On 9/21/15 4:08 PM, Cliff Hill wrote:
So, I have some code that is constructing a declarative table
definition on the fly, built from a YAML file. It works great.
However, when I try to do *help(tablename)*, I get the following error:
Python 3.4.3 (default, May 14 2015, 09:48:44)
[GCC 4.3.4 [gcc-4_3-branch revision 152973]] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from pampas.sql import tables
>>> tables.pampas.Plugin
<class 'pampas.sql.tables.pampas.Plugin'>
>>> help(tables.pampas.Plugin)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/_sitebuiltins.py", line
103, in __call__
return pydoc.help(*args, **kwds)
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/pydoc.py", line 1827,
in __call__
self.help(request)
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/pydoc.py", line 1877,
in help
else: doc(request, 'Help on %s:', output=self._output)
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/pydoc.py", line 1614,
in doc
pager(render_doc(thing, title, forceload))
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/pydoc.py", line 1607,
in render_doc
return title % desc + '\n\n' + renderer.document(object, name)
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/pydoc.py", line 372, in
document
if inspect.isclass(object): return self.docclass(*args)
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/pydoc.py", line 1272,
in docclass
for name, kind, cls, value in classify_class_attrs(object)
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/pydoc.py", line 205, in
classify_class_attrs
for (name, kind, cls, value) in
inspect.classify_class_attrs(object):
File "/lfs/opt/qqpamp0/pkg/lib/python3.4/inspect.py", line 405,
in classify_class_attrs
obj = get_obj or dict_obj
File
"/lfs/opt/qqpamp0/venv/cliff/lib/python3.4/site-packages/sqlalchemy/sql/elements.py",
line 539, in __bool__
raise TypeError("Boolean value of this clause is not defined")
TypeError: Boolean value of this clause is not defined
Mind you, this only happens if I try to do help() on my table. It
doesn't come up otherwise. I have warned people using my API in my
company away from using help(), and instead relying on the table
definition in the database and/or the YAML files.
this is a bug in Python itself which I provided a fix for in
http://bugs.python.org/issue23898. It appeared in 3.4 and is fixed in
Python 3.5.
Essentially, what is done is:
1. Code looks up a YAML definition locally for the table.
2. if the file doesn't exist, it uses DB reflection to construct the
file.
3. Load the YAML file
4. Use the YAML file to construct the required namespace dict to pass
to sqlalchemy.ext.declarative.api.DeclarativeMeta(), using my
existing declarative_base instance for the base class of my new class.
5. Returns the class definition constructed
from sqlalchemy.ext.declarative.api.DeclarativeMeta(). (note, I
have used type() rather
than sqlalchemy.ext.declarative.api.DeclarativeMeta() with the
same results).
This all works well, and I am very happy it does. However, where it is
lacking is that help() is broken. When I make a table statically, I
have no such problem. When I am attempting to construct it
dynamically, I run into this problem.
I'm not sure why the issue doesn't appear under some circumstances; if
you have a DeclarativeMeta type of structure, it should consist of the
same elements no matter how the class structure was generated.
I have considered constructing the class definition completely as a
string, and then exec() the whole thing, as this could work for me.
However I'd hope there was a solution that wouldn't involve such a
heavy-handed alteration to my code.
I am running SQLAlchemy 1.0.8, on Python 3.4.3.
Thanks for any help.
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