I'm writing a program which uses its own config files (in /etc/program/ and ~/.config/program/)
The problem is sometimes it needs to be executed as root (so most of the time with sudo) and in this case ~/.config/program is actually in root's directory (/root in my case)
I want my program to use the user's config file even if used with sudo. My idea is to first get the parent program's PID (in my case BASH) -- which is normally executed by a normal user -- and then get the name of the user executing the parent program. I easly found out how to get the PPID, but I didn't found how to find the name of the user.
I imagine a script like :
parents_name=$(echo $PPID | *command_which_gives_the_name_from_a_PID*)
. /home/$parents_name/.config/program # includes the user's config file
I hope someone have an awnser !
sudo adds the name of the invoking user to the environment of the command.
. /home/$SUDO_USER/.config/program
$HOME might already be the same as /home/$SUDO_USER, but there are too many ways to configure sudo to make it the home directory of the target user.
However, that makes your script dependent on being called by sudo. Better to be explicit and pass the path of the desired configuration to the script as an argument. Something like
sudo theScript "$HOME/.config/program"
and
. "$1"
To get the user of a process
$ ps u -p 2813
To strip it out
$ ps u -p 2813 | awk '{print $1}'
USER
john
Or
$ ps -o user= -p 2813
john
https://unix.stackexchange.com/questions/284934/return-owner-of-process-given-pid/284938 has a similar question. The accepted answer was to use ps -o user= -p PIDHERE. So, in your case, you'd do something like:
parents_name=$(ps -o user= -p $PPID)
In this case, we don't want column headers, so we add --no-headers.
I write a script for the network dispatcher in /etc/NetworkDispatcher/dispatcher.d/.
How can I get the current logged user?
I already tried these commands :
$USER
$LOGNAME
Thank you for help.
If running a script with /usr/bin/sudo you can access the original user from the $SUDO_USER environment variable.
For example if this is the contents of a script test.sh:
#!/bin/sh
echo "USER: $USER"
echo "SUDO_USER: $SUDO_USER"
Then if you run it as "alex":
alex#yuzu:~$ ./test.sh
USER: alex
SUDO_USER:
And if you run it via sudo:
alex#yuzu:~$ sudo ./test.sh
USER: root
SUDO_USER: alex
You can use whoami. Or perhaps you could get a list of users logged-in with users. It really depends on what you need specifically.
You could also use "last":
$ last | grep "logged in"
I needed this for a postins script in a deb file, nothing of the above worked, but this did on LinuxMint 19.3 (tricia):
`users`
I need to execute part of a bash script as a different user, and inside that user's $HOME directory. However, I'm not sure how to determine this variable. Switching to that user and calling $HOME does not provide the correct location:
# running script as root, but switching to a different user...
su - $different_user
echo $HOME
# returns /root/ but should be /home/myuser
Update:
It appears that the issue is with the way that I am trying to switch users in my script:
$different_user=deploy
# create user
useradd -m -s /bin/bash $different_user
echo "Current user: `whoami`"
# Current user: root
echo "Switching user to $different_user"
# Switching user to deploy
su - $different_user
echo "Current user: `whoami`"
# Current user: root
echo "Current user: `id`"
# Current user: uid=0(root) gid=0(root) groups=0(root)
sudo su $different_user
# Current user: root
# Current user: uid=0(root) gid=0(root) groups=0(root)
What is the correct way to switch users and execute commands as a different user in a bash script?
Update: Based on this question's title, people seem to come here just looking for a way to find a different user's home directory, without the need to impersonate that user.
In that case, the simplest solution is to use tilde expansion with the username of interest, combined with eval (which is needed, because the username must be given as an unquoted literal in order for tilde expansion to work):
eval echo "~$different_user" # prints $different_user's home dir.
Note: The usual caveats regarding the use of eval apply; in this case, the assumption is that you control the value of $different_user and know it to be a mere username.
By contrast, the remainder of this answer deals with impersonating a user and performing operations in that user's home directory.
Note:
Administrators by default and other users if authorized via the sudoers file can impersonate other users via sudo.
The following is based on the default configuration of sudo - changing its configuration can make it behave differently - see man sudoers.
The basic form of executing a command as another user is:
sudo -H -u someUser someExe [arg1 ...]
# Example:
sudo -H -u root env # print the root user's environment
Note:
If you neglect to specify -H, the impersonating process (the process invoked in the context of the specified user) will report the original user's home directory in $HOME.
The impersonating process will have the same working directory as the invoking process.
The impersonating process performs no shell expansions on string literals passed as arguments, since no shell is involved in the impersonating process (unless someExe happens to be a shell) - expansions by the invoking shell - prior to passing to the impersonating process - can obviously still occur.
Optionally, you can have an impersonating process run as or via a(n impersonating) shell, by prefixing someExe either with -i or -s - not specifying someExe ... creates an interactive shell:
-i creates a login shell for someUser, which implies the following:
someUser's user-specific shell profile, if defined, is loaded.
$HOME points to someUser's home directory, so there's no need for -H (though you may still specify it)
The working directory for the impersonating shell is the someUser's home directory.
-s creates a non-login shell:
no shell profile is loaded (though initialization files for interactive nonlogin shells are; e.g., ~/.bashrc)
Unless you also specify -H, the impersonating process will report the original user's home directory in $HOME.
The impersonating shell will have the same working directory as the invoking process.
Using a shell means that string arguments passed on the command line MAY be subject to shell expansions - see platform-specific differences below - by the impersonating shell (possibly after initial expansion by the invoking shell); compare the following two commands (which use single quotes to prevent premature expansion by the invoking shell):
# Run root's shell profile, change to root's home dir.
sudo -u root -i eval 'echo $SHELL - $USER - $HOME - $PWD'
# Don't run root's shell profile, use current working dir.
# Note the required -H to define $HOME as root`s home dir.
sudo -u root -H -s eval 'echo $SHELL - $USER - $HOME - $PWD'
What shell is invoked is determined by "the SHELL environment variable if it is set or the shell as specified in passwd(5)" (according to man sudo). Note that with -s it is the invoking user's environment that matters, whereas with -i it is the impersonated user's.
Note that there are platform differences regarding shell-related behavior (with -i or -s):
sudo on Linux apparently only accepts an executable or builtin name as the first argument following -s/-i, whereas OSX allows passing an entire shell command line; e.g., OSX accepts sudo -u root -s 'echo $SHELL - $USER - $HOME - $PWD' directly (no need for eval), whereas Linux doesn't (as of sudo 1.8.95p).
Older versions of sudo on Linux do NOT apply shell expansions to arguments passed to a shell; for instance, with sudo 1.8.3p1 (e.g., Ubuntu 12.04), sudo -u root -H -s echo '$HOME' simply echoes the string literal "$HOME" instead of expanding the variable reference in the context of the root user. As of at least sudo 1.8.9p5 (e.g., Ubuntu 14.04) this has been fixed. Therefore, to ensure expansion on Linux even with older sudo versions, pass the the entire command as a single argument to eval; e.g.: sudo -u root -H -s eval 'echo $HOME'. (Although not necessary on OSX, this will work there, too.)
The root user's $SHELL variable contains /bin/sh on OSX 10.9, whereas it is /bin/bash on Ubuntu 12.04.
Whether the impersonating process involves a shell or not, its environment will have the following variables set, reflecting the invoking user and command: SUDO_COMMAND, SUDO_USER, SUDO_UID=, SUDO_GID.
See man sudo and man sudoers for many more subtleties.
Tip of the hat to #DavidW and #Andrew for inspiration.
In BASH, you can find a user's $HOME directory by prefixing the user's login ID with a tilde character. For example:
$ echo ~bob
This will echo out user bob's $HOME directory.
However, you say you want to be able to execute a script as a particular user. To do that, you need to setup sudo. This command allows you to execute particular commands as either a particular user. For example, to execute foo as user bob:
$ sudo -i -ubob -sfoo
This will start up a new shell, and the -i will simulate a login with the user's default environment and shell (which means the foo command will execute from the bob's$HOME` directory.)
Sudo is a bit complex to setup, and you need to be a superuser just to be able to see the shudders file (usually /etc/sudoers). However, this file usually has several examples you can use.
In this file, you can specify the commands you specify who can run a command, as which user, and whether or not that user has to enter their password before executing that command. This is normally the default (because it proves that this is the user and not someone who came by while the user was getting a Coke.) However, when you run a shell script, you usually want to disable this feature.
For the sake of an alternative answer for those searching for a lightweight way to just find a user's home dir...
Rather than messing with su hacks, or bothering with the overhead of launching another bash shell just to find the $HOME environment variable...
Lightweight Simple Homedir Query via Bash
There is a command specifically for this: getent
getent passwd someuser | cut -f6 -d:
getent can do a lot more... just see the man page. The passwd nsswitch database will return the user's entry in /etc/passwd format. Just split it on the colon : to parse out the fields.
It should be installed on most Linux systems (or any system that uses GNU Lib C (RHEL: glibc-common, Deb: libc-bin)
The title of this question is How to get $HOME directory of different user in bash script? and that is what people are coming here from Google to find out.
There is a safe way to do this!
on Linux/BSD/macOS/OSX without sudo or root
user=pi
user_home=$(bash -c "cd ~$(printf %q $USER) && pwd")
NOTE: The reason this is safe is because bash (even versions prior to 4.4) has its own printf function that includes:
%q quote the argument in a way that can be reused as shell input
See: help printf
Compare the how other answers here respond to code injection
# "ls /" is not dangerous so you can try this on your machine
# But, it could just as easily be "sudo rm -rf /*"
$ user="root; ls /"
$ printf "%q" "$user"
root\;\ ls\ /
# This is what you get when you are PROTECTED from code injection
$ user_home=$(bash -c "cd ~$(printf "%q" "$user") && pwd"); echo $user_home
bash: line 0: cd: ~root; ls /: No such file or directory
# This is what you get when you ARE NOT PROTECTED from code injection
$ user_home=$(bash -c "cd ~$user && pwd"); echo $user_home
bin boot dev etc home lib lib64 media mnt ono opt proc root run sbin srv sys tmp usr var /root
$ user_home=$(eval "echo ~$user"); echo $user_home
/root bin boot dev etc home lib lib64 media mnt ono opt proc root run sbin srv sys tmp usr var
on Linux/BSD/macOS/OSX as root
If you are doing this because you are running something as root then you can use the power of sudo:
user=pi
user_home=$(sudo -u $user sh -c 'echo $HOME')
on Linux/BSD (but not modern macOS/OSX) without sudo or root
If not, the you can get it from /etc/passwd. There are already lots of examples of using eval and getent, so I'll give another option:
user=pi
user_home=$(awk -v u="$user" -v FS=':' '$1==u {print $6}' /etc/passwd)
I would really only use that one if I had a bash script with lots of other awk oneliners and no uses of cut. While many people like to "code golf" to use the fewest characters to accomplish a task, I favor "tool golf" because using fewer tools gives your script a smaller "compatibility footprint". Also, it's less man pages for your coworker or future-self to have to read to make sense of it.
You want the -u option for sudo in this case. From the man page:
The -u (user) option causes sudo to run the specified command as a user other than root.
If you don't need to actually run it as them, you could move to their home directory with ~<user>. As in, to move into my home directory you would use cd ~chooban.
So you want to:
execute part of a bash script as a different user
change to that user's $HOME directory
Inspired by this answer, here's the adapted version of your script:
#!/usr/bin/env bash
different_user=deploy
useradd -m -s /bin/bash "$different_user"
echo "Current user: $(whoami)"
echo "Current directory: $(pwd)"
echo
echo "Switching user to $different_user"
sudo -u "$different_user" -i /bin/bash - <<-'EOF'
echo "Current user: $(id)"
echo "Current directory: $(pwd)"
EOF
echo
echo "Switched back to $(whoami)"
different_user_home="$(eval echo ~"$different_user")"
echo "$different_user home directory: $different_user_home"
When you run it, you should get the following:
Current user: root
Current directory: /root
Switching user to deploy
Current user: uid=1003(deploy) gid=1003(deploy) groups=1003(deploy)
Current directory: /home/deploy
Switched back to root
deploy home directory: /home/deploy
This works in Linux. Not sure how it behaves in other *nixes.
getent passwd "${OTHER_USER}"|cut -d\: -f 6
I was also looking for this, but didn't want to impersonate a user to simply acquire a path!
user_path=$(grep $username /etc/passwd|cut -f6 -d":");
Now in your script, you can refer to $user_path in most cases would be /home/username
Assumes: You have previously set $username with the value of the intended users username.
Source: http://www.unix.com/shell-programming-and-scripting/171782-cut-fields-etc-passwd-file-into-variables.html
Quick and dirty, and store it in a variable:
USER=somebody
USER_HOME="$(echo -n $(bash -c "cd ~${USER} && pwd"))"
I was struggling with this question because I was looking for a way to do this in a bash script for OS X, hence /etc/passwd was out of the question, and my script was meant to be executed as root, therefore making the solutions invoking eval or bash -c dangerous as they allowed code injection into the variable specifying the username.
Here is what I found. It's simple and doesn't put a variable inside a subshell. However it does require the script to be ran by root as it sudos into the specified user account.
Presuming that $SOMEUSER contains a valid username:
echo "$(sudo -H -u "$SOMEUSER" -s -- "cd ~ && pwd")"
I hope this helps somebody!
If the user doesn't exist, getent will return an error.
Here's a small shell function that doesn't ignore the exit code of getent:
get_home() {
local result; result="$(getent passwd "$1")" || return
echo $result | cut -d : -f 6
}
Here's a usage example:
da_home="$(get_home missing_user)" || {
echo 'User does NOT exist!'; exit 1
}
# Now do something with $da_home
echo "Home directory is: '$da_home'"
The output of getent passwd username can be parsed with a Bash regular expression
OTHER_HOME="$(
[[ "$(
getent \
passwd \
"${OTHER_USER}"
)" =~ ([^:]*:){5}([^:]+) ]] \
&& echo "${BASH_REMATCH[2]}"
)"
If you have sudo active, just do:
sudo su - admin -c "echo \$HOME"
NOTE: replace admin for the user you want to get the home directory
We can run something like chsh -s /usr/local/bin/zsh to set a new default shell. Is there a command we can run to know what that shell is?
I don’t mean having a terminal open and running a command to know which shell we’re in. I mean like in the example above, if I’m in a terminal with /bin/bash open, what should I run to get /usr/local/bin/zsh if it’s the current default shell?
You can use the following command:
echo $SHELL
For macOS:
dscl . -read /Users/username UserShell
For the current macOS user:
dscl . -read ~/ UserShell
To parse the path inline using sed:
dscl . -read ~/ UserShell | sed 's/UserShell: //'
Using $SHELL will report the current login shell, not the default login shell. In certain cases, these are not the same. For example, when working in an IDE such as Visual Studio Code which opens an integrated terminal without consulting the default shell.
In addition, as pointed out by Martin C. Martin, $SHELL is a constant that will not change after chsh changes the default login shell.
You can grep in the /etc/passwd file for current username, and use cut to extract the appropriate column of information:
grep ^$(id -un): /etc/passwd | cut -d : -f 7-
$(id -un) is a safer than $USER to get user name. Using ^ in front of user name and : after makes sure you don't get a false match if your user name is a sub section of someone else user name.
$SHELL can also be used, as suggested. However it won't work if chsh was used in current shell, as the variable is not updated. Also the variable is not protected against being changed, so it can theoretically be set to something completely different.
Update to attempt an OS X compatible solution. Probably not optimal regexp:
grep ^.*:.*:$(id -u): /etc/passwd | cut -d : -f 7-
This is based on user id's. If the whole user entry is missing, not only user name, then osx must store this somewhere else.
If you want to get the default shell of a user, you could grep file /etc/passwd. like:
grep "$USER" /etc/passwd
# kent:x:1000:1000::/home/kent:/bin/zsh
telling me that the current user (kent) has the default shell /bin/zsh.
If you just want to catch the shell part:
awk -F: -v u="$USER" 'u==$1&&$0=$NF' /etc/passwd
# /bin/zsh
If you want to get the default shell of other user, just replace the $USER part.
In OS X, using the command env | grep -i 'SHELL' produces an output such as: SHELL=/bin/sh
(as root, however regular users tend to have /bin/bash as default shell) with a little parsing, the path the shell (and thus the shell itself) could be easily identified and extracted from there..
I have a sh script that needs to be run as root, however it is run by the end user using sudo. How can I get the users home directory when ~/ points to /root when running with sudo?
Try to avoid eval. Especially with root perms.
You can do:
USER_HOME=$(getent passwd $SUDO_USER | cut -d: -f6)
Update:
here is why to avoid eval.
The user's home directory would be ~$SUDO_USER. You can use eval as follows:
USER_HOME=$(eval echo ~${SUDO_USER})
echo ${USER_HOME}
$ sudo env |grep USER
USER=root
USERNAME=root
SUDO_USER=glglgl
So you can access $SUDO_USER and ask the system for his homedir with getent passwd $SUDO_USER | cut -d: -f6.
Try accessing the environment variable $SUDO_USER
Unless I misunderstood the question, when the user runs the script with sudo, the $HOME environment variable does not change. To test out, I created this script:
#!/bin/bash
#sudo_user.sh
env | grep -e USER -e HOME
... and run it:
sudo ./sudo_user.sh
Output:
USER=root
HOME=/home/haiv
USERNAME=root
SUDO_USER=haiv
The output tells me that $HOME is still pointing to the user's home (in this case, /home/haiv).
It comes to a little improvement for the eval solution.
Before sending the username variable to eval, test it with the user existence, so if there's arbitrary code, it won't pass the check.
#? Description:
#? Get the user's home directory by username regardless of whether is running with `sudo`.
#? The username is checked, it must exist, to avoid the arbitrary code execution of `eval`.
#? The returned home directory's existence is not checked.
#?
#? Usage:
#? #home USERNAME
function home () {
if id "$1" >/dev/null 2>&1; then
eval echo "~$1"
else
return 255
fi
}
Run the code on macOS:
$ home root
Output:
/var/root
If you run it with something like:
home 'root ; ls'
It gives nothing and returns an error, the ls won't be executed.
You can use Hai's incomplete answer to very simply construct the user's home directory. He and others correctly state that you can get the original user while in the sudo shell. You can use that to construct the home directory into a new variable called user_home (or whatever you like):
#user_home="/home/${$SUDO_USER}";