[SQLObject] bug in sqlobject-admin

[Imri G. <lor...@gm...>]

Heya
I've worked a bit with sqlobject-admin, and there's a bug in the
implementation of the status command.
Steps to reproduce:

1. Create a simple table with a decimal column, for example:

class MyThing(sqlobject.SQLObject):
    bla1 = sqlobject.DecimalCol(size = 10, precision = 2)

2. Use sqlobject-admin create --egg=... -c sqlite://....
3. Use sqlobject-admin status --egg=... -c sqlite://...

You should get the following exception:
[snip]
  File
"D:\applic\program\python25\lib\site-packages\sqlobject-0.12.1-py2.5.egg\
sqlobject\manager\command.py", line 765, in command
    col = col.withClass(soClass)
  File
"D:\applic\program\python25\lib\site-packages\sqlobject-0.12.1-py2.5.egg\
sqlobject\col.py", line 408, in withClass
    **self._kw)
  File
"D:\applic\program\python25\lib\site-packages\sqlobject-0.12.1-py2.5.egg\
sqlobject\col.py", line 1329, in __init__
    "You must give a size argument"
AssertionError: You must give a size argument

After researching this issue a bit, I still haven't figured out whether this
is the result of a bug in the command implementation, or the way that
keyword arguments are handled in the column class hierarchy.

Cheers,
Imri

-- 
Imri Goldberg
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