According to a document, it says:
When an interactive shell that is not a login shell is started, Bash reads and executes commands from ~/.bashrc, if that file exists.
I did a quick test:
At my server,
[USER#MYSERVER ~]$ cat .bashrc
...
echo 'I am in a bashrc file of my server'
...
At a remote server,
# unquoted
[USER#REMOTESERVER ~]$ ssh MYSERVER echo $-
I am in a bashrc file of my server
himBH
#quoted
[USER#REMOTESERVER ~]$ ssh MYSERVER 'echo $-'
I am in a bashrc file of my server
hBc
When command is unquoted, it seems to be run in an interactive shell, and when quoted, it seems to be run in a non-interactive shell.
Why is this so?
And both read the bashrc file of MYSERVER, which doesn't follow the rule in the document.
Any link or comment appreciated.
EDITED:
And it seems to be a non-login shell.
[USER#REMOTESERVER ~]$ ssh MYSERVER 'shopt -q login_shell && echo 1 || echo 2'
2
In the bash document, there says:
Invoked by remote shell daemon
Bash attempts to determine when it is being run with its standard input connected to a network connection, as when executed by the remote shell daemon, usually rshd, or the secure shell daemon sshd. If Bash determines it is being run in this fashion, it reads and executes commands from ~/.bashrc, if that file exists and is readable.
I missed this part...
Therefore, calling from ssh should read .bashrc file.
And ssh remote command is a non-interactive shell, as comments to the question explain.
The remote bash is indeed not started as an interactive shell (as we can see from the output from $-), so somewhat else must be sourcing your .bashrc. For sure, it is run as a login shell. Could it be that you have a ~/.bash_profile or ~.bash_login or ~/.profile, which explicitly sources .bashrc?
I'm very confused about how my shell is reading bash_profile.
In root, my ~/.bash_profile looks like so
# .bash_profile
# Get the aliases and functions
if [-f ~/.bashrc ]; then
.~/.bashrc
fi
PATH=$PATH:$HOME/bin:$HOME/sbin:$HOME/usr/sbin:$HOME/usr/bin:/usr/sbin
LD_LIBRARY_PATH=$LD_LIBRARY_PATH:/usr/local/lib
export PATH=$PATH
export LD_LIBRARY_PATH=$LD_LIBRARY_PATH
unset USERNAME
There is no ~/.profile file.
In a user called maruhan, my ~/.bash_profile looks like so
# .bash_profile
# Get the aliases and functions
if [-f ~/.bashrc ]; then
.~/.bashrc
fi
PATH=$PATH:$HOME/bin:$HOME/sbin:$HOME/usr/sbin:$HOME/usr/bin:/usr/sbin
LD_LIBRARY_PATH=$LD_LIBRARY_PATH:/home/maruhan/Desktop/issac:/usr/local/lib
ASDF=$ASDF:/home
export PATH=$PATH
export LD_LIBRARY_PATH=$LD_LIBRARY_PATH
export ASDF=$ASDF
unset USERNAME
And my ~/.profile looks like so
LD_LIBRARY_PATH=/home/maruhan/Desktop/issac:/usr/local/lib:$LD_LIBRARY_PATH
ASDF=/home:$ASDF
export ASDF=$ASDF
export LD_LIBRARY_PATH=$LD_LIBRARY_PATH
You can clearly see that ASDF is not defined in root's bash_profile.
However when I call export, I get this in root.
declare -x ASDF=":/home"
but nothing about LD_LIBRARY_PATH.
Strangely in maruhan, running export shows both ASDF and LD_LIBRARY_PATH.
Also, nothing about ASDF or LD_LIBRARY_PATH exist in /etc/environment. I also don't have a /etc/bash_profile file.
Running echo $0 gives me bash for both root and maruhan.
How come LD_LIBRARY_PATH disappeared in root while ASDF is there?
The rules are a bit complicated. According to bash's man page:
INVOCATION
A login shell is one whose first character of argument zero is a -, or one
started with the --login option.
An interactive shell is one started without non-option arguments (unless -s is
specified) and without the -c option whose standard input and error are both
connected to terminals (as determined by isatty(3)), or one started with the -i
option. PS1 is set and $- includes i if bash is interactive, allowing a shell
script or a startup file to test this state.
... ...
When bash is invoked as an interactive login shell, or as a non-interactive
shell with the --login option, it first reads and executes commands from the
file /etc/profile, if that file exists. After reading that file, it looks for
~/.bash_profile, ~/.bash_login, and ~/.profile, in that order, and reads and
executes commands from the first one that exists and is readable. The --noprofile
option may be used when the shell is started to inhibit this behavior.
... ...
When an interactive shell that is not a login shell is started, bash reads and
executes commands from ~/.bashrc, if that file exists. This may be inhibited by
using the --norc option. The --rcfile file option will force bash to read and
execute commands from file instead of ~/.bashrc.
... ...
Note that on some systems bash may be customized so that it would also execute a system wide rc file (e.g. /etc/bash.bashrc) before sourcing ~/.bashrc for an interactive shell that's not a login shell.
Shells started by a login mechanism (usually with a username/password prompt, like console login, telnet, ssh, ...) are usually login shells. For a login shell, $0 is usually -bash.
[local] % ssh user#host <-- The user is trying to login
Password: P#ssw0rd
[remote] % echo $0
-bash <-- This is a login shell
[remote] % bash <-- This is not a login (no username/password)
[remote] % echo $0
bash <-- Not a login shell
[remote] %
To make life easier I would put all rc things in ~/.bashrc and source ~/.bashrc in ~/.bash_profile. For example:
% cat ~/.bash_profile
[[ -f ~/.bashrc ]] && source ~/.bashrc
% cat ~/.bashrc
# return immediately if not in an interactive shell
[[ $- != *i* ]] && return 0
export FOO=bar
PATH=$PATH:/my/path
%
Can anyone please tell me what shell has installed in my system?
Because when I am logging into my system using my username it is initially showing bash shell but later it is showing korn shell after doing sudo.
Please see below for details.
-bash-3.2$ pwd
/home/w4x2spxt
-bash-3.2$ echo $SHELL
/bin/bash
-bash-3.2$ su - XXXXXXX
Password:
You have new mail.
The Oracle base remains unchanged with value /apps/oracle
abc0300ab123:/a30/home/XXXXXXX >> echo $SHELL
/bin/ksh
SHELL environment variable gives you your login shell.
check the shell path mentioned(last column) against your username or XXXXXXX(for su) in /etc/passwd like this: grep ^XXXXXXX /etc/passwd.
The shell mentioned in that file will be your default shell when you login or su to that user.
To check all installed shells on your system use this: cat /etc/shells
Your user is using /bin/bash as default shell.
root is using /bin/ksh as its default shell.
The default shell is a user-specific setting, so there is nothing picky in having different ones among users. Just check the last column in /etc/passwd and surprise yourself with a variety of values.
Note by the way that when you do su, you log in as root. If you add the dash and say su - you are loading the root profile, so that you have its environment.
The environment variable SHELL always contains the login shell of the user logged in, defined in /etc/passwd.
If the user changes his/her shell after login by e.g. exec bash (bash), the SHELL will still expand to the login shell.
In your case, the user XXXXXXX has the login shell /bin/ksh, do:
grep '<user_name>' /etc/passwd
to match the results.
To find the current shell:
echo $0
Or
ps -p $$
When you do this:
-bash-3.2$ su - XXXXXXX
you are starting whatever shell is assigned to user XXXXXXX. This is usually a good thing for su -, since that runs their shell as a login process so you get their normal shell startup initialization (profile, *shrc, etc). If you run a different shell from the one their account is set up for, you probably miss out on all their customization.
You can see what shell is associated with an account by looking them up in the password database. This is pretty reliable across different types of systems and authentication schemes:
perl -MUser::pwent -le 'print( (getpwnam "XXXXXXX")->shell || "/bin/sh" )'
You can always run a shell explicitly as the other user if you have one in mind that you want:
su XXXXXXX -c "bash --login"
or
sudo -u XXXXXXX bash --login # if you have sudo privs
To see what shell you're currently running, look at $0:
echo $0
To see what shell you get by default as you, look at $SHELL.
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
Lets assume i am normal user, the i will switch to root:
user ~ $ su - root
Password:
root ~ #
So once i logged in as root, i want to run following command automatically:
source .bash_profile
How can i have that above command run automatically please?
According to the bash man page, .bash_profile is executed for login shells, while .bashrc is executed for interactive non-login shells.
In your case, you don't need to source .bash_profile like this.
You just need to put source .bash_profile in your root's .bashrc file
if [ -f ~/.bash_profile ]; then
source ~/.bash_profile
fi
Read me for better understanding of .bash_profile and .bashrc
Update
Example:
[root#sgeorge-ld ~]# cat .bashrc | tail -1
echo "Testing .bashrc for a stack query"
[root#sgeorge-ld ~]# exit
logout
[sgeorge#sgeorge-ld ~]$ su - root
Password:
Testing .bashrc for a stack query
[root#sgeorge-ld ~]#
First of all, when you switch to root user, you will be still your regular user's home directory. Which .bash_profile you want to execute? /Users/myuser/.bash_profile or root's /var/root/.bash_profile?
Regardless of what you would like to execute, you can edit /var/root/.bashrc (if you don't have it, create one) and add your command there.