Python-Oracle connection issue with Pyodbc

Question

Tried to connect to the Oracle database using Pyodbc:

try:
    db_connection = pyodbc.connect('DSN=OraDev; PWD=%s' % Key.dbKeys['password'])
except pyodbc.Error, err:
        print >> debug_file, "Database connection failed: %s" %err 
        sys.exit()

and it keeps giving me this error message:

'[IM014] [Microsoft][ODBC Driver Manager] The specified DSN contains an architecture mismatch between the Driver and Application (0) (SQLDriverConnect)

So I am wondering if the culprit is that I am not having the "matching" components. Currently, I have:

Python 2.7 32bit

Pyodbc 3.0.6 win32-py2.7

Oracle ODBC driver: 32bit

Windows server 2008 64bit

Oracle 11.2.0 64bit

Are there anything wrong with the versions here? Thanks.

Nowadays I would use Oracle's python-oracledb driver which has a lot more functionality than ODBC, and also doesn't need all the extra Oracle client libraries or ODBC driver manager that an ODBC installation needs. — Christopher Jones
– Christopher Jones, Commented May 3 at 2:37

BigHandsome · Accepted Answer · 2012-09-25 02:24:31Z

1

The issue you are having is that 64 bit Windows does not play well with 32 bit ODBC. You can read more about it here:

The 32-bit version of the ODBC Administrator tool and the 64-bit version of the ODBC Administrator tool display both the 32-bit user DSNs and the 64-bit user DSNs in a 64-bit version of the Windows operating system.

The work around is to specifically target the ODBC driver for the architecture you are looking to implement.

A side note, trying to implement ODBC for oracle on a mixed architecture platform became a headache for us. So, we implemented access to oracle through CX_Oracle

answered Sep 25, 2012 at 2:24

BigHandsome

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BigHandsome Over a year ago

You need to find out whether your Oracle client is 32-bit or 64-bit. Everything must be 64-bit or everything must be 32-bit you cannot mix and match. Sometimes the 64-bit client also has 32-bit libraries but they will be in an alternative directory.

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