* 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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