* ktb ([EMAIL PROTECTED]) [010830 21:14]:
> On Thu, Aug 30, 2001 at 08:45:29PM -0700, Vineet Kumar wrote:
> >
> > * Brian Schramm ([EMAIL PROTECTED]) [010830 19:41]:
> > > Is there a way that I can take a passwd file and compare the full name
> > > data
> > > in it to the email ldap server and give a a list of what it finds and
> > > what it
> > > misses? I am doing this manually but with the number of users that there
> > > are
> > > involved it is going to be really time consuming.
> >
> > I don't really know what I'm talking about, but this should probably
> > help you get started:
> >
> > awk -F : '{print $5}' /etc/passwd | sed -e "s/^\([^,]*\).*$/\1/"
> >
> > That will give you a list of just the full names. Pipe that into
> > something else that will look each one up in the directory service.
> >
> > Not a complete answer, but it's a start...
>
> BTW what does [ sed -e "s/^\([^,]*\).*$/\1/" ] accomplish? I'm just
> grooving on one liners lately and am curious. It seems like -
> awk -F : '{print $5}' /etc/passwd is all you need to spit out the full
> names.That will print out the full GECOS field. The sed part strips out everything but what's before the first comma (leaving just the name). -- Vineet http://www.anti-dmca.org Unauthorized use of this .sig may constitute violation of US law. Qba\'g gernq ba zr\! |tr 'a-zA-Z' 'n-za-mN-ZA-M'
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