Hi Michael,
OK, that's why.
Below are the stack traces. They are not the same for A.b and A_b. If i use
other labels this error does not happen.
cheers, lars
with expr.label( "%s_%s" % (cls.__name__, attr_name)):
File "C:\python27\lib\site-packages\bottle.py", line 781, in _handle
return
On Mar 20, 2014, at 9:22 AM, lars van gemerden wrote:
> I solved it (again):
>
> for future reference:
>
> say you have the expression Person.name (Person is SQLA class) to use in a
> query, to have a label show up under that name in the resulting KeyedTuple's:
>
> to do:
>
> Person.name.la
On Mar 19, 2014, at 2:49 PM, lars van gemerden wrote:
> Hi all,
>
> I solved this before, but i took me half a day and a lot of guessing and i
> forgot.
>
> If i add (sometimes calculated) columns to a query and i want to set the
> names manually (via label()?), how do i do that? How will i
Hi all,
I solved this before, but i took me half a day and a lot of guessing and i
forgot.
If i add (sometimes calculated) columns to a query and i want to set the
names manually (via label()?), how do i do that? How will i make them show
up in a resulting KeyedTuple/column_descriptions()?
Ch
Thomas Heller wrote:
> I'm doing my first steps with sqlalchemy - a really great package?
>
> Here is my question:
>
> I know already how to label columns or functions that I'm querying for,
> like this:
>
> session.query(count().label("foo"), MyClass.id.label("my_id"))
>
> How can I label the resu
I'm doing my first steps with sqlalchemy - a really great package?
Here is my question:
I know already how to label columns or functions that I'm querying for,
like this:
session.query(count().label("foo"), MyClass.id.label("my_id"))
How can I label the result when I search for a class? This d
