NetNews Usenet Archive 1992 #19/NN_1992_19.iso/spool/comp/unix/question/10366

Internet Message Format | 1992-08-25 | 2.2 KB

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From: [e-mail] (Larry Wall)
Newsgroups: comp.unix.questions
Subject: Re: I want a program that prints out base gid's of any user
Message-ID: <[e-mail]>
Date: 25 Aug 92 19:30:28 GMT
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In article <[e-mail]> [e-mail] (Benjamin Z. Goldsteen) writes:
:
:    I have been trying to write a program that prints out the
: base gid of a user (given on the command line).  I have tried
: using awk and perl, but I can not get anything decent.
: Can show me how to do this.  One thing I have tried
: [ gid username ]
: #! /usr/bin/awk -f
: BEGIN { FS=":"; FILENAME="/etc/passwd" }
: ARGV[1] == $1 {print $4}
:
: I am not sure if 1 if right for ARGV, but whatever...anyway, it
: won't read from /etc/passwd - it only reads from stdin.  I can;t
: make it do otherwise.

You can't make a sow's purse out of a silk ear.  Or something like that... :-)

: I don't really understand perl...

You don't have to Understand Perl(TM).  You only have to know enough
of it to get the job done.

The way to do this in Perl is to use the standard system routine for it:

    #!/usr/bin/perl -l
    print((getpwnam($ARGV[0]))[3], "\n");

or, more readably,

    #!/usr/bin/perl
    ($login,$passwd,$uid,$gid) = getpwnam($ARGV[0]);
    print "$gid\n";

This will even work on YP systems, er, that is, NIS system.

Now, if you *wanted* to parse /etc/passwd yourself in Perl, you'd probably
do it something like this:

    #!/usr/bin/perl
    open(PASSWD, "/etc/passwd") || die "Can't open /etc/passwd: $!\n";
    while (<PASSWD)) {
    ($login, $passwd, $uid, $gid, $gcos, $home, $shell) = split(/:/);
    print "$gid\n" if $login eq $ARGV[0];
    }

But ordinarily you should use the provided system calls.  It's probably
a mistake in the original design of Unix that system calls and shells were
kept so far apart from each other...

Have you ever read The Origin of Perl in the Breakdown of the Bicameral Unix?

:-)

Larry Wall
[e-mail]