Create a file to automatically check user list for users specified
up vote
1
down vote
favorite
I am realitively new to bash scripting, but have created this from my intense googling
I am attempting to create a bash script to run on Ubuntu that will check the user list and find any users that aren’t in a specified list.
I have written this so far:
Cat /etc/passwd | grep -o -P ‘.{0,40}:1[0-9][0-9][0-9].{0,0}.’ | cut -d: -f1
views passwd file, then finds only the lines with a user ID of 1000 or above, and everything before the userID of each line is piped into cut which removes all except the username of each user.
I then want to have the script check a file for usernames I specify (probably copied from a list) and compare each to the output of the above. Removing all usernames specified.
So for instance:
I have users John, Ben and Tom on my computer
If I put John and Ben in the file the script is accessing, it should output Tom since he is not specified
How would I go about doing this?
bash scripts
asked Dec 4 at 3:46
Tech_Person
83
add a comment |
up vote
1
down vote
favorite
I am realitively new to bash scripting, but have created this from my intense googling
I am attempting to create a bash script to run on Ubuntu that will check the user list and find any users that aren’t in a specified list.
I have written this so far:
Cat /etc/passwd | grep -o -P ‘.{0,40}:1[0-9][0-9][0-9].{0,0}.’ | cut -d: -f1
views passwd file, then finds only the lines with a user ID of 1000 or above, and everything before the userID of each line is piped into cut which removes all except the username of each user.
I then want to have the script check a file for usernames I specify (probably copied from a list) and compare each to the output of the above. Removing all usernames specified.
So for instance:
I have users John, Ben and Tom on my computer
If I put John and Ben in the file the script is accessing, it should output Tom since he is not specified
How would I go about doing this?
bash scripts
asked Dec 4 at 3:46
Tech_Person
83
1
Have you heard of awk. I'm not very good at it but it's a well powerful programing language. ` awk -F: '{print $1}' /etc/passwd` lists all user id
– rhubarbdog
Dec 4 at 3:55
I'll experiment more with it, but when I run that, it also includes system users such as root, I'm only looking for user accounts that you can login as, such as a personal user account
– Tech_Person
Dec 4 at 4:00
Yes but this is where awk becomes powerful because you can precede the print program with a pattern based on $2 the user number. I think you can putifstatements in your program
– rhubarbdog
Dec 4 at 4:06
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am realitively new to bash scripting, but have created this from my intense googling
I am attempting to create a bash script to run on Ubuntu that will check the user list and find any users that aren’t in a specified list.
I have written this so far:
Cat /etc/passwd | grep -o -P ‘.{0,40}:1[0-9][0-9][0-9].{0,0}.’ | cut -d: -f1
views passwd file, then finds only the lines with a user ID of 1000 or above, and everything before the userID of each line is piped into cut which removes all except the username of each user.
I then want to have the script check a file for usernames I specify (probably copied from a list) and compare each to the output of the above. Removing all usernames specified.
So for instance:
I have users John, Ben and Tom on my computer
If I put John and Ben in the file the script is accessing, it should output Tom since he is not specified
How would I go about doing this?
bash scripts
asked Dec 4 at 3:46
Tech_Person
83
I am realitively new to bash scripting, but have created this from my intense googling
I am attempting to create a bash script to run on Ubuntu that will check the user list and find any users that aren’t in a specified list.
I have written this so far:
Cat /etc/passwd | grep -o -P ‘.{0,40}:1[0-9][0-9][0-9].{0,0}.’ | cut -d: -f1
views passwd file, then finds only the lines with a user ID of 1000 or above, and everything before the userID of each line is piped into cut which removes all except the username of each user.
I then want to have the script check a file for usernames I specify (probably copied from a list) and compare each to the output of the above. Removing all usernames specified.
So for instance:
I have users John, Ben and Tom on my computer
If I put John and Ben in the file the script is accessing, it should output Tom since he is not specified
How would I go about doing this?
bash scripts
bash scripts
asked Dec 4 at 3:46
Tech_Person
83
asked Dec 4 at 3:46
Tech_Person
83
asked Dec 4 at 3:46
Tech_Person
83
asked Dec 4 at 3:46
Tech_Person
83
asked Dec 4 at 3:46
Tech_Person
83
83
1
Have you heard of awk. I'm not very good at it but it's a well powerful programing language. ` awk -F: '{print $1}' /etc/passwd` lists all user id
– rhubarbdog
Dec 4 at 3:55
I'll experiment more with it, but when I run that, it also includes system users such as root, I'm only looking for user accounts that you can login as, such as a personal user account
– Tech_Person
Dec 4 at 4:00
Yes but this is where awk becomes powerful because you can precede the print program with a pattern based on $2 the user number. I think you can putifstatements in your program
– rhubarbdog
Dec 4 at 4:06
add a comment |
1
Have you heard of awk. I'm not very good at it but it's a well powerful programing language. ` awk -F: '{print $1}' /etc/passwd` lists all user id
– rhubarbdog
Dec 4 at 3:55
I'll experiment more with it, but when I run that, it also includes system users such as root, I'm only looking for user accounts that you can login as, such as a personal user account
– Tech_Person
Dec 4 at 4:00
Yes but this is where awk becomes powerful because you can precede the print program with a pattern based on $2 the user number. I think you can putifstatements in your program
– rhubarbdog
Dec 4 at 4:06
1
1
Have you heard of awk. I'm not very good at it but it's a well powerful programing language. ` awk -F: '{print $1}' /etc/passwd` lists all user id
– rhubarbdog
Dec 4 at 3:55
Have you heard of awk. I'm not very good at it but it's a well powerful programing language. ` awk -F: '{print $1}' /etc/passwd` lists all user id
– rhubarbdog
Dec 4 at 3:55
I'll experiment more with it, but when I run that, it also includes system users such as root, I'm only looking for user accounts that you can login as, such as a personal user account
– Tech_Person
Dec 4 at 4:00
I'll experiment more with it, but when I run that, it also includes system users such as root, I'm only looking for user accounts that you can login as, such as a personal user account
– Tech_Person
Dec 4 at 4:00
Yes but this is where awk becomes powerful because you can precede the print program with a pattern based on $2 the user number. I think you can put
if statements in your program– rhubarbdog
Dec 4 at 4:06
Yes but this is where awk becomes powerful because you can precede the print program with a pattern based on $2 the user number. I think you can put
if statements in your program– rhubarbdog
Dec 4 at 4:06
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
This code should work for you.
for username in $( awk -F':' '$3 >= 1000 {print $1}' /etc/passwd )
do
if ! grep -q "$username" list.txt; then
echo "$username"
fi
done
Keep the usernames to check in list.txt as one user per line
answered Dec 4 at 4:05
Kishor V
1362
1
This works almost perfectly, when i run it, i get the "nobody" user in the output aswell, is there any easy way to get around this?
– Tech_Person
Dec 4 at 4:14
Simply change the awk condition to $3 >= 1000 && $1 != "nobody"
– Kishor V
Dec 4 at 10:36
ok, this entire thing is exactly what i was looking for thankyou!
– Tech_Person
Dec 5 at 0:17
add a comment |
up vote
1
down vote
grep allows specifying patterns from file. Thus you can use grep -f option for that purpose and -v for inverse pattern matching, i.e. print lines not matching the pattern. Thus,
printf "^%sn" 'John' 'Ben' > /tmp/list.txt
grep -v -f /tmp/list.txt /etc/passwd | cut -d ':' -f 1
rm l/tmp/ist.txt
We could also take into account UID greater than 999, since by default that refers to human users unless changed by your sysadmin.
awk -F ':' '$1 !~ /John|Ben|nobody/ && $3 > 999' /etc/passwd
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This code should work for you.
for username in $( awk -F':' '$3 >= 1000 {print $1}' /etc/passwd )
do
if ! grep -q "$username" list.txt; then
echo "$username"
fi
done
Keep the usernames to check in list.txt as one user per line
answered Dec 4 at 4:05
Kishor V
1362
1
This works almost perfectly, when i run it, i get the "nobody" user in the output aswell, is there any easy way to get around this?
– Tech_Person
Dec 4 at 4:14
Simply change the awk condition to $3 >= 1000 && $1 != "nobody"
– Kishor V
Dec 4 at 10:36
ok, this entire thing is exactly what i was looking for thankyou!
– Tech_Person
Dec 5 at 0:17
add a comment |
up vote
2
down vote
accepted
This code should work for you.
for username in $( awk -F':' '$3 >= 1000 {print $1}' /etc/passwd )
do
if ! grep -q "$username" list.txt; then
echo "$username"
fi
done
Keep the usernames to check in list.txt as one user per line
answered Dec 4 at 4:05
Kishor V
1362
1
This works almost perfectly, when i run it, i get the "nobody" user in the output aswell, is there any easy way to get around this?
– Tech_Person
Dec 4 at 4:14
Simply change the awk condition to $3 >= 1000 && $1 != "nobody"
– Kishor V
Dec 4 at 10:36
ok, this entire thing is exactly what i was looking for thankyou!
– Tech_Person
Dec 5 at 0:17
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This code should work for you.
for username in $( awk -F':' '$3 >= 1000 {print $1}' /etc/passwd )
do
if ! grep -q "$username" list.txt; then
echo "$username"
fi
done
Keep the usernames to check in list.txt as one user per line
answered Dec 4 at 4:05
Kishor V
1362
This code should work for you.
for username in $( awk -F':' '$3 >= 1000 {print $1}' /etc/passwd )
do
if ! grep -q "$username" list.txt; then
echo "$username"
fi
done
Keep the usernames to check in list.txt as one user per line
answered Dec 4 at 4:05
Kishor V
1362
edited Dec 4 at 4:10
answered Dec 4 at 4:05
Kishor V
1362
answered Dec 4 at 4:05
Kishor V
1362
answered Dec 4 at 4:05
Kishor V
1362
1362
1
This works almost perfectly, when i run it, i get the "nobody" user in the output aswell, is there any easy way to get around this?
– Tech_Person
Dec 4 at 4:14
Simply change the awk condition to $3 >= 1000 && $1 != "nobody"
– Kishor V
Dec 4 at 10:36
ok, this entire thing is exactly what i was looking for thankyou!
– Tech_Person
Dec 5 at 0:17
add a comment |
1
This works almost perfectly, when i run it, i get the "nobody" user in the output aswell, is there any easy way to get around this?
– Tech_Person
Dec 4 at 4:14
Simply change the awk condition to $3 >= 1000 && $1 != "nobody"
– Kishor V
Dec 4 at 10:36
ok, this entire thing is exactly what i was looking for thankyou!
– Tech_Person
Dec 5 at 0:17
1
1
This works almost perfectly, when i run it, i get the "nobody" user in the output aswell, is there any easy way to get around this?
– Tech_Person
Dec 4 at 4:14
This works almost perfectly, when i run it, i get the "nobody" user in the output aswell, is there any easy way to get around this?
– Tech_Person
Dec 4 at 4:14
Simply change the awk condition to $3 >= 1000 && $1 != "nobody"
– Kishor V
Dec 4 at 10:36
Simply change the awk condition to $3 >= 1000 && $1 != "nobody"
– Kishor V
Dec 4 at 10:36
ok, this entire thing is exactly what i was looking for thankyou!
– Tech_Person
Dec 5 at 0:17
ok, this entire thing is exactly what i was looking for thankyou!
– Tech_Person
Dec 5 at 0:17
add a comment |
up vote
1
down vote
grep allows specifying patterns from file. Thus you can use grep -f option for that purpose and -v for inverse pattern matching, i.e. print lines not matching the pattern. Thus,
printf "^%sn" 'John' 'Ben' > /tmp/list.txt
grep -v -f /tmp/list.txt /etc/passwd | cut -d ':' -f 1
rm l/tmp/ist.txt
We could also take into account UID greater than 999, since by default that refers to human users unless changed by your sysadmin.
awk -F ':' '$1 !~ /John|Ben|nobody/ && $3 > 999' /etc/passwd
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
add a comment |
up vote
1
down vote
grep allows specifying patterns from file. Thus you can use grep -f option for that purpose and -v for inverse pattern matching, i.e. print lines not matching the pattern. Thus,
printf "^%sn" 'John' 'Ben' > /tmp/list.txt
grep -v -f /tmp/list.txt /etc/passwd | cut -d ':' -f 1
rm l/tmp/ist.txt
We could also take into account UID greater than 999, since by default that refers to human users unless changed by your sysadmin.
awk -F ':' '$1 !~ /John|Ben|nobody/ && $3 > 999' /etc/passwd
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
add a comment |
up vote
1
down vote
up vote
1
down vote
grep allows specifying patterns from file. Thus you can use grep -f option for that purpose and -v for inverse pattern matching, i.e. print lines not matching the pattern. Thus,
printf "^%sn" 'John' 'Ben' > /tmp/list.txt
grep -v -f /tmp/list.txt /etc/passwd | cut -d ':' -f 1
rm l/tmp/ist.txt
We could also take into account UID greater than 999, since by default that refers to human users unless changed by your sysadmin.
awk -F ':' '$1 !~ /John|Ben|nobody/ && $3 > 999' /etc/passwd
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
grep allows specifying patterns from file. Thus you can use grep -f option for that purpose and -v for inverse pattern matching, i.e. print lines not matching the pattern. Thus,
printf "^%sn" 'John' 'Ben' > /tmp/list.txt
grep -v -f /tmp/list.txt /etc/passwd | cut -d ':' -f 1
rm l/tmp/ist.txt
We could also take into account UID greater than 999, since by default that refers to human users unless changed by your sysadmin.
awk -F ':' '$1 !~ /John|Ben|nobody/ && $3 > 999' /etc/passwd
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
edited Dec 4 at 4:24
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
answered Dec 4 at 4:04
Sergiy Kolodyazhnyy
68.9k9143303
68.9k9143303
add a comment |
add a comment |
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1
Have you heard of awk. I'm not very good at it but it's a well powerful programing language. ` awk -F: '{print $1}' /etc/passwd` lists all user id
– rhubarbdog
Dec 4 at 3:55
I'll experiment more with it, but when I run that, it also includes system users such as root, I'm only looking for user accounts that you can login as, such as a personal user account
– Tech_Person
Dec 4 at 4:00
Yes but this is where awk becomes powerful because you can precede the print program with a pattern based on $2 the user number. I think you can put
ifstatements in your program– rhubarbdog
Dec 4 at 4:06