Mac Auth and post-auth logging to SQL
Jason Antman
jantman at oit.rutgers.edu
Fri Mar 25 20:39:28 CET 2011
I'm referencing the Mac-Auth wiki page at:
http://wiki.freeradius.org/Mac-Auth
Alan DeKok wrote:
> Jason Antman wrote:
>
>> I'm running FreeRADIUS 2.1.7 on CentOS 5, and trying to configure MAC
>> Auth Bypass. I got everything functioning correctly using the Mac-Auth
>> Wiki page as a guide, including placement of the actual CSID
>> authentication code in the post-auth section. However, I just enabled
>> SQL in the post-auth section, and everything is getting logged to SQL
>> with reply Access-Accept, even if it matched the "reject" statement.
>>
>
> I don't see how that is possible. Are you sure you know what it's
> doing? Have you run the server in debugging mode?
>
Yes, I have, and am.
As per the wiki page... I have in authenticate {}:
### snip ###
Auth-Type CSID {
if(Chap-Password){
update control {
Cleartext-Password := "%{User-Name}"
}
chap
}
else{
ok
}
}
### end snip###
which ALWAYS returns OK. Period.
And in post-auth{}:
### snip ###
if(control:Auth-Type == 'CSID'){
# Authorization happens here
authorized_macs.authorize
if(!ok){
reject
}
}
### end snip ###
If I put a "sql" line before this, it always logs with Access-Accept,
since that's what authenticate{} ALWAYS returns, and the sql module is
being called before . If I put a "sql" line after this, it never gets
executed for "reject" statements...
>
>> It seems to me that it's pretty logical that post-auth would be entered
>> with Auth-Type == Access-Accept, the SQL log would happen, and *then*
>> the "reject" statement would get executed
>>
>
> That makes no sense. "If it's accept, it runs reject" ?
>
See above.
>
>> . What I don't understand is
>> why I shouldn't move the actual authentication
>> (authorized_macs.authorize) to the auth { } section, or else how I go
>> about logging rejected requests.
>>
>
> I have no idea what that means.
>
Why is the authorize statement in the post-auth { } section? That seems
to be the cause of these problems...
> Alan DeKok.
> -
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