it appears here the "anon_2" is a label being given to your otherwise unnamed 
FirstThing.moved_by.any() call, which is a subquery.   

you're not showing me the full query being rendered but I would imagine the 
important bits are:

SELECT anon_1.anon_2 AS anon_1_anon_2 FROM
(SELECT EXISTS (...) AS anon_2) AS anon_1

which is valid.  The query would fail to execute if it weren't.

NoSuchColumnError here would likely be alleviated if you just said 
FirstThing.moved_by.any().label("some_label").

I'll look into seeing why an anonymous any() subquery doesn't get targeted by 
Query correctly here.



On Feb 28, 2012, at 11:02 AM, naktinis wrote:

> I should have pointed out that I got a NoSuchColumnError because of 
> "anon_1.anon_2". There is no column "anon_2" in any of the tables. It's just 
> an alias name of a derived table.
> 
> Is "table_name_1.table_name_2" supposed to mean anything?
> 
> On Tuesday, February 28, 2012 5:53:42 PM UTC+2, Michael Bayer wrote:
> 
> On Feb 28, 2012, at 9:40 AM, naktinis wrote:
> 
>> Column "anon_1.anon_2" is generated in the following scenario:
>> 
>> dbsession.query(FirstThing, FirstThing.moved_by.any(User.id == 
>> user_id)).options(joinedload_all('some_property'))
>> query = query.join(SecondThing, SecondThing.first_thing_id == FirstThing.id)
>> query = query.order_by(OneThing.ordering_field).limit(count)
>> 
>> Also, it is important that both FirstThing and SecondThing polymorphically 
>> inherit from Thing.
>> 
>> Effectively, query.all() generates a query like
>> 
>> SELECT ... anon_1.anon_2 AS anon_1_anon_2 ...
>> FROM
>> (SELECT first_thing.id AS first_thing.id, EXISTS (SELECT 1 FROM 
>> first_thing_moves, users ...) AS anon_2
>>  FROM thing JOIN first_thing ON ... JOIN (SELECT ... FROM thing JOIN 
>> second_thing) AS anon_3 ON ... ORDER BY ... LIMIT ...) AS anon_1 ORDER BY ...
>> 
>> Why would "anon_1.anon_2" column be generated there - it is, I think, not 
>> even a valid syntax?
> 
> it's valid, "anon_1" is the label applied to a subquery, you can see where it 
> has "(SELECT .... ) AS anon_1".  "anon_1" becomes what we sometimes call a 
> "derived table" in the query and is then valid like any other alias name.
> 
> The join is because when we have a joined inheritance class B inherits from 
> A, then we join to it from C, we are effectively joining:
> 
> SELECT * FROM C JOIN (A JOIN B ON A.id=B.id) ON C.x=A.y
> 
> That is valid SQL, however, it doesn't work on SQLite, and also doesn't work 
> on MySQL versions before 5.   It also may or may not have issues on some 
> other backends.    So SQLAlchemy turns "A JOIN B" into a subquery:
> 
> SELECT * FROM C JOIN (SELECT * FROM A JOIN B ON A.id=B.id) AS anon_1 ON 
> C.x=anon_1.y
> 
> as it turns out, this approach generalizes much more nicely than just putting 
> "A JOIN B" in there.  Suppose classes B1 and B2 inherit from A in a concrete 
> fashion, using tables "B1" and "B2" to represent the full row.   Then you 
> wanted to join from C to A.    SQLAlchemy would have you doing a "polymorphic 
> union" which means you select from the UNION of B1 and B2:
> 
> SELECT * FROM C JOIN (SELECT * FROM B1 UNION SELECT * FROM B2) AS anon_1 ON 
> C.x=anon_1.y
> 
> where "anon_1.y" here would be "y" from B1 unioned to "y" from B2.   
> 
> Anyway, SQLAlchemy is very quick to wrap up a series of rows in a subquery, 
> applying an alias to it, since that syntax works the most consistently across 
> not only all backends but across a really wide range of scenarios.
> 
> 
> 
> 
> 
> -- 
> You received this message because you are subscribed to the Google Groups 
> "sqlalchemy" group.
> To view this discussion on the web visit 
> https://groups.google.com/d/msg/sqlalchemy/-/g5juNMWd4moJ.
> To post to this group, send email to sqlalchemy@googlegroups.com.
> To unsubscribe from this group, send email to 
> sqlalchemy+unsubscr...@googlegroups.com.
> For more options, visit this group at 
> http://groups.google.com/group/sqlalchemy?hl=en.

-- 
You received this message because you are subscribed to the Google Groups 
"sqlalchemy" group.
To post to this group, send email to sqlalchemy@googlegroups.com.
To unsubscribe from this group, send email to 
sqlalchemy+unsubscr...@googlegroups.com.
For more options, visit this group at 
http://groups.google.com/group/sqlalchemy?hl=en.

Reply via email to