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I have a simple one-column list of users. It comes from the lastlog command. I need to be able to tell if these accounts are disabled and I do not have root access. I can only view with more and execute the lastlog command. I need to see if users that havent logged in within the last 90 days are still active.

So far I execute

sudo lastlog -b 90

Which gives me the users that logged in..in the last 90 days. Now I just run it through awk printing the first field and that is my simple user list. Now Id like to basically see if in /etc/passwd, their shell is /bin/bash and if in /etc/shadow, there is a ! in their line. How can I do this. I tried for loops and while loops and cant seem to get it right? Any help would be appreciated.

  • can you clarify what output you're looking to get exactly? That would be helpful when writing a script. Also, can you even read /etc/shadow without root access? – AtomHeartFather Sep 4 '14 at 9:53
  • yes i have access to the more command as root and also grep as root. I think I found what I was looking for using grep -f users /etc/passwd (actually /etc/passwd had to be formatted to grep only /bin/bash out of it since I only want active users from that so I redirected it to another file). So in the end it was grep -f users bash_users. Does that sound right? – unixpipe Sep 4 '14 at 15:41
  • see the answer for my approach. hope that helps. – AtomHeartFather Sep 6 '14 at 13:24
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I am not sure if I understand you properly. Is that what you are after?

lastlog -b 90 | awk '{if (NR > 1) print $1 }' | grep -f - /etc/passwd | grep "/bin/bash$"

Here's the breakdown:

  • awk '{if (NR > 1) print $1 }' - print the first column skipping the first line (lastlog's header)
  • grep -f - /etc/passwd - use the standard input as a file to obtain patterns from and search for patterns in /etc/passwd
  • grep "/bin/bash$" - grep for /bin/bash , $ means EOL.

A few notes:

  • lastlog -b 90 - that gives you records older than 90 days. If you want users that logged in within the last 90 days use lastlog -t 90.
  • I only went with (NR > 1) solution because we're using awk anyway. More succinct way of skipping the first line of a file would be sed 1d.

EDIT

You can use a similar idiom if you also want to filter out the records with ! from /etc/shadow. So the complete command would be:

lastlog -b 90 | awk '{if (NR > 1) print $1 }' | grep -f - /etc/passwd | grep "/bin/bash$" | cut -d ':' -f 1 | grep -f - /etc/shadow | grep -v ":\!:"
  • cut -d':' -f 1 splits the line using : as a delimiter and returns the first field
  • grep -v means invert match (select non-matching lines).
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The exclamation point as you have noticed needs to be 'escaped out' -- try entering searches for it with a backslash first: \! instead of ! so it doesn't execute it like a command instead of being treated as text. You could also in many cases quote it with single quotes (not double) like so: '!' -- depending on the command you choose one of those should do it for you.

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