On 04/06/2016 06:55 PM, Brian Leach wrote:
Hi everyone!
I have asked this question on StackOverflow, please see it for full
detail of how the models are laid
out:
http://stackoverflow.com/questions/36463623/optimally-access-an-sqlalchemy-model-paramater-n-number-of-relationships-away
I have some "test" models that all have a relationship back to a
"location" model, each a varying number of relationships away. When I
want to get the name of the "location" associated with a "test", there
is a ton of SQL emitted. Is it possible to build in some sort of
shortcut to the location relationship?
yes, however you have to decide what kind of SQL you'd want to emit for
this. The examples at
http://docs.sqlalchemy.org/en/rel_1_0/orm/join_conditions.html#composite-secondary-joins
and
http://docs.sqlalchemy.org/en/rel_1_0/orm/join_conditions.html#relationship-to-non-primary-mapper
illustrate some advanced techniques for building relationships that
directly load across many tables, without loading the intermediary objects.
Otherwise, for garden-variety reduction of SQL, the joinedload()
directive as was pointed out in the SO answer is typically the more
expedient approach to loading a series of object relationships, but in
this case you're also loading all the intermediate collections into memory.
I have tried setting the 'lazy' parameter on the relationship to
'select', 'joined', and 'subquery' although these do not seem to make a
significant difference when it comes to tying a test model to its location.
The way I have things defined also make it difficult for me to filter a
group of tests based on their location, since the relationship is so
different with each. I do not even know how to join these tables using
raw SQL or how to begin to approach trying to emulate what the ORM is
constructing for me.
I would imagine that this has been nailed before and am hoping that one
of you out there can fill me in on the missing concept.
- Brian
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