I am facing a problem with my script.
I have to found the primary group of users .
So first I selected all the groups and users register from a specific user : ONE
Then I am making a file with all groups attached to the user : ONE
Then I am making a file with all users from each group of that user : ONE Then for each user I have to found his primary group
Ex : Groups of USER ONE
GRP1 ::111 :ONE,TWO,THRE
GRP2 ::2848 :SIX ;ONE,SEVEN
and so one
Here is the script that i have writen :
My problem is the number of users listed is by far beyond of the userlist.
I've got 151 users but at the end the list contain 298 entries .... I can't find what is wrong.
When users login, they are directed to menu (aix script). The menu enables the user to choose an environment to work in. Each environment has a different group id. When a user chooses a menu option, I want to change his primary group to that specific environment's group id. Is this at all possible... (3 Replies)
Is there a command or better combination of cmds that will give me the list of Unix users in a particular Unix group whether their primary group is that group in question (information stored in /etc/passwd) or they are in a secondary group (information stored in /etc/group).
So far all I got... (5 Replies)
HI
I need to know what is the primary group name of a particular user.
How to do this ?
Maybe with groups cmd ? (first group name in line, is the primary group)
thx for help. (2 Replies)
Hi Experts,
I am new to scripting. We have around 400 Linux servers in our environment. I want to add a new user to a perticular group on all the servers using SSH.
Requirements:
1) Need to take the server names from a text file.
2) Login into each server and check whether perticular... (1 Reply)
Hi,
I have a file with usernames, and the comment section, e.g :
Data removed by request of sanchitadutta91, 20 May 2020
I need to add these users into a server. Is it possible to use a script to create the users, together with the comment ?
From the commandline to add one user, the... (2 Replies)