cx_Oracle doesn't connect when using SID instead of service name on connection string

Andy picture Andy · Jun 10, 2014 · Viewed 40.4k times · Source

I have a connection string that looks like this

con_str = "myuser/[email protected]:1521/ora1"

Where ora1 is the SID of my database. Using this information in SQL Developer works fine, meaning that I can connect and query without problems.

However, if I attempt to connect to Oracle using this string, it fails.

cx_Oracle.connect(con_str)

DatabaseError:  ORA-12514:  TNS:listener  does  not  currently  know  of  service  requested  in  connect  descriptor

This connection string format works if the ora1 is a service name, though.

I have seen other questions that seem to have the reverse of my problem (it works with SID, but not Service name)

What is the proper way to connect to Oracle, using cx_Oracle, using an SID and not a service name? How do I do this without the need to adjust the TNSNAMES.ORA file? My application is distributed to many users internally and making changes to the TNSNAMES file is less than ideal when dealing with users without administrator privileges on their Windows machines. Additionally, when I use service name, I don't need to touch this file at all and would like it keep it that way.

Answer

Andreas Fester picture Andreas Fester · Sep 22, 2014

I a similar scenario, I was able to connect to the database by using cx_Oracle.makedsn() to create a dsn string with a given SID (instead of the service name):

dsnStr = cx_Oracle.makedsn("oracle.sub.example.com", "1521", "ora1")

This returns something like

(DESCRIPTION=(ADDRESS_LIST=(ADDRESS=(PROTOCOL=TCP)(HOST=oracle.sub.example.com)(PORT=1521)))(CONNECT_DATA=(SID=ora1)))

which can then be used with cx_Oracle.connect() to connect to the database:

con = cx_Oracle.connect(user="myuser", password="mypass", dsn=dsnStr)
print con.version
con.close()