Teradata pypyodbc SQL_INVALID_HANDLE

0

While connecting pypyodbc with teradata i am getting SQL_INVALID_HANDLE

OS - Red Hat 4.4.7-17

Python Version - 2.6.6

I am not having administrative privilages to install the pypyodbc so i have downloaded the module, set pythonpath and imported it.

Done the required exports to point the python module to find the correct libodbc.so

export LD_LIBRARY_PATH=/opt/teradata/client/15.10/odbc_64/lib/

export ODBCINST=/opt/teradata/client/ODBC_64/odbcinst.ini

export ODBCINI=/opt/teradata/client/ODBC_64/odbc.ini

used the below code to create a db connection

pypyodbc.connect("Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>")

The odbc connection string i have used is correct. I have tested using the below approach

/opt/teradata/client/15.10/bin/tdxodbc64 -C "Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>"

**Note:**I am not using teradata OOTB python module because it works with a higher version of python which i cannot install.

Error

  File "<stdin>", line 1, in <module>

  File "<LOCAL_PATH>/pypyodbc.py", line 2421, in __init__

    AllocateEnv()

  File "<LOCAL_PATH>/pypyodbc.py", line 1000, in AllocateEnv

    check_success(SQL_NULL_HANDLE, ret)

  File "<LOCAL_PATH>/pypyodbc.py", line 990, in check_success

    ctrl_err(SQL_HANDLE_ENV, ODBC_obj, ret, False)

  File "<LOCAL_PATH>/pypyodbc.py", line 970, in ctrl_err

    raise ProgrammingError('', 'SQL_INVALID_HANDLE')

pypyodbc.ProgrammingError: ('', 'SQL_INVALID_HANDLE')

any guidance on the matter is helpful

python teradata pyodbc python-2.6 pypyodbc

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asked Nov 21 '18 at 7:12

CharityCharity

115114

add a comment | 

0

While connecting pypyodbc with teradata i am getting SQL_INVALID_HANDLE

OS - Red Hat 4.4.7-17

Python Version - 2.6.6

I am not having administrative privilages to install the pypyodbc so i have downloaded the module, set pythonpath and imported it.

Done the required exports to point the python module to find the correct libodbc.so

export LD_LIBRARY_PATH=/opt/teradata/client/15.10/odbc_64/lib/

export ODBCINST=/opt/teradata/client/ODBC_64/odbcinst.ini

export ODBCINI=/opt/teradata/client/ODBC_64/odbc.ini

used the below code to create a db connection

pypyodbc.connect("Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>")

The odbc connection string i have used is correct. I have tested using the below approach

/opt/teradata/client/15.10/bin/tdxodbc64 -C "Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>"

**Note:**I am not using teradata OOTB python module because it works with a higher version of python which i cannot install.

Error

  File "<stdin>", line 1, in <module>

  File "<LOCAL_PATH>/pypyodbc.py", line 2421, in __init__

    AllocateEnv()

  File "<LOCAL_PATH>/pypyodbc.py", line 1000, in AllocateEnv

    check_success(SQL_NULL_HANDLE, ret)

  File "<LOCAL_PATH>/pypyodbc.py", line 990, in check_success

    ctrl_err(SQL_HANDLE_ENV, ODBC_obj, ret, False)

  File "<LOCAL_PATH>/pypyodbc.py", line 970, in ctrl_err

    raise ProgrammingError('', 'SQL_INVALID_HANDLE')

pypyodbc.ProgrammingError: ('', 'SQL_INVALID_HANDLE')

any guidance on the matter is helpful

python teradata pyodbc python-2.6 pypyodbc

share|improve this question

asked Nov 21 '18 at 7:12

CharityCharity

115114

add a comment | 

0

0

0

While connecting pypyodbc with teradata i am getting SQL_INVALID_HANDLE

OS - Red Hat 4.4.7-17

Python Version - 2.6.6

I am not having administrative privilages to install the pypyodbc so i have downloaded the module, set pythonpath and imported it.

Done the required exports to point the python module to find the correct libodbc.so

export LD_LIBRARY_PATH=/opt/teradata/client/15.10/odbc_64/lib/

export ODBCINST=/opt/teradata/client/ODBC_64/odbcinst.ini

export ODBCINI=/opt/teradata/client/ODBC_64/odbc.ini

used the below code to create a db connection

pypyodbc.connect("Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>")

The odbc connection string i have used is correct. I have tested using the below approach

/opt/teradata/client/15.10/bin/tdxodbc64 -C "Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>"

**Note:**I am not using teradata OOTB python module because it works with a higher version of python which i cannot install.

Error

  File "<stdin>", line 1, in <module>

  File "<LOCAL_PATH>/pypyodbc.py", line 2421, in __init__

    AllocateEnv()

  File "<LOCAL_PATH>/pypyodbc.py", line 1000, in AllocateEnv

    check_success(SQL_NULL_HANDLE, ret)

  File "<LOCAL_PATH>/pypyodbc.py", line 990, in check_success

    ctrl_err(SQL_HANDLE_ENV, ODBC_obj, ret, False)

  File "<LOCAL_PATH>/pypyodbc.py", line 970, in ctrl_err

    raise ProgrammingError('', 'SQL_INVALID_HANDLE')

pypyodbc.ProgrammingError: ('', 'SQL_INVALID_HANDLE')

any guidance on the matter is helpful

python teradata pyodbc python-2.6 pypyodbc

share|improve this question

asked Nov 21 '18 at 7:12

CharityCharity

115114

While connecting pypyodbc with teradata i am getting SQL_INVALID_HANDLE

OS - Red Hat 4.4.7-17

Python Version - 2.6.6

I am not having administrative privilages to install the pypyodbc so i have downloaded the module, set pythonpath and imported it.

Done the required exports to point the python module to find the correct libodbc.so

export LD_LIBRARY_PATH=/opt/teradata/client/15.10/odbc_64/lib/

export ODBCINST=/opt/teradata/client/ODBC_64/odbcinst.ini

export ODBCINI=/opt/teradata/client/ODBC_64/odbc.ini

used the below code to create a db connection

pypyodbc.connect("Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>")

The odbc connection string i have used is correct. I have tested using the below approach

/opt/teradata/client/15.10/bin/tdxodbc64 -C "Driver={Teradata};DBCNAME=<DB>;UID=<UNAME>;PWD=<PASS>"

**Note:**I am not using teradata OOTB python module because it works with a higher version of python which i cannot install.

Error

  File "<stdin>", line 1, in <module>

  File "<LOCAL_PATH>/pypyodbc.py", line 2421, in __init__

    AllocateEnv()

  File "<LOCAL_PATH>/pypyodbc.py", line 1000, in AllocateEnv

    check_success(SQL_NULL_HANDLE, ret)

  File "<LOCAL_PATH>/pypyodbc.py", line 990, in check_success

    ctrl_err(SQL_HANDLE_ENV, ODBC_obj, ret, False)

  File "<LOCAL_PATH>/pypyodbc.py", line 970, in ctrl_err

    raise ProgrammingError('', 'SQL_INVALID_HANDLE')

pypyodbc.ProgrammingError: ('', 'SQL_INVALID_HANDLE')

any guidance on the matter is helpful

python teradata pyodbc python-2.6 pypyodbc

python teradata pyodbc python-2.6 pypyodbc

share|improve this question

asked Nov 21 '18 at 7:12

CharityCharity

115114

share|improve this question

asked Nov 21 '18 at 7:12

CharityCharity

115114

share|improve this question

share|improve this question

asked Nov 21 '18 at 7:12

CharityCharity

115114

asked Nov 21 '18 at 7:12

CharityCharity

115114

asked Nov 21 '18 at 7:12

CharityCharity

115114

115114

add a comment | 

add a comment | 

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