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1 Odgovori
5037 Prikazi

I installed OpenERP on Debian Squeeze via the nightly apt repository.

The main user interface and the underlying bits all seem to work after I fixed the postgresql template, but the /xmlrpc/ interface is throwing 404 (Not Found) errors when I try to point a browser at http://[servername]:8069/xmlrpc/ or http://[servername]:8069/xmlrpc/common etc.

Are the xmlrpc files not included in the apt repository? From what I understand, it's removed from the SaaS version, but this is one I'm running myself.

Here is my configuration:

[options]
admin_passwd = xxx
db_host = False
db_port = False
db_user = openerp
db_password = False
xmlrpc = True
xmlrpc_interface = x.x.x.x
xmlrpc_port = 8069
proxy_mode = True

(The bottom bit was suggested as it will be running behind an nginx proxy to handle SSL)

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Best Answer

I have been using XMLRPC for several weeks on V7, but indirectly through https://pypi.python.org/pypi/openerp-client-lib/1.0.0

I connect with :

    connection = openerplib.get_connection
        (
              login = creds['user_id']
            , database = creds['db_name']
            , hostname = creds['host_name']
            , password = creds['user_pwd']
            , protocol = 'xmlrpc'
            , port = 8069
        )

However, I also get a "Not Found" if I try to point my browser at http://[servername]:8069/xmlrpc/ I think the problem you are seeing is a protocol clash; not missing code. In your browser you are specifying http, not xmlrpc. It doesn't work for me either, but I know that XMLRPC is working.

The openerplib package makes many things much easier. I'd suggest you give it a try. I chose to use it, precisely because I didn't want to have to fiddle with the nuts'n bolts of XMLRPC.

The following little script should show you if XMLRPC is open for business on your server.

import openerplib

connection = openerplib.get_connection(hostname="localhost", database="my_db", \
    login="my_user", password="xxx")
user_model = connection.get_model("res.users")
ids = user_model.search([("login", "=", "admin")])
user_info = user_model.read(ids[0], ["name"])
print user_info["name"]
# will print "Administrator"
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