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Old 09-03-2007, 01:49 AM   #1
narainhere
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Registered: Jul 2007
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Post Finger command to list all the user names


Hi all,
I am having a perl script which should mail a report.I need to figure out the Real names for the user-id's .The problem is I need to retrieve the real-names for all the user-id's on the server.Is there a way to do this?
 
Old 09-03-2007, 02:38 AM   #2
adam_blackice
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hello

i think the proper way is to go to the /etc/passwd file and then extract the fields you want and then pass that to variable and use it to get the list if the real name

here is the code that u will use

Code:
#Author: Black-Ice
#Date: 20/6/2007
#!/usr/bin/perl -w
use strict;

my $file = "/etc/passwd";

open (han1, "$file") || die  "error opening file: $!";


my @newrecords = ;

foreach (@newrecords) {

my @columns = split /:/;

my $username = $columns[0];
my $x = $columns[1];
my $userid = $columns[2];
my $groupid = $columns[3];
my $realname = $columns[4];
my $homedir = $columns[5];
my $shellpath = $columns[6];

#then u can echo the field that you want to disply
either u can do it throw a shell script and that will be good too


Code:
cat /etc/passwd | cut -d : -f "the field of the real name in the /etc/passwd file "
 
1 members found this post helpful.
Old 09-03-2007, 04:05 AM   #3
ghostdog74
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Quote:
Originally Posted by adam_blackice View Post
Code:
cat /etc/passwd | cut -d : -f "the field of the real name in the /etc/passwd file "
no need for cat
Code:
cut -d":" -f5 /etc/passwd
 
Old 09-03-2007, 04:56 AM   #4
adam_blackice
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you are right but i used to do it like that
 
  


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