Red Hat

Hi everyone. I have a quesiton about awk. I am running the following line of code under bash.

Code:

awk -F: '{print $7}' /etc/passwd

When I run that it returns the list of each shell for each account on the machine. If there is a blank spot, however, it denotes that there is no shell. I need to have a way to fill in any blank spot with:

Code:

/sbin/nologin

I know there's a way to do this, just haven't been able to find out anywhere. Thanks in advance!

EDIT:

Let me be more clear. I need /etc/passwd to actually be modified. I don't want to just output /sbin/login in place of blanks, I want to change those blanks in the file to /sbin/login.

Code:

awk -F: '{print ($7)?$7:"/sbin/nologin"}' /etc/passwd

awk allows you to alter the fields themselves, which is handy. For if, you can use if. For the field, you use the field.

Code:

awk -F: '{ if(!$7) $7="/sbin/nologin"; print $7; }' < /etc/passwd

There's probably a way to do all that in three keystrokes but I prefer my awk code to be readable...

Something like:

Code:

awk -F: -vOFS=: '{$7=($7)?$7:"/sbin/nologin"}1' /etc/passwd

?

These are great everyone. Is there a way to make it only verbose?

---------- Post updated at 11:36 AM ---------- Previous update was at 11:15 AM ----------

This one works best, is there a way to only have it echo lines that were changed?

depending on your definition of 'changed', but I think you mean this....
many ways to skin that 'cat':

Code:

awk -F: '!$7{print "/sbin/nologin"}' /etc/passwd

Well I mean after it changes the blank line to /sbin/nologin...I tried your code

Code:

awk -F: '!$7{print "/sbin/nologin"}' /etc/passwd

and it echoed "/sbin/nologin" but it didn't actually change /etc/passwd at all.