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

Question

I have a connection string that looks like this

con_str = \"myuser/mypass@oracle.sub.example.com:1521/ora1\"

Where ora1 is the

Answer 1

I also met this issue. The solution is:

1: get the service name at tnsnames.ora
2: put the service name in
con_str = "myuser/mypass@oracle.sub.example.com:1521/ora1"

Answer 2

SID's may not be easily accessible or you might not have it created for your database.

In my case, I'm working from the client side requesting access to a cloud database so creating an SID didn't really make sense.

Instead, you might have a string that looks similar to this:

"(DESCRIPTION = (ADDRESS = (PROTOCOL = TCP)(HOST = something.cloud.company)
(PORT = 12345)) (ADDRESS = (PROTOCOL = TCP)(HOST = something.cloud.company)
(PORT = 12345)) (CONNECT_DATA = (SERVER = DEDICATED) (SERVICE_NAME = 
something.company)))"

You can use it in replacement of the SID.

connection = cx_Oracle.connect("username", "pw", "(DESCRIPTION = (ADDRESS = 
                (PROTOCOL = TCP)(HOST = something.cloud.company)(PORT = 12345)) (ADDRESS = 
                (PROTOCOL = TCP)(HOST = something.cloud.company)(PORT = 12345)) 
                (CONNECT_DATA = (SERVER = DEDICATED) (SERVICE_NAME = something.company)))")

Answer 3

I thought during a while that I would not be able to use Magic SQL (%sql, %%sql) because of service name issue in connection that would force to use the alternative way described above with cx_Oracle.connect(), cx_Oracle.makedsn()... I finally found a solution working for me: declare and set a variable for the service name first and then use it in the command (since not working if literal string for service name put in the command !)

import cx_Oracle

user='youruser'
pwd='youruserpwd'
dbhost='xx.xx.xx.xx'
service='yourservice'

%load_ext sql
%sql oracle+cx_oracle://$user:$pwd@$dbhost:1521/?service_name=$service

output (what you get in successful connection):

u'Connected: youruser@'

Answer 4

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()

Answer 5

It still may not work. You need to take the output of dsnStr and modify the string by replacing SID with SERVICE_NAME and use that variable in the con string. This procedure worked for me.

Answer 6

For those looking for how to specify service_name instead of SID.

From changelog for SQLAlchemy 1.0.0b1 (released on March 13, 2015):

[oracle] [feature] Added support for cx_oracle connections to a specific service name, as opposed to a tns name, by passing ?service_name=<name> to the URL. Pull request courtesy Sławomir Ehlert.

The change introduces new, Oracle dialect specific option service_name which can be used to build connect string like this:

from sqlalchemy import create_engine
from sqlalchemy.engine import url

connect_url = url.URL(
    'oracle+cx_oracle',
    username='some_username',
    password='some_password',
    host='some_host',
    port='some_port',
    query=dict(service_name='some_oracle_service_name'))

engine = create_engine(connect_url)