[sqlalchemy] Re: result as the dict()
MyBase = type(MyBase, (Base, MyMixin), {})
Is this any different than just doing
class MyBase(Base, MyMixin): pass
?
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[sqlalchemy] Re: result as the dict()
On Jul 6, 2008, at 1:16 AM, Empty wrote:
You might want to look into the new instrument_declarative function in
0.5. It allows you to wrap a class with the declarative functionality
without using the metaclass. This should permit you to work with a
subclass, although I haven't tried it.
it also accepts cls as a keyword argument which is used as the base
class:
Base = declarative_base(..., cls=MyMixin)
you can also add a mixin to any class in Python like:
Base = declarative_base(...)
MyBase = type(MyBase, (Base, MyMixin), {})
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[sqlalchemy] Re: result as the dict()
it also accepts cls as a keyword argument which is used as the base
class:
Base = declarative_base(..., cls=MyMixin)
you can also add a mixin to any class in Python like:
Base = declarative_base(...)
MyBase = type(MyBase, (Base, MyMixin), {})
Thanks! Both are good to know and better than mixing each time I
create a class. I didn't know about the second use of type()... that
could be very useful.
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[sqlalchemy] Re: result as the dict()
Spoke a bit to soon... looks like that cls argument is only available
in 0.5, and the type() mixin trick still complains about not having a
mapped_table specifed. SQLA 0.4.6 is unhappy with the snippet
below...
import sqlalchemy as sa
from sqlalchemy.ext.declarative import declarative_base
class _BaseORMMixin():
pass
_OrmBaseBase = declarative_base()
_ORM_BASE = type(_ORM_BASE, (_OrmBaseBase, _BaseORMMixin), {})
Even trying to add a bogus __tablename__ doesn't seem to work.
Dropping back to mixing the mixin with every class derived from the
declarative base restores functionality.
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[sqlalchemy] Re: result as the dict()
I was just looking for the solution to a similar problem...
I think the way you're thinking about it actually does work...at
least, it does for me! See below:
y = select([table]).execute().fetchall()
for x in y:
... print x.__dict__
...
{'_RowProxy__parent': sqlalchemy.engine.base.ResultProxy object at
0x18e5c90, '_RowProxy__row': ('data', 'more_data',
datetime.datetime(2008, 7, 2, 14, 20, 22, 999023), 1)}
mylist = [dict(r) for r in y]
mylist
[{u'description': 'data', u'other': 'more_data', u'number' : 1,
u'changed' : datetime.datetime(2008, 7, 2, 14, 20, 22, 999023)}]
On Jul 5, 2:39 pm, zipito [EMAIL PROTECTED] wrote:
Good day community,
there was similar theme - which ended without answer.
How can I convert mine query results to list of dicts. I.e. I want the
following
select = session.select(MineTableObject)
res = select.fetchall()
res_list_dict = [dict(r) for r in res]
but that doesn't works.
dict(r) method returns the array of column_number-column_value not
the dictionary column_name-column_value.
Is there a possibility to have this ??
best regards, Ilya Dyoshin
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[sqlalchemy] Re: result as the dict()
I also had this issue and searched around for an answer, but found
nothing. In my case I;m using the orm and wanted to have those object
properties that map to a table be wrapped up in a dict. The new SQLA
book is enroute and I figured I'd look in there when I get it, but in
the mean time I just added a cheap/crude method for conversion as per
below.
import sqlalchemy as sa
from sqlalchemy.ext.declarative import declarative_base
_ORM_BASE = declarative_base()
class _BaseORMMixin():
def ToDict(self, ExclusionList = []):
Converts fields in object to a dict, unless field names are
in ExclusionList.
ret = {}
for c in self.__mapper__.columns.keys():
if not c in ExclusionList:
ret[c] = getattr(self, c)
return ret
class ORM_User(_BaseORMMixin, _ORM_BASE):
__tablename__ = user
snip
I figure there must be a better way.. I had to dip into __mapper__,
which didn't seem right, but __dict__ obviously has more than what I
wanted. In addition, I had to use a mixin because with
ext.declaritive (which I like because it is less verbose) I couldn't
make a base class to derive from because it complained that I needed
__tablename__ in the base class.
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[sqlalchemy] Re: result as the dict()
Hi Russell, On Sun, Jul 6, 2008 at 12:32 AM, Russell Warren [EMAIL PROTECTED] wrote: I also had this issue and searched around for an answer, but found nothing. In my case I;m using the orm and wanted to have those object properties that map to a table be wrapped up in a dict. The new SQLA book is enroute and I figured I'd look in there when I get it, but in the mean time I just added a cheap/crude method for conversion as per below. I figure there must be a better way.. I had to dip into __mapper__, which didn't seem right, but __dict__ obviously has more than what I wanted. You could use: class_mapper(self).columns.keys() but I don't really know how that's any better. In addition, I had to use a mixin because with ext.declaritive (which I like because it is less verbose) I couldn't make a base class to derive from because it complained that I needed __tablename__ in the base class. You might want to look into the new instrument_declarative function in 0.5. It allows you to wrap a class with the declarative functionality without using the metaclass. This should permit you to work with a subclass, although I haven't tried it. Michael Trier blog.michaeltrier.com --~--~-~--~~~---~--~~ 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: result as the dict()
On Jul 5, 2008, at 4:39 PM, zipito wrote:
Good day community,
there was similar theme - which ended without answer.
How can I convert mine query results to list of dicts. I.e. I want the
following
select = session.select(MineTableObject)
res = select.fetchall()
res_list_dict = [dict(r) for r in res]
but that doesn't works.
dict(r) method returns the array of column_number-column_value not
the dictionary column_name-column_value.
Is there a possibility to have this ??
dict(r) works for me -
from sqlalchemy import *
e = create_engine('sqlite://')
r = e.execute(select 1 as foo, 2 as bar).fetchall()
for x in r:
... print dict(x)
...
{u'foo': 1, u'bar': 2}
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