Freeradius proxy support

Ilavajuthy Palanisamy ilavajuthy at gmail.com
Mon Apr 7 22:43:30 CEST 2014


Hi Alan,

I have set it up Freeradius as proxy with JRadius plugin. I'm using JRadius
to modify the Access_Accept packets.
With my setup I'm able to receive the radius packets through JRadius plugin
during post-auth. In JRadius I have a handler to modify the Access_Accept
and it is working fine.

Now I need a help in setting up various sets of Radius servers in
proxy.conf based on some VSA value in Access_Request. The requirement is to
redirect the requests to appropriate Radius servers based on the VSA value.
What I can understand from the proxy.conf is to define various home_servers
and home_server pool linked to a realm.
Now I'm not sure how to link the VSA value in Access_Request to these
home_servers.
Could you please provide some example to achieve my above requirement.

Thanks,
Ila.



On Fri, Mar 21, 2014 at 5:53 AM, Alan DeKok <aland at deployingradius.com>wrote:

> Ilavajuthy Palanisamy wrote:
> > I'm new to Freeradius. I would like to get some pointers and
> > documentation details about setting up Freeradius as proxy.
>
>   raddb/proxy.conf.
>
> > Our requirement is to have a Radius proxy and modify the Access Accept
> > response based on certain criteria.
>
>   raddb/sites-available/default.  Look for the "post-proxy" section.
>
> > Few questions regarding Freeradius proxy support.
> > Does Freeradius proxy has support for a plugin probably in Java to snoop
> > the Radius request calls?
>
>   No.  If you want to snoop on radius packets, use pcap.  If you want to
> look at what the server is doing, use "radiusd -X".  If you want to
> modify the servers behavior, use "unlang".
>
> > Does Freeradius proxy setup has option to specify which Radius server to
> > pickup based on some attribute value in Access Request?
>
>   Yes.  See raddb/proxy.conf
>
> > Our requirement
> > is to have multiple Radius Servers for certain groups. This group
> > information will be available in one of the VSA in Access Request.
>
>   That will work.
>
>   Alan DeKok.
>
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