on Thu, Aug 30, 2001 at 11:09:07PM -0500, ktb (x.y.f@home.com) wrote:
> On Thu, Aug 30, 2001 at 08:45:29PM -0700, Vineet Kumar wrote:
> >
> > * Brian Schramm (brian.schramm@ncmail.net) [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.
Not quite the same thing:
$ awk -F : '/karsten/ {print $5}' /etc/passwd
Karsten M. Self,,,
$ awk -F : '{print $5}' /etc/passwd | sed -e "s/^\([^,]*\).*$/\1/"
Karsten M. Self
In the original pattern:
sed -e "s/^\([^,]*\).*$/\1/"
We have:
-e: expression to evaluate.
s: create a substitution using the following pattern.
/ start of expression
^ beginning of line (actually, beginning of fifth field
\( start a substitution
[^,]* match zero or more instances of any character other than ','
\) end substitution
.*$ match to end of line
/ end of expression
\1 replace with contents of first substitution (the \([^,]*\)
pattern)
/ end expression
sed is for people who think Perl's too easy to understand.
--
Karsten M. Self <kmself@ix.netcom.com> http://kmself.home.netcom.com/
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