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#1
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Number of users in passwd
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? Last edited by vbe; 03-10-2010 at 05:26 AM.. Reason: code tags |
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#2
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Try this :- Code:
awk -F':' '{ print $3}' /etc/passwd | sort | uniq -cLast edited by pludi; 03-10-2010 at 05:34 AM.. Reason: code tags, please... |
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#3
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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] |
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#4
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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 Last edited by Nila; 03-10-2010 at 05:44 AM.. |
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#5
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editing my last post
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.
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#6
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4Th field is the group id Code:
awk -F':' '{ print $4}' /etc/passwd |sort -nr | uniq -c |
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