How to find sudo rights in shell script
I am struggling to work out how to achieve this, as it does not support -a parameter on CentOS/RHEL.
My question is : I will get all fields via passwd -S <user> on each username and parse the output per your original script. How can I modify script ?
#!/bin/bash
passwd -Sa | while read LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS; do
# Is this account disabled?
[[ $PASS_TYPE == LK ]] && DIS="disabled" || DIS=""
# Can this account sudo?
#sudo -l -U "$LOGIN" | grep -q not allowed && CAN_SUDO="" || CAN_SUDO="sudo"
sudo -l -U "$LOGIN" | grep -q "(ALL) NOPASSWD: ALL|(ALL) ALL" && CAN_SUDO="sudo" || CAN_SUDO=""
# Grab misc. info from passwd entry
IFS=: read _ PASS USERID GROUPID FULLNAME HOMEDIR SH < <(getent passwd "$LOGIN")
# Grab login time of the latest entry in the lastlog output
LATEST="$(last ${LOGIN} | head -1 | cut -c40-)"
echo "${LOGIN}|${FULLNAME}|${USERID}|${LAST_CHANGE}|${MIN_AGE}|${MAX_AGE}|${WARN_DAYS}|${INACTIVE_DAYS}|${DIS}|${CAN_SUDO}|${LATEST}"
done
linux shell-script
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
asked Dec 17 at 6:33
Cell-o
2741414
add a comment |
I am struggling to work out how to achieve this, as it does not support -a parameter on CentOS/RHEL.
My question is : I will get all fields via passwd -S <user> on each username and parse the output per your original script. How can I modify script ?
#!/bin/bash
passwd -Sa | while read LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS; do
# Is this account disabled?
[[ $PASS_TYPE == LK ]] && DIS="disabled" || DIS=""
# Can this account sudo?
#sudo -l -U "$LOGIN" | grep -q not allowed && CAN_SUDO="" || CAN_SUDO="sudo"
sudo -l -U "$LOGIN" | grep -q "(ALL) NOPASSWD: ALL|(ALL) ALL" && CAN_SUDO="sudo" || CAN_SUDO=""
# Grab misc. info from passwd entry
IFS=: read _ PASS USERID GROUPID FULLNAME HOMEDIR SH < <(getent passwd "$LOGIN")
# Grab login time of the latest entry in the lastlog output
LATEST="$(last ${LOGIN} | head -1 | cut -c40-)"
echo "${LOGIN}|${FULLNAME}|${USERID}|${LAST_CHANGE}|${MIN_AGE}|${MAX_AGE}|${WARN_DAYS}|${INACTIVE_DAYS}|${DIS}|${CAN_SUDO}|${LATEST}"
done
linux shell-script
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
asked Dec 17 at 6:33
Cell-o
2741414
add a comment |
I am struggling to work out how to achieve this, as it does not support -a parameter on CentOS/RHEL.
My question is : I will get all fields via passwd -S <user> on each username and parse the output per your original script. How can I modify script ?
#!/bin/bash
passwd -Sa | while read LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS; do
# Is this account disabled?
[[ $PASS_TYPE == LK ]] && DIS="disabled" || DIS=""
# Can this account sudo?
#sudo -l -U "$LOGIN" | grep -q not allowed && CAN_SUDO="" || CAN_SUDO="sudo"
sudo -l -U "$LOGIN" | grep -q "(ALL) NOPASSWD: ALL|(ALL) ALL" && CAN_SUDO="sudo" || CAN_SUDO=""
# Grab misc. info from passwd entry
IFS=: read _ PASS USERID GROUPID FULLNAME HOMEDIR SH < <(getent passwd "$LOGIN")
# Grab login time of the latest entry in the lastlog output
LATEST="$(last ${LOGIN} | head -1 | cut -c40-)"
echo "${LOGIN}|${FULLNAME}|${USERID}|${LAST_CHANGE}|${MIN_AGE}|${MAX_AGE}|${WARN_DAYS}|${INACTIVE_DAYS}|${DIS}|${CAN_SUDO}|${LATEST}"
done
linux shell-script
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
asked Dec 17 at 6:33
Cell-o
2741414
I am struggling to work out how to achieve this, as it does not support -a parameter on CentOS/RHEL.
My question is : I will get all fields via passwd -S <user> on each username and parse the output per your original script. How can I modify script ?
#!/bin/bash
passwd -Sa | while read LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS; do
# Is this account disabled?
[[ $PASS_TYPE == LK ]] && DIS="disabled" || DIS=""
# Can this account sudo?
#sudo -l -U "$LOGIN" | grep -q not allowed && CAN_SUDO="" || CAN_SUDO="sudo"
sudo -l -U "$LOGIN" | grep -q "(ALL) NOPASSWD: ALL|(ALL) ALL" && CAN_SUDO="sudo" || CAN_SUDO=""
# Grab misc. info from passwd entry
IFS=: read _ PASS USERID GROUPID FULLNAME HOMEDIR SH < <(getent passwd "$LOGIN")
# Grab login time of the latest entry in the lastlog output
LATEST="$(last ${LOGIN} | head -1 | cut -c40-)"
echo "${LOGIN}|${FULLNAME}|${USERID}|${LAST_CHANGE}|${MIN_AGE}|${MAX_AGE}|${WARN_DAYS}|${INACTIVE_DAYS}|${DIS}|${CAN_SUDO}|${LATEST}"
done
linux shell-script
linux shell-script
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
asked Dec 17 at 6:33
Cell-o
2741414
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
asked Dec 17 at 6:33
Cell-o
2741414
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
edited Dec 17 at 8:49
ctrl-alt-delor
10.8k41957
10.8k41957
asked Dec 17 at 6:33
Cell-o
2741414
asked Dec 17 at 6:33
Cell-o
2741414
asked Dec 17 at 6:33
Cell-o
2741414
2741414
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The RHEL variant of passwd supports the -S option, but you have to use it with a single username at a time.
Therefore, start with a list of users from /etc/passwd and then use the passwd command on all of them in turn.
#!/bin/bash
for user in $(cut -d: -f1 /etc/passwd); do
read -r LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS CRYPT < <(passwd -S $user)
# Rest of the processing
done
Note that I've added an additional variable named CRYPT to capture the cryptographic information from the end of the line. The INACTIVE_DAYS variable does not get a numeric value without that.
answered Dec 17 at 8:11
Haxiel
1,144310
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The RHEL variant of passwd supports the -S option, but you have to use it with a single username at a time.
Therefore, start with a list of users from /etc/passwd and then use the passwd command on all of them in turn.
#!/bin/bash
for user in $(cut -d: -f1 /etc/passwd); do
read -r LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS CRYPT < <(passwd -S $user)
# Rest of the processing
done
Note that I've added an additional variable named CRYPT to capture the cryptographic information from the end of the line. The INACTIVE_DAYS variable does not get a numeric value without that.
answered Dec 17 at 8:11
Haxiel
1,144310
add a comment |
The RHEL variant of passwd supports the -S option, but you have to use it with a single username at a time.
Therefore, start with a list of users from /etc/passwd and then use the passwd command on all of them in turn.
#!/bin/bash
for user in $(cut -d: -f1 /etc/passwd); do
read -r LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS CRYPT < <(passwd -S $user)
# Rest of the processing
done
Note that I've added an additional variable named CRYPT to capture the cryptographic information from the end of the line. The INACTIVE_DAYS variable does not get a numeric value without that.
answered Dec 17 at 8:11
Haxiel
1,144310
add a comment |
The RHEL variant of passwd supports the -S option, but you have to use it with a single username at a time.
Therefore, start with a list of users from /etc/passwd and then use the passwd command on all of them in turn.
#!/bin/bash
for user in $(cut -d: -f1 /etc/passwd); do
read -r LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS CRYPT < <(passwd -S $user)
# Rest of the processing
done
Note that I've added an additional variable named CRYPT to capture the cryptographic information from the end of the line. The INACTIVE_DAYS variable does not get a numeric value without that.
answered Dec 17 at 8:11
Haxiel
1,144310
The RHEL variant of passwd supports the -S option, but you have to use it with a single username at a time.
Therefore, start with a list of users from /etc/passwd and then use the passwd command on all of them in turn.
#!/bin/bash
for user in $(cut -d: -f1 /etc/passwd); do
read -r LOGIN PASS_TYPE LAST_CHANGE MIN_AGE MAX_AGE WARN_DAYS INACTIVE_DAYS CRYPT < <(passwd -S $user)
# Rest of the processing
done
Note that I've added an additional variable named CRYPT to capture the cryptographic information from the end of the line. The INACTIVE_DAYS variable does not get a numeric value without that.
answered Dec 17 at 8:11
Haxiel
1,144310
answered Dec 17 at 8:11
Haxiel
1,144310
answered Dec 17 at 8:11
Haxiel
1,144310
answered Dec 17 at 8:11
Haxiel
1,144310
1,144310
add a comment |
add a comment |
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