On 07/13/2016 02:22 PM, Seth P wrote:
That works!
Obviously I need to know that the joining field is called a_id, and I
can live with that, since in practice it's uniform. But I'm just curious
if there's an automated way to figure out which entity/column is related
to A. (There could be more
That works!
Obviously I need to know that the joining field is called a_id, and I can
live with that, since in practice it's uniform. But I'm just curious if
there's an automated way to figure out which entity/column is related to A.
(There could be more than one entity in q, though just one
On 07/13/2016 01:04 PM, Seth P wrote:
Thank you, as always, for the quick and detailed response.
With the join to the subquery that's on func.max(A.id), once you use
that function, the column loses it's "A.id-ness", because SQLA doesn't
know anything about func.max() and for all
Thank you, as always, for the quick and detailed response.
With the join to the subquery that's on func.max(A.id), once you use
> that function, the column loses it's "A.id-ness", because SQLA doesn't
> know anything about func.max() and for all it knows it could be turning
> it into anything.
On 07/13/2016 02:29 AM, Seth P wrote:
[Apologies for posting an incomplete version of this post earlier.
Please ignore it.]
If B has a ForeignKey (and relationship) to A (e.g. B.a_id -> A.id),
then I can write query(B.b_num).join(A) without specifying the
condition, and SQLAlchemy will figure
On Wednesday, July 13, 2016 at 4:25:24 PM UTC+1, Jonathan Vanasco wrote:
>
>
>
> On Wednesday, July 13, 2016 at 11:01:08 AM UTC-4, Jonathan Underwood wrote:
>
> Thanks - you're right. That's weird though, as it contradicts the
>> documentation. Anyway, I've found it all works as expected,
Actually, taking a closer look, the sql generated for query 5 doesn't look
correct (or at least not what I want), since it isn't joining max_a_id with
anything.
On Wednesday, July 13, 2016 at 2:29:34 AM UTC-4, Seth P wrote:
>
> [Apologies for posting an incomplete version of this post earlier.
On Wednesday, July 13, 2016 at 4:01:08 PM UTC+1, Jonathan Underwood wrote:
>
>
>
> On Wednesday, July 13, 2016 at 2:35:06 PM UTC+1, Антонио Антуан wrote:
>>
>> `cteq_alias.union_all(...`
>>
>> Also, you do not need to create cteq_alias, you can use cteq, like in
>> your query, but you have to
On Wednesday, July 13, 2016 at 11:01:08 AM UTC-4, Jonathan Underwood wrote:
Thanks - you're right. That's weird though, as it contradicts the
> documentation. Anyway, I've found it all works as expected, simply by not
> bothering to create the aliases:
>
I think I've run into this problem
On Wednesday, July 13, 2016 at 2:35:06 PM UTC+1, Антонио Антуан wrote:
>
> `cteq_alias.union_all(...`
>
> Also, you do not need to create cteq_alias, you can use cteq, like in
> your query, but you have to replace cteq_alias.c.id with cteq.c.id
>
>
Thanks - you're right. That's weird though, as
`cteq_alias.union_all(...`
Also, you do not need to create cteq_alias, you can use cteq, like in your
query, but you have to replace cteq_alias.c.id with cteq.c.id
ср, 13 июл. 2016 г. в 15:20, Jonathan Underwood <
jonathan.underw...@gmail.com>:
> Hi,
>
> I am struggling to get a simple
Hi,
I am struggling to get a simple recursive CTE query to work with sqlalchemy
1.0.14, sqlite backend (3.8.10.2) and pysqlite 2.8.2, python 2.7.12. Below
is a reproducer:
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, String
from sqlalchemy import orm
Amazing! :)
Big thanks.
ср, 13 июл. 2016 г. в 12:59, Simon King :
> OK, I have to admit that I haven't fully understood the details of
> what you are doing. However, it sounds like you are implementing
> horizontal sharding, which SQLAlchemy has a bit of support for:
>
>
OK, I have to admit that I haven't fully understood the details of
what you are doing. However, it sounds like you are implementing
horizontal sharding, which SQLAlchemy has a bit of support for:
http://docs.sqlalchemy.org/en/latest/orm/extensions/horizontal_shard.html
examples:
from sqlalchemy import func
project_list = ['platform', 'vip']
content = 'this is test text'
r = session.query(Project).filter(Project.name.in_(project_list))
r.update({'change_log': func.concat(Project.change_log, content)},
synchronize_session='fetch')
session.commit()
[Apologies for posting an incomplete version of this post earlier. Please
ignore it.]
If B has a ForeignKey (and relationship) to A (e.g. B.a_id -> A.id), then I
can write query(B.b_num).join(A) without specifying the condition, and
SQLAlchemy will figure out the join automatically. [See query
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