Bash: inserting output from `id` into print statement

Question

I'm trying to achieve the following: List my Linux users, that have set bin/bash in /etc/passwd AND append each user's info from id [username] into the print statement, something in the lines of:

cat /etc/passwd | grep bin/bash | awk -F\: '{print $1";"$(id $1)}'

(which obviously does not work ;) )

I've got the feeling that I'm pretty close; what am I missing?

Output is supposed to be (see DavDav's comment):

User1;uid=1000(User1) gid=1000(User1) groups=1000(User1) 
User2;uid=1001(User2) gid=1001(User2) groups=1001(User2) 
User3;uid=1002(User3) gid=1002(User3) groups=1002(User3) 

Answer 1

How about

 </etc/passwd grep bin/bash | cut -d: -f1 | xargs -n1 id

Answer 2

Try this,

 awk -F ':' '/bin\/bash/ {printf ($1";"); system("id "$1)}' /etc/passwd

• system will executes the command and returns the value to awk

Answer 3

Here's another option:

for i in $(grep /bin/bash /etc/passwd | awk -F: '{ print $1 }'); do
    echo "${i};$(id ${i})"
done

The grep ... awk ... will generate the list of user names corresponding to lines in /etc/passwd that contain /bin/bash. On each iteration of the for loop, i will be one of those usernames. The echo then prints the username followed by a semi-colon, followed by the output of id for that username.