Shell Programming and Scripting

This command prints out username/users in /etc/passwd:

Code:

cut -d ':' -f '1,5' /etc/passwd | sort

I wonder if I also, after above commands output, can get an output that lists number of users in the group? I need to use uniq to get rid of duplicates.
I´ve tried this, but cant get it right, can anyone help please?

Try this :-

Code:

awk -F':' '{ print $3}' /etc/passwd   | sort  | uniq -c

Ok, thank you, but I dont get the output, first 13 and then 1 and -2(?) up till 99.
Please help me to understand
Output:

[ 1 -2
1 0
1 1
1 26
1 27
1 4
1 54
1 55
1 56
1 58
1 59
1 60
1 65
1 67
1 70
1 71
1 72
1 73
1 74
1 75
1 76
1 77
1 78
1 79
1 82
1 83
1 84
1 85
1 86
1 87
1 88
1 89
1 91
1 92
1 93
1 94
1 95
1 96
1 97
1 99]

I think Jairaj ,it should be $4 instead of $3.We can achieve it easily by cut command as follows.
This will give output as number of users in each group. First field is a no of users and second field is s group_id.

Code:

cut -d ':' -f 4 /etc/passwd | sort | uniq -c

Dont know why but the post was missing 13 first in the output I wrote, adn dont mind the [] just for showing the oouput start and end.

4Th field is the group id

Code:

awk -F':' '{ print $4}' /etc/passwd |sort -nr | uniq -c