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Shell Programming and Scripting

find information about logins

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# 1  
Old 03-06-2010
find information about logins
Hi, all
I want to make a bash script that print all users from a system using last command.
I want to print the number of user's login in the format (descending order):

Code:
5   user1    address1
4   user2    address2


I am trying the command
Code:
last | awk '{print $1 " " $3}' | sort | uniq

but it is not run as I want


Could you help me ??
Last edited by Scott; 03-06-2010 at 02:42 PM.. Reason: Code tags
# 2  
Old 03-06-2010
MySQL
use this command

Code:
last | sed -r 's/[ \t]+/ /g'  | cut -d ' ' -f 1,3 | sort | uniq

# 3  
Old 03-06-2010
Awk version:

Code:
 last | awk '/./ {$0=++c " " $1 "  " $2}1;' | tac

# 4  
Old 03-06-2010
Thanks a lot for your quick answer.

The problem I have is that I have to count the number of logins and sort by number then.
# 5  
Old 03-06-2010
Code:
last | tr -s " " | cut -d ' ' -f 1,3 | sort | uniq

# 6  
Old 03-06-2010
Thanks for your help

I check the sort and uniq parameters and I found it.
Last edited by peter20; 03-06-2010 at 10:47 AM..
# 7  
Old 03-06-2010
This awk command might work for you:

Code:
last -d | awk '/./{print $1, $3}' | sort -n -k1 | uniq -w1 -c
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