CODEWITHSUNDEEP
linuxbashshellloops
San
San

Reputation: 233

How to print al list of user one by one with complete information in linux using shell

I have started writing a small piece of code to print all the list of users available in the linux box. But I want to pass one by one user into my command to display each user details together.

to list all users

root@bt# getent passwd | grep /home/ | cut -d ':' -f 1

root
san
postgres

Now I want to pass one by user in to the below command to display each user details together.

root@bt# chage -l ${user1} ; chage -l ${user2} etcc.

should I need to user for loop or while loop here? can any one help me in suggesting how to write the same?

Upvotes: 3

Views: 3069

Answers (2)

Sicco
Sicco

Reputation: 6271

I would use xargs, which runs a command on each output item of the previous pipe:

getent passwd | grep /home/ | cut -d ':' -f 1 | sudo xargs -I % sh -c '{ echo "User: %"; chage -l %; echo;}'
  • sudo is used to get information about all users, if you don't have access to this information then you can remove sudo
  • -I % is used to specify that % is a placeholder for the input item (in your case a user)
  • sh -c '{ command1; command2; ...;}' is the command executed by xargs on every % item; in turn, the command sh -c allows multiple shell commands to be executed
  • '{ echo "User: %"; chage -l %; echo;}' echoes the current user in %, then runs chage -l on this user and finished with a final empty echo to format the ouput

Upvotes: 1

choroba
choroba

Reputation: 241768

You can use the while loop:

getent passwd | grep /home/ | cut -d ':' -f 1 | \
  while read user ; do
    chage -l "$user"
  done

or the for loop:

for user in $(getent passwd | grep /home/ | cut -d ':' -f 1) ; do
    chage -l "$user"
done

or xargs:

getent passwd | grep /home/ | cut -d ':' -f 1 | \
  xargs -n1 chage -l

Upvotes: 1

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