On Dec 7, 2011, at 11:08 PM, kris wrote:
Hmm, I think I want the opposite behavior. I was the defualt mapper to
Resource unless a more specific mapper has been defined.
I never defined a 'C' or 'D' and I would like simple get a Resource when no
better mapper is found.
i.e. I get
the recipe is broken, please wait until i have time to fix it, thanks.
On Dec 8, 2011, at 11:27 AM, Michael Bayer wrote:
On Dec 7, 2011, at 11:08 PM, kris wrote:
Hmm, I think I want the opposite behavior. I was the defualt mapper to
Resource unless a more specific mapper has been
OK it works (not yet in tip though).
On Dec 8, 2011, at 11:43 AM, Michael Bayer wrote:
the recipe is broken, please wait until i have time to fix it, thanks.
On Dec 8, 2011, at 11:27 AM, Michael Bayer wrote:
On Dec 7, 2011, at 11:08 PM, kris wrote:
Hmm, I think I want the opposite
Does this mean I can use this in 7.3? or Do I need to download tip?
Also could you show me how this would work with old style mapper(...) s
instead of declarative_base?
Thanks,
Kris
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works in 0.7.3. will also work in 0.7.4 Just not the current tip at the
moment.
the regular mapper() setup is equivalent (see
http://www.sqlalchemy.org/docs/orm/mapper_config.html#classical-mappings) :
discriminator_expr = case(...)
mapper(Person, ..., polymorphic_on=discriminator_expr,
On Dec 7, 2011, at 2:53 AM, kris wrote:
I am using a single table scheme to store for a set of resource types.
I would like to load a specific class if a mapper is defined and use the base
class if no
more specific version can be found. Is there a way to do this?
i.e.
resource =
Hmm, I think I want the opposite behavior. I was the defualt mapper to
Resource unless a more specific mapper has been defined.
I never defined a 'C' or 'D' and I would like simple get a Resource when no
better mapper is found.
i.e. I get Resource class for C, D and get a Resource_A for A and
I am using a single table scheme to store for a set of resource types.
I would like to load a specific class if a mapper is defined and use the
base class if no
more specific version can be found. Is there a way to do this?
i.e.
resource = Table('resource',
