Re: [sqlalchemy] accessing base class on hybrid expressions from the class side

[Mike Bayer]

On 05/10/2016 04:14 PM, Brian Cherinka wrote:

Ok, thanks for the response.  What's the timeframe for the 1.1 release?


it will bethis year :)   hopefully before the summer is over.   I 
try to get one major version each year, and I'm pretty busy as this is 
the first full year in a long time I'm employed full time.   1.1 is 
mostly ready for betas and I just need to find the time to close out a 
few more things and put some out.





  In the meantime, I will have a look into adding my own class_
attribute, or using the Comparator.

I tried something like

setattr(datadb.Cube.plateifu, "class_", datadb.Cube.id.class_)

but it didn't seem to work.  But I'll dig a bit deeper.  If I can't get
something working with a 1.0X release, I'll try the 1.1 in bitbucket.


On Tuesday, May 10, 2016 at 2:32:22 PM UTC-4, Mike Bayer wrote:

in 1.1 these hybrids will have the class_ attribute like other
attributes.

Until then you can probably add your own class_ attribute to the object
which you are returning.   Also, using a custom Comparator class (see
the example in the hybrid docs) will also return an instrumented
attribute that should have a class_ attribute.

Or you could try using the 1.1 hybrid_property class yourself, it
should
be compatible with 1.0.   The commits are illustrated in
https://bitbucket.org/zzzeek/sqlalchemy/issues/3653
 but you can
probably
just use the hybrid.py straight from the git repository with 1.0.




On 05/10/2016 02:01 PM, Brian Cherinka wrote:
 >
 > I'm trying to build a query system where given a filter parameter
name,
 > I can figure out which DeclarativeBase class it is attached to.
  I need
 > to do this for a mix of standard InstrumentedAttributes and Hybrid
 > Properties/Expressions. I have several Declarative Base classes with
 > hybrid properties / expressions defined, in addition to the standard
 > InstrumentedAttributes from the actual table.
 >   mydb.dataModelClasses.Cube for example.
 >
 > For a standard attribute, I can access the class using the class_
variable.
 >
 > Standard Attribute on the DeclarativeBase class Cube
 > |
 > type(datadb.Cube.id )
 > sqlalchemy.orm.attributes.InstrumentedAttribute
 >
 > printdatadb.Cube.id.class_
 > mydb.DataModelClasses.Cube
 > |
 >
 > What's the best way to retrieve this same information for a hybrid
 > expression?  My expressions are other types, thus don't have the
class_
 > attribute.  One example of my hybrid property defined in the Cube
class
 >
 > |
 > @hybrid_property
 > defplateifu(self):
 > return'{0}-{1}'.format(self.plate,self.ifu.name
)
 >
 > @plateifu.expression
 > defplateifu(cls):
 > returnfunc.concat(Cube.plate,'-',IFUDesign.name)
 > |
 >
 > |
 > type(datadb.Cube.plateifu)
 > sqlalchemy.sql.functions.concat
 > |
 >
 > Since this property is now a function concat, what's the best way to
 > retrieve the name of the class that this property is attached to,
namely
 > 'mydb.DataModelClasses.Cube'?  It doesn't seem to have a .class_ or
 > .parent attribute.  Is there a way to add a new attribute onto my
hybrid
 > columns that let me access the parent class?
 >
 > I need to do this for a variety of hybrid properties/expressions,
that
 > are all constructed in unique ways.  This particular example is a
 > function concat, however I have others that are of type
 > sqlalchemy.sql.elements.BinaryExpression.
 >
 > Is there a way to generically do this no matter the type of hybrid
 > expression I define?
 >
 > Thanks.
 >
 > --

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Re: [sqlalchemy] accessing base class on hybrid expressions from the class side

[Brian Cherinka]

Ok, thanks for the response.  What's the timeframe for the 1.1 release?  In 
the meantime, I will have a look into adding my own class_ attribute, or 
using the Comparator.  

I tried something like

setattr(datadb.Cube.plateifu, "class_", datadb.Cube.id.class_)

but it didn't seem to work.  But I'll dig a bit deeper.  If I can't get 
something working with a 1.0X release, I'll try the 1.1 in bitbucket. 


On Tuesday, May 10, 2016 at 2:32:22 PM UTC-4, Mike Bayer wrote:
>
> in 1.1 these hybrids will have the class_ attribute like other attributes. 
>
> Until then you can probably add your own class_ attribute to the object 
> which you are returning.   Also, using a custom Comparator class (see 
> the example in the hybrid docs) will also return an instrumented 
> attribute that should have a class_ attribute. 
>
> Or you could try using the 1.1 hybrid_property class yourself, it should 
> be compatible with 1.0.   The commits are illustrated in 
> https://bitbucket.org/zzzeek/sqlalchemy/issues/3653 but you can probably 
> just use the hybrid.py straight from the git repository with 1.0. 
>
>
>
>
> On 05/10/2016 02:01 PM, Brian Cherinka wrote: 
> > 
> > I'm trying to build a query system where given a filter parameter name, 
> > I can figure out which DeclarativeBase class it is attached to.  I need 
> > to do this for a mix of standard InstrumentedAttributes and Hybrid 
> > Properties/Expressions. I have several Declarative Base classes with 
> > hybrid properties / expressions defined, in addition to the standard 
> > InstrumentedAttributes from the actual table. 
> >   mydb.dataModelClasses.Cube for example. 
> > 
> > For a standard attribute, I can access the class using the class_ 
> variable. 
> > 
> > Standard Attribute on the DeclarativeBase class Cube 
> > | 
> > type(datadb.Cube.id) 
> > sqlalchemy.orm.attributes.InstrumentedAttribute 
> > 
> > printdatadb.Cube.id.class_ 
> > mydb.DataModelClasses.Cube 
> > | 
> > 
> > What's the best way to retrieve this same information for a hybrid 
> > expression?  My expressions are other types, thus don't have the class_ 
> > attribute.  One example of my hybrid property defined in the Cube class 
> > 
> > | 
> > @hybrid_property 
> > defplateifu(self): 
> > return'{0}-{1}'.format(self.plate,self.ifu.name) 
> > 
> > @plateifu.expression 
> > defplateifu(cls): 
> > returnfunc.concat(Cube.plate,'-',IFUDesign.name) 
> > | 
> > 
> > | 
> > type(datadb.Cube.plateifu) 
> > sqlalchemy.sql.functions.concat 
> > | 
> > 
> > Since this property is now a function concat, what's the best way to 
> > retrieve the name of the class that this property is attached to, namely 
> > 'mydb.DataModelClasses.Cube'?  It doesn't seem to have a .class_ or 
> > .parent attribute.  Is there a way to add a new attribute onto my hybrid 
> > columns that let me access the parent class? 
> > 
> > I need to do this for a variety of hybrid properties/expressions, that 
> > are all constructed in unique ways.  This particular example is a 
> > function concat, however I have others that are of type 
> > sqlalchemy.sql.elements.BinaryExpression. 
> > 
> > Is there a way to generically do this no matter the type of hybrid 
> > expression I define? 
> > 
> > Thanks. 
> > 
> > -- 
>
>

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Re: [sqlalchemy] accessing base class on hybrid expressions from the class side

[Mike Bayer]

in 1.1 these hybrids will have the class_ attribute like other attributes.

Until then you can probably add your own class_ attribute to the object 
which you are returning.   Also, using a custom Comparator class (see 
the example in the hybrid docs) will also return an instrumented 
attribute that should have a class_ attribute.


Or you could try using the 1.1 hybrid_property class yourself, it should 
be compatible with 1.0.   The commits are illustrated in 
https://bitbucket.org/zzzeek/sqlalchemy/issues/3653 but you can probably 
just use the hybrid.py straight from the git repository with 1.0.





On 05/10/2016 02:01 PM, Brian Cherinka wrote:


I'm trying to build a query system where given a filter parameter name,
I can figure out which DeclarativeBase class it is attached to.  I need
to do this for a mix of standard InstrumentedAttributes and Hybrid
Properties/Expressions. I have several Declarative Base classes with
hybrid properties / expressions defined, in addition to the standard
InstrumentedAttributes from the actual table.
  mydb.dataModelClasses.Cube for example.

For a standard attribute, I can access the class using the class_ variable.

Standard Attribute on the DeclarativeBase class Cube
|
type(datadb.Cube.id)
sqlalchemy.orm.attributes.InstrumentedAttribute

printdatadb.Cube.id.class_
mydb.DataModelClasses.Cube
|

What's the best way to retrieve this same information for a hybrid
expression?  My expressions are other types, thus don't have the class_
attribute.  One example of my hybrid property defined in the Cube class

|
@hybrid_property
defplateifu(self):
return'{0}-{1}'.format(self.plate,self.ifu.name)

@plateifu.expression
defplateifu(cls):
returnfunc.concat(Cube.plate,'-',IFUDesign.name)
|

|
type(datadb.Cube.plateifu)
sqlalchemy.sql.functions.concat
|

Since this property is now a function concat, what's the best way to
retrieve the name of the class that this property is attached to, namely
'mydb.DataModelClasses.Cube'?  It doesn't seem to have a .class_ or
.parent attribute.  Is there a way to add a new attribute onto my hybrid
columns that let me access the parent class?

I need to do this for a variety of hybrid properties/expressions, that
are all constructed in unique ways.  This particular example is a
function concat, however I have others that are of type
sqlalchemy.sql.elements.BinaryExpression.

Is there a way to generically do this no matter the type of hybrid
expression I define?

Thanks.

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[sqlalchemy] accessing base class on hybrid expressions from the class side

[Brian Cherinka]

I'm trying to build a query system where given a filter parameter name, I 
can figure out which DeclarativeBase class it is attached to.  I need to do 
this for a mix of standard InstrumentedAttributes and Hybrid 
Properties/Expressions. I have several Declarative Base classes with 
hybrid properties / expressions defined, in addition to the standard 
InstrumentedAttributes from the actual table.  mydb.dataModelClasses.Cube 
for example.

For a standard attribute, I can access the class using the class_ variable. 
 

Standard Attribute on the DeclarativeBase class Cube
type(datadb.Cube.id)
sqlalchemy.orm.attributes.InstrumentedAttribute

print datadb.Cube.id.class_
mydb.DataModelClasses.Cube

What's the best way to retrieve this same information for a hybrid 
expression?  My expressions are other types, thus don't have the class_ 
attribute.  One example of my hybrid property defined in the Cube class

@hybrid_property
def plateifu(self):
return '{0}-{1}'.format(self.plate, self.ifu.name)

@plateifu.expression
def plateifu(cls):
return func.concat(Cube.plate, '-', IFUDesign.name)

type(datadb.Cube.plateifu)
sqlalchemy.sql.functions.concat

Since this property is now a function concat, what's the best way to 
retrieve the name of the class that this property is attached to, namely '
mydb.DataModelClasses.Cube'?  It doesn't seem to have a .class_ or .parent 
attribute.  Is there a way to add a new attribute onto my hybrid columns 
that let me access the parent class?

I need to do this for a variety of hybrid properties/expressions, that are 
all constructed in unique ways.  This particular example is a function 
concat, however I have others that are of type 
sqlalchemy.sql.elements.BinaryExpression. 

Is there a way to generically do this no matter the type of hybrid 
expression I define?

Thanks. 
  

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