ArkProject.client))
.filter(ArkClient.id == id)
.all())
However, if you are actually going to be working with the client,
project and invoice objects after this query, you may find it easier to
start from the client:
client = (session.query(ArkClient)
.options(joinedload_all('projects.invoices'))
.filter(ArkClient.id == id)
.one())
After this query, you could access client.projects and
client.projects[n].invoices without further database queries.
See http://www.sqlalchemy.org/docs/orm/loading.html for a description of
joinedload_all.
I hope that helps,
Simon
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