Comments (3)
is this related to pypy only or pypyodbc has the same problem on cpython?
Original comment by [email protected]
on 19 Mar 2014 at 5:22
from pypyodbc.
Has the same problem on cpython:
==================
$ python
Python 2.7.2 (default, Jun 19 2012, 12:13:20)
[GCC 4.6.1 20110627 (Mandriva)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import pyodbc
>>> import pypyodbc
>>> pyodbc.connect("DSN=pyProgress")
<pyodbc.Connection object at 0x7f61f93b4c60>
>>> pypyodbc.connect("DSN=pyProgress")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/site-packages/pypyodbc-1.3.1-py2.7.egg/pypyodbc.py", line 2402, in __init__
AllocateEnv()
File "/usr/lib/python2.7/site-packages/pypyodbc-1.3.1-py2.7.egg/pypyodbc.py", line 995, in AllocateEnv
check_success(SQL_NULL_HANDLE, ret)
File "/usr/lib/python2.7/site-packages/pypyodbc-1.3.1-py2.7.egg/pypyodbc.py", line 989, in check_success
ctrl_err(SQL_HANDLE_ENV, ODBC_obj, ret, False)
File "/usr/lib/python2.7/site-packages/pypyodbc-1.3.1-py2.7.egg/pypyodbc.py", line 969, in ctrl_err
raise ProgrammingError('', 'SQL_INVALID_HANDLE')
pypyodbc.ProgrammingError: ('', 'SQL_INVALID_HANDLE')
>>>
Original comment by [email protected]
on 19 Mar 2014 at 8:35
from pypyodbc.
Have just run into this same error.
Python 3.4.1 (Win32)
pypyodbc 1.3.3
Have executed several queries, but then this one I've been fighting with, and
thought I'd finally got working, results in this [very unhelpful] error message.
Original comment by [email protected]
on 18 Jul 2014 at 3:33
from pypyodbc.
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