Real users (users with a valid login shell)











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How can I print the real users(in a script if possible)? I know that the real users have a valid login shell and i know that the valid login shell can be found in /etc/shells. But I don't know how to take the users (from /etc/passwd) that have the respective shell.










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  • Is this an isolated system or do you need to consider users in LDAP, NIS, Active Directory, etc.?
    – roaima
    Nov 22 at 18:09










  • It's an isolated system
    – A.C.JKH
    Nov 22 at 18:13










  • grep -F "$(grep -v '^#' /etc/shells)" /etc/passwd
    – mosvy
    Nov 22 at 19:44

















up vote
1
down vote

favorite












How can I print the real users(in a script if possible)? I know that the real users have a valid login shell and i know that the valid login shell can be found in /etc/shells. But I don't know how to take the users (from /etc/passwd) that have the respective shell.










share|improve this question







New contributor




A.C.JKH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Is this an isolated system or do you need to consider users in LDAP, NIS, Active Directory, etc.?
    – roaima
    Nov 22 at 18:09










  • It's an isolated system
    – A.C.JKH
    Nov 22 at 18:13










  • grep -F "$(grep -v '^#' /etc/shells)" /etc/passwd
    – mosvy
    Nov 22 at 19:44















up vote
1
down vote

favorite









up vote
1
down vote

favorite











How can I print the real users(in a script if possible)? I know that the real users have a valid login shell and i know that the valid login shell can be found in /etc/shells. But I don't know how to take the users (from /etc/passwd) that have the respective shell.










share|improve this question







New contributor




A.C.JKH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











How can I print the real users(in a script if possible)? I know that the real users have a valid login shell and i know that the valid login shell can be found in /etc/shells. But I don't know how to take the users (from /etc/passwd) that have the respective shell.







shell-script






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share|improve this question







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A.C.JKH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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asked Nov 22 at 16:43









A.C.JKH

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New contributor





A.C.JKH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






A.C.JKH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Is this an isolated system or do you need to consider users in LDAP, NIS, Active Directory, etc.?
    – roaima
    Nov 22 at 18:09










  • It's an isolated system
    – A.C.JKH
    Nov 22 at 18:13










  • grep -F "$(grep -v '^#' /etc/shells)" /etc/passwd
    – mosvy
    Nov 22 at 19:44




















  • Is this an isolated system or do you need to consider users in LDAP, NIS, Active Directory, etc.?
    – roaima
    Nov 22 at 18:09










  • It's an isolated system
    – A.C.JKH
    Nov 22 at 18:13










  • grep -F "$(grep -v '^#' /etc/shells)" /etc/passwd
    – mosvy
    Nov 22 at 19:44


















Is this an isolated system or do you need to consider users in LDAP, NIS, Active Directory, etc.?
– roaima
Nov 22 at 18:09




Is this an isolated system or do you need to consider users in LDAP, NIS, Active Directory, etc.?
– roaima
Nov 22 at 18:09












It's an isolated system
– A.C.JKH
Nov 22 at 18:13




It's an isolated system
– A.C.JKH
Nov 22 at 18:13












grep -F "$(grep -v '^#' /etc/shells)" /etc/passwd
– mosvy
Nov 22 at 19:44






grep -F "$(grep -v '^#' /etc/shells)" /etc/passwd
– mosvy
Nov 22 at 19:44












1 Answer
1






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up vote
0
down vote













awk -F: 'NR == FNR {shells[$0]; next} $NF in shells' /etc/{shells,passwd}


or, grep but it's less precise



grep -Ff /etc/shells /etc/passwd





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    1 Answer
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    1 Answer
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    up vote
    0
    down vote













    awk -F: 'NR == FNR {shells[$0]; next} $NF in shells' /etc/{shells,passwd}


    or, grep but it's less precise



    grep -Ff /etc/shells /etc/passwd





    share|improve this answer

























      up vote
      0
      down vote













      awk -F: 'NR == FNR {shells[$0]; next} $NF in shells' /etc/{shells,passwd}


      or, grep but it's less precise



      grep -Ff /etc/shells /etc/passwd





      share|improve this answer























        up vote
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        down vote










        up vote
        0
        down vote









        awk -F: 'NR == FNR {shells[$0]; next} $NF in shells' /etc/{shells,passwd}


        or, grep but it's less precise



        grep -Ff /etc/shells /etc/passwd





        share|improve this answer












        awk -F: 'NR == FNR {shells[$0]; next} $NF in shells' /etc/{shells,passwd}


        or, grep but it's less precise



        grep -Ff /etc/shells /etc/passwd






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        share|improve this answer



        share|improve this answer










        answered Nov 22 at 19:07









        glenn jackman

        49.4k469106




        49.4k469106






















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