unix home directories without entries in /etc/passwd - shell

Question

I am able to get both listings ( /etc/passwd and /home ) but how to script something like read line of /etc/passwd, parse home directory, then look for that in /home . If it doesn't exist, throw an error, if it does exist, move along.
/etc/passwd home dir listing for users
cut -d":" -f6 /etc/passwd | grep home | sort
user listing from /home
ls -1 /home | (while read line; do echo "/home/"$line; done)
Maybe right out output from first command to a file, then read each line into a find command and...or, test with
if [ -d "$DIRECTORY" ]; then
echo "directory found for user that doesn't exist"
fi
Now how to put it all together...
EDIT: isedev had exactly what I needed. I may have mis-worded my original message...we have been cleaning up users, but not cleaning up their /home directory. So I want to know what /home directories still exist that don't have /etc/passwd entries.
this is what worked to a T
for name in /home/*; do
if [ -d "$name" ]; then
cut -d':' -f6 /etc/passwd | egrep -q "^$name$"
if [ $? -ne 0 ]; then
echo "directory $name does not correspond to a valid user"
fi
fi
done
from now on, we will be running
userdel -r login

Answer 1

This will report all home directories from /etc/passwd that should be in /home but aren't:
cut -d":" -f6 /etc/passwd | grep home | sort |
while read dir; do [ -e "$dir" ] || echo Missing $dir; done
And this one reports all that don't exist:
cut -d":" -f6 /etc/passwd | while read dir; do
[ -e "$dir" ] || echo Missing $dir
done

Answer 2

as 1st approximation:
perl -F: -lane 'next if m/^#/;print "$F[5] for user $F[0] missing\n" unless(-d $F[5])' /etc/passwd
if you want find the differences between the /etc/passwd and the /home
comm <(find /home -type d -maxdepth 1 -mindepth 1 -print|sort) <(grep -v '^#' /etc/passwd | cut -d: -f6| grep '/home' | sort)
in an narrow form
comm <(
find /home -type d -maxdepth 1 -mindepth 1 -print |sort
) <(
grep -v '^#' /etc/passwd |cut -d: -f6 |grep /home |sort
)
if you will use
comm ... (without args as above) will show 3 colums 1.) only in /home 2.)only in /etc/passwd 3.) common
comm -23 .... - will show directories what are only in the /home (and not in the /etc/passwd)
comm -13 .... - will show dirs what are only in the /etc/passwd and not in the /home
comm -12 .... - will show correct directories (exists in the /etc/passwd and the /home too)
I'm not sure with the -{max|min}depth on the AIX..

Answer 3

So, assuming you want to know if there are directories under /home which do not correspond to existing users:
for name in /home/*; do
if [ -d "$name" ]; then
cut -d':' -f6 /etc/passwd | egrep -q "^$name$"
if [ $? -ne 0 ]; then
echo "directory $name does not correspond to a valid user"
fi
fi
done
Then again, this assumes you are not using a name service such as LDAP or NIS, in which case, change the line starting with cut to:
getent passwd | cut -d':' -f6 | egrep -q "^$name$"