Re: SASL2 auth hack percent in login

This is a discussion on Re: SASL2 auth hack percent in login within the mailing.postfix.users forums, part of the Mail Servers and Related category; Dear Patrick, I think not! The information is passed from postfix, I don't question how can I auth and ...


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Old 05-24-2005
Junix
 
Posts: n/a
Default Re: SASL2 auth hack percent in login

Dear Patrick,

I think not!

The information is passed from postfix, I don't question how can I auth
and my question is how the postfix can past the login information right.

Thank you.

[]'s=20


Junix

Em Ter, 2005-05-24 =C3=A0s 19:31 +0200, Patrick Ben Koetter escreveu:
> * Junix <junixbr@gmail.com>:
> > How could I do to the postfix "hack percent" the login in sasl
> > authentication, where some requests arrive with character "%" instead=

of
> > "@", so the sasl don't validate the attempt.

>=20
> This is a question for the cyrus-sasl mailing list. Please contact them=

..
>=20
> p@rick
>=20
>=20
>=20
>=20
> >=20
> > The server receive requests with "@" and "%" and I need work both.
> >=20
> > I tried use the allow_percent_hack, but, doesn't work.
> >=20
> > I done the alternative resolution and work fine, but, This is not
> > correct way.
> >=20
> > Look my resolution in smtpd.conf:
> > pwcheck_method: auxprop
> > auxprop_plugin: sql
> >=20
> > mech_list: plain login
> >=20
> > sql_engine: pgsql
> > sql_user: postfix
> > sql_passwd: xxxxxx
> > sql_hostnames: 200.200.200.200
> > sql_database: postfix_db
> > sql_select: SELECT password FROM mailbox_table WHERE LOWER(username) =

=3D
> > CASE POSITION('%%' IN '%u@%r') WHEN 0 THEN '%u@%r' ELSE
> > replace(replace('%u@%r@domain.com.br', '@domain.com.br', ''), '%%', '=

@')
> > END
> >=20
> >=20
> > Sorry my english.
> >=20
> > Thanks for all.
> >=20
> > Regards,
> >=20
> > []'s
> >=20
> > Junix
> >=20
> >=20

>=20


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