[sqlalchemy] Re: Reading in single columns as numpy arrays
2009/6/5 Thomas thomas.robitai...@gmail.com Hi, I'm trying to read in single columns from an SQL database as 1D numpy arrays with the correct types. So a FLOAT column would be returned as a numpy.float32 array, etc. Is there an easy way to do this? Maybe you have to define your own Float datatype using NumPy and modify all table definistions to use this datatype. -- XUE Can --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sqlalchemy] Re: ForeignKey not saved - sqlite3, TG1.1
Looks like the culprit is: user.blogs = [Blog(title=input['blog_title'] this replaces your user.blogs relation with a single entry list containing the last Blog try this to add a Blog to the relation list user.blogs.append(Blog(title=input['blog_title')) -- Mike Conley On Fri, Jun 5, 2009 at 11:46 PM, Adam Yee adamj...@gmail.com wrote: I've got a simple mapping between User and Blog: # Parent user_table = Table('user', metadata, Column('id', Integer, primary_key=True), Column('username', String(50)), Column('password', Unicode(50)), ) # Child blog_table = Table('blog', metadata, Column('id', Integer, primary_key=True), Column('title', String(100)), Column('date_created', DateTime()), Column('user_id', Integer, ForeignKey('user.id')) ) ... mapper(User, user_table, properties = { 'blogs': relation(Blog, backref='user') }) mapper(Blog, blog_table) My foreignkeys aren't being saved/stored properly. No matter how many 'children' I create, the foreignkey only stays saved with the most recently added: sqlite select * from blog; 1|blog1|2009-06-05 19:41:33.555387| 2|blog2|2009-06-05 19:41:42.551838| 3|blog3|2009-06-05 19:42:28.280046| 4|blog4|2009-06-05 19:44:19.180090| 5|blog5|2009-06-05 19:46:38.580777|1 sqlite Here's how they are being saved. My controller saveblog(): @expose() def saveblog(self, **input): if request.method == 'POST': if not input['blog_title']: msg = 'You must give a title name.' redirect('/createblog', message=msg) try: session_user = cherrypy.session['username'] user = session.query(User).filter_by (username=session_user).one() user.blogs = [Blog(title=input['blog_title'], date_created=datetime.datetime.now())] msg = 'Successfully created blog %s.' % input ['blog_title'] redirect('/', message=msg) except KeyError: redirect('/', message='Session lost or timed out') except NoResultFound: redirect('/', message='Error: database corrupt.') Why is it doing this? Please help, thanks. --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sqlalchemy] Re: union with two different orders
On Saturday 06 June 2009 14.18:33 naktinis wrote: I want to use union on two queries, which have different order: q1 = Thing.query().order_by(Thing.a) q2 = Thing.query().order_by(Thing.b) q = q1.union(q2).all() SQL doesn't work as you think it does here. A UNION does not concatenate the results of the two queries, but is allowed to return the result in any order. ORDER BY can *then* be applied to the end result of your union. So even if you use subqueries, the order by in the subqueries might just be ignored. This is to allow the SQL query planner to be clever while building the union (perhaps a large union over two queries over the same table: if both queries require a table scan over the large table, the planner might decide to build the union by scanning the table only once while running both queries in parallel, so the table is loaded from disk once insead of twice. The UNION would then contain the resulting rows in more or less random order.) But I digress. What you want to do is something like: SELECT 1 as COL1, ... FROM ... UNION SELECT 2 as COL1, ... FROM ... ORDER BY COL1, ... cheers -- vbi But after this query I get MySQL error message Incorrect usage of UNION and ORDER BY. I guess that this could be because having SELECT ... UNION SELECT ... ORDER BY B, it is not clear whether the second subquery or both queries should be ordered using B criteria. I think this can be solved by adding brackets to each of the subquery: (SELECT ...) UNION (SELECT ...). Is there any way to create this query using SQLAlchemy ORM? I am using SQLAlchemy 0.5.4. -- Vertrauen ist gut. Anwalt ist saugeil. --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sqlalchemy] Re: Session is already flushing
usually it means mutiple threads are hitting a single Session and/or connection. keep in mind all objects attached to a session are an extension of that session's state and similarly cannot be shared among threads unless detached. On Jun 6, 2009, at 1:03 AM, Michael Mileusnich wrote: I am using a scoped session in my sql alchemy app. Can anybody help me with what this error means: Session is already flushing _mysql_exceptions.ProgrammingError: (2014, Commands out of sync; you can't run this command now5 --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sqlalchemy] Re: union with two different orders
On Jun 6, 2009, at 11:39 AM, naktinis wrote: I think this was not the case, since I didn't expect the merged result to be ordered. To be more precise, the query looks like: q1 = Thing.query().filter(...).order_by(Thing.a.desc()).limit(1) q2 = Thing.query().filter(...).order_by(Thing.a.asc()).limit(1) q = q1.union(q2).order_by(Thing.id).all() The q1 returns first filtered element with largest 'a' column, q2 - first with smallest 'a'. So, I guess my question is still valid. if youre using limit with order by, you would have to wrap those queries within subqueries in order for UNION to accept them as encapsulated relations. On 6 Bir, 17:49, Adrian von Bidder avbid...@fortytwo.ch wrote: On Saturday 06 June 2009 14.18:33 naktinis wrote: I want to use union on two queries, which have different order: q1 = Thing.query().order_by(Thing.a) q2 = Thing.query().order_by(Thing.b) q = q1.union(q2).all() SQL doesn't work as you think it does here. A UNION does not concatenate the results of the two queries, but is allowed to return the result in any order. ORDER BY can *then* be applied to the end result of your union. So even if you use subqueries, the order by in the subqueries might just be ignored. This is to allow the SQL query planner to be clever while building the union (perhaps a large union over two queries over the same table: if both queries require a table scan over the large table, the planner might decide to build the union by scanning the table only once while running both queries in parallel, so the table is loaded from disk once insead of twice. The UNION would then contain the resulting rows in more or less random order.) But I digress. What you want to do is something like: SELECT 1 as COL1, ... FROM ... UNION SELECT 2 as COL1, ... FROM ... ORDER BY COL1, ... cheers -- vbi But after this query I get MySQL error message Incorrect usage of UNION and ORDER BY. I guess that this could be because having SELECT ... UNION SELECT ... ORDER BY B, it is not clear whether the second subquery or both queries should be ordered using B criteria. I think this can be solved by adding brackets to each of the subquery: (SELECT ...) UNION (SELECT ...). Is there any way to create this query using SQLAlchemy ORM? I am using SQLAlchemy 0.5.4. -- Vertrauen ist gut. Anwalt ist saugeil. --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sqlalchemy] Re: ForeignKey not saved - sqlite3, TG1.1
Thanks Mike, it's working now. But I'm curious, how was the foreignkey erased from the previous entry upon each newly added blog? What does .append do that keeps the foreignkeys saved? On Jun 6, 5:24 am, Mike Conley mconl...@gmail.com wrote: Looks like the culprit is: user.blogs = [Blog(title=input['blog_title'] this replaces your user.blogs relation with a single entry list containing the last Blog try this to add a Blog to the relation list user.blogs.append(Blog(title=input['blog_title')) -- Mike Conley On Fri, Jun 5, 2009 at 11:46 PM, Adam Yee adamj...@gmail.com wrote: I've got a simple mapping between User and Blog: # Parent user_table = Table('user', metadata, Column('id', Integer, primary_key=True), Column('username', String(50)), Column('password', Unicode(50)), ) # Child blog_table = Table('blog', metadata, Column('id', Integer, primary_key=True), Column('title', String(100)), Column('date_created', DateTime()), Column('user_id', Integer, ForeignKey('user.id')) ) ... mapper(User, user_table, properties = { 'blogs': relation(Blog, backref='user') }) mapper(Blog, blog_table) My foreignkeys aren't being saved/stored properly. No matter how many 'children' I create, the foreignkey only stays saved with the most recently added: sqlite select * from blog; 1|blog1|2009-06-05 19:41:33.555387| 2|blog2|2009-06-05 19:41:42.551838| 3|blog3|2009-06-05 19:42:28.280046| 4|blog4|2009-06-05 19:44:19.180090| 5|blog5|2009-06-05 19:46:38.580777|1 sqlite Here's how they are being saved. My controller saveblog(): @expose() def saveblog(self, **input): if request.method == 'POST': if not input['blog_title']: msg = 'You must give a title name.' redirect('/createblog', message=msg) try: session_user = cherrypy.session['username'] user = session.query(User).filter_by (username=session_user).one() user.blogs = [Blog(title=input['blog_title'], date_created=datetime.datetime.now())] msg = 'Successfully created blog %s.' % input ['blog_title'] redirect('/', message=msg) except KeyError: redirect('/', message='Session lost or timed out') except NoResultFound: redirect('/', message='Error: database corrupt.') Why is it doing this? Please help, thanks. --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---