dovecot - Aug 2013 - Getting default uid/gid of users via socket for virtual user support for sendmail

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Hi,

I'm trying to make my life easier with sendmail and virtual users. For 
that I'd like to pass sendmail the home directory and uid/gid for each 
user. The user data is storred in a LDAP,& I retrieve it from the 
auth-userdb socket fine.

If I add uid / gid to each user, I get them via that socket as well, but 
is there a way I get the default uid/gid? I mean, without calling an 
external program, such as doveadm. There is an "initialize" phase, so
the
default information is retrieved just once per lifetime of the demon, but 
each time the sendmail binary is invoked.

Kind regards,

- -- 
Steffen Kaiser
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On 1.8.2013, at 17.52, Steffen Kaiser <skdovecot at smail.inf.fh-brs.de>
wrote:
> I'm trying to make my life easier with sendmail and virtual users. For
that I'd like to pass sendmail the home directory and uid/gid for each user.
The user data is storred in a LDAP,& I retrieve it from the auth-userdb
socket fine.
> 
> If I add uid / gid to each user, I get them via that socket as well, but is
there a way I get the default uid/gid? I mean, without calling an external
program, such as doveadm. There is an "initialize" phase, so the
default information is retrieved just once per lifetime of the demon, but each
time the sendmail binary is invoked.
You can't get the mail_uid/gid/home, because they really don't belong to
auth process.. But what you could do with v2.2 is:

userdb {
  driver = ldap
  args = ..
  default_fields = uid=1234 gid=1234 home=/home/user/%u
}

But anyway, what is sendmail going to be doing with these fields? For Postfix
I've been trying to remove its need to know anything at all about users
(only domains and aliases).