[sqlalchemy] Re: unexpected behaviour of in_
You seem to have stumbled into the same bug as I have while i fixed the IN SQL generation code for oracle to work inside the column list, not just in the where-clause. I ended up hacking SQLAlchemy/trunk/lib/sqlalchemy/sql/expression.py like this: 2166 https://projects.freecode.no/internal/projects/browser/div4/div4c/software/SQLAlchemy/trunk/lib/sqlalchemy/sql/expression.py#L2166 *class* _BinaryExpression(ColumnElement): ... 2181 https://projects.freecode.no/internal/projects/browser/div4/div4c/software/SQLAlchemy/trunk/lib/sqlalchemy/sql/expression.py#L2181 *def* *_get_from_objects*(self, **modifiers): 2182 https://projects.freecode.no/internal/projects/browser/div4/div4c/software/SQLAlchemy/trunk/lib/sqlalchemy/sql/expression.py#L2182 res = self.left._get_from_objects(**modifiers) 2183 https://projects.freecode.no/internal/projects/browser/div4/div4c/software/SQLAlchemy/trunk/lib/sqlalchemy/sql/expression.py#L2183 *if* self.operator *is* operators.in_op: *return* res 2184 https://projects.freecode.no/internal/projects/browser/div4/div4c/software/SQLAlchemy/trunk/lib/sqlalchemy/sql/expression.py#L2184 *return* res + self.right._get_from_objects(**modifiers) That seems to do the trick :) Best regards, Egil casbon wrote: Hi All, I am seeing something I didn't expect using in_. Here is a simple example, exactly as I expect: In [13]: col = Trade.c.TradeId.in_([1,2]) In [14]: sel = select([col]) In [15]: print col Trade.TradeId IN (?, ?) In [16]: print sel SELECT Trade.TradeId IN (?, ?) AS anon_1 FROM Trade But now, if I use a subselect, I see a problem: In [17]: col = Trade.c.TradeId.in_(select([Trade.c.TradeId])) In [18]: sel = select([col]) In [19]: print col Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) In [20]: print sel SELECT Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) AS anon_1 FROM Trade, (SELECT Trade.TradeId AS TradeId FROM Trade) The column definition (col) is as expected, but the select definition (sel) is strange. It selects two things and generates n^2 rows. How can I get the select I expect: SELECT Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) AS anon_1 FROM Trade thanks, James --~--~-~--~~~---~--~~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~--- signature.asc Description: OpenPGP digital signature
[sqlalchemy] Re: unexpected behaviour of in_
this is ticket #1074. A slightly klunky workaround for now is: col = t1.c.c1.in_([select([t1.c.c1]).as_scalar()]) On Jun 12, 2008, at 10:04 AM, casbon wrote: Hi All, I am seeing something I didn't expect using in_. Here is a simple example, exactly as I expect: In [13]: col = Trade.c.TradeId.in_([1,2]) In [14]: sel = select([col]) In [15]: print col Trade.TradeId IN (?, ?) In [16]: print sel SELECT Trade.TradeId IN (?, ?) AS anon_1 FROM Trade But now, if I use a subselect, I see a problem: In [17]: col = Trade.c.TradeId.in_(select([Trade.c.TradeId])) In [18]: sel = select([col]) In [19]: print col Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) In [20]: print sel SELECT Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) AS anon_1 FROM Trade, (SELECT Trade.TradeId AS TradeId FROM Trade) The column definition (col) is as expected, but the select definition (sel) is strange. It selects two things and generates n^2 rows. How can I get the select I expect: SELECT Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) AS anon_1 FROM Trade --~--~-~--~~~---~--~~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
[sqlalchemy] Re: unexpected behaviour of in_
Ah, thanks. Should have searched the bug reports as well as the list. On Jun 12, 3:27 pm, Michael Bayer [EMAIL PROTECTED] wrote: this is ticket #1074. A slightly klunky workaround for now is: col = t1.c.c1.in_([select([t1.c.c1]).as_scalar()]) On Jun 12, 2008, at 10:04 AM, casbon wrote: Hi All, I am seeing something I didn't expect using in_. Here is a simple example, exactly as I expect: In [13]: col = Trade.c.TradeId.in_([1,2]) In [14]: sel = select([col]) In [15]: print col Trade.TradeId IN (?, ?) In [16]: print sel SELECT Trade.TradeId IN (?, ?) AS anon_1 FROM Trade But now, if I use a subselect, I see a problem: In [17]: col = Trade.c.TradeId.in_(select([Trade.c.TradeId])) In [18]: sel = select([col]) In [19]: print col Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) In [20]: print sel SELECT Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) AS anon_1 FROM Trade, (SELECT Trade.TradeId AS TradeId FROM Trade) The column definition (col) is as expected, but the select definition (sel) is strange. It selects two things and generates n^2 rows. How can I get the select I expect: SELECT Trade.TradeId IN (SELECT Trade.TradeId FROM Trade) AS anon_1 FROM Trade --~--~-~--~~~---~--~~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
[sqlalchemy] Re: unexpected behaviour of in_
On Jun 12, 2008, at 10:51 AM, casbon wrote: Ah, thanks. Should have searched the bug reports as well as the list. no no, I just created that ticket :) --~--~-~--~~~---~--~~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
