Has anyone ever tried adding a RHEL subscription to a Fedora 11/12 box
before?
I attempted to use rhn_register, but after giving it my RHN credentials,
which are valid, I receive a Python traceback[1] and it quits.
[1] Attached.
[Mon Feb 22 10:34:20 2010] rhn_register
Traceback (most recent call last):
File "/usr/sbin/rhn_register", line 83, in <module>
app.run()
File "/usr/share/rhn/up2date_client/rhncli.py", line 66, in run
sys.exit(self.main() or 0)
File "/usr/sbin/rhn_register", line 65, in main
ui.main()
File "/usr/share/rhn/up2date_client/tui.py", line 1722, in main
tui.run()
File "/usr/share/rhn/up2date_client/tui.py", line 1609, in run
if self._show_subscription_window() == False:
File "/usr/share/rhn/up2date_client/tui.py", line 1544, in _show_subscription_window
self._activate_hardware()
File "/usr/share/rhn/up2date_client/tui.py", line 1523, in _activate_hardware
hardwareInfo, self.other['org_id'])
File "/usr/share/rhn/up2date_client/rhnreg.py", line 799, in activateHardwareInfo
hardwareInfo, other)
File "/usr/share/rhn/up2date_client/rhnserver.py", line 50, in __call__
return rpcServer.doCall(method, *args, **kwargs)
File "/usr/share/rhn/up2date_client/rpcServer.py", line 200, in doCall
ret = method(*args, **kwargs)
File "/usr/lib64/python2.6/xmlrpclib.py", line 1199, in __call__
return self.__send(self.__name, args)
File "/usr/share/rhn/up2date_client/rpcServer.py", line 39, in _request1
ret = self._request(methodname, params)
File "/usr/lib/python2.6/site-packages/rhn/rpclib.py", line 318, in _request
request = self._req_body(params, methodname)
File "/usr/lib/python2.6/site-packages/rhn/rpclib.py", line 222, in _req_body
return xmlrpclib.dumps(params, methodname, encoding=self._encoding)
File "/usr/lib64/python2.6/xmlrpclib.py", line 1132, in dumps
data = m.dumps(params)
File "/usr/lib64/python2.6/xmlrpclib.py", line 677, in dumps
dump(v, write)
File "/usr/lib64/python2.6/xmlrpclib.py", line 699, in __dump
f(self, value, write)
File "/usr/lib64/python2.6/xmlrpclib.py", line 778, in dump_struct
raise TypeError, "dictionary key must be string"
<type 'exceptions.TypeError'>: dictionary key must be string
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