I am trying to set up a cron on several AWS EC2 machines and would like to run a command on all of them at once, with the following shell script:
#!/bin/sh
cd /etc/cron.daily
touch ecs.sh
echo '#!/bin/sh' > /etc/cron.daily/ecs.sh
echo 'sudo yum update -y ecs-init' >> /etc/cron.daily/ecs.sh
echo 'sudo yum update -y' >> /etc/cron.daily/ecs.sh
sudo chmod 755 /etc/cron.daily/ecs.sh
cd ~
(crontab -u root -l; echo '0 0 * * * /etc/cron.daily/ecs.sh') | crontab -u root -
sudo yum update -y
The part that does not work is: chmod 755 /etc/cron.daily/ecs.sh
I am not sure, what am I missing.
If you can (have sufficient rights to) create a file, you do not need to sudo to change its permissions to 0755. Which would also likely prompt you to input your password and run non-interactively could be the reason why the action did not take place.
On the other hand, if the user running this did not have the necessary (write) permission, preceding lines creating the file would not happen either.
You also do not need to touch a file, because that > redirection will create it (always a new one).
You also should not cd somewhere and and continue performing actions without checking directory was changed as expected. But since on all action but the unnecessary touch you use absolute path names, you can just as well leave out both cd lines.
If you clean-up the script and it still does not perform expected action, it might be useful (assuming non-interactive execution) to save its output (redirect both standard > (or 1>) and error (2>) output to a file) and examine it for errors.
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
I have a bash script that partially needs to be running with default user rights, but there are some parts that involve using sudo (like copying stuff into system folders) I could just run the script with sudo ./script.sh, but that messes up all file access rights, if it involves creating or modifying files in the script.
So, how can I run script using sudo for some commands? Is it possible to ask for sudo password in the beginning (when the script just starts) but still run some lines of the script as a current user?
You could add this to the top of your script:
while ! echo "$PW" | sudo -S -v > /dev/null 2>&1; do
read -s -p "password: " PW
echo
done
That ensures the sudo credentials are cached for 5 minutes. Then you could run the commands that need sudo, and just those, with sudo in front.
Edit: Incorporating mklement0's suggestion from the comments, you can shorten this to:
sudo -v || exit
The original version, which I adapted from a Python snippet I have, might be useful if you want more control over the prompt or the retry logic/limit, but this shorter one is probably what works well for most cases.
Each line of your script is a command line. So, for the lines you want, you can simply put sudo in front of those lines of your script. For example:
#!/bin/sh
ls *.h
sudo cp *.h /usr/include/
echo "done" >>log
Obviously I'm just making stuff up. But, this shows that you can use sudo selectively as part of your script.
Just like using sudo interactively, you will be prompted for your user password if you haven't done so recently.
I try to run the following bash script to create a bunch of users, groups, home dirs for the users and correct permissions for all of these. The OS is CentOS.
When I try to run the following, which I though should work, it returns "command not found" when running via terminal. it only gets as far as creating the /homedirs directory, nothing more. I'm a total noob at bash scripting so forgive me if this looks ugly.
mkdir /homedirs; chmod 775 /homedirs;
for iYear in {1..3} do
sYear = $iYear"ti"
sYearDir = "/homerirs/"$sYear
groupadd $sYear; mkdir $sYearDir; chgrp $sYear $sYearDir; chmod 750 $sYearDir
for sClass in {a,b} do
sClassDir = $sYearDir/$sClass
mkdir $sClassDir
sClassGrp = $sYear$sClass
groupadd $sClassGrp; chgrp $sClassGrp $sClassDir; chmod 750 $sClassDir
for iUser in {1..3} do
sUserName = "i"$iYear$sClass"g"$iUser
sUserDir = $sClassDir/$sUserName
useradd -d $sUserDir -g $sClassGrp -G $sYear -m $sUserName
chown $sUserName $sUserDir; chmod 750 $sUserDir
done
done
done
The error message is caused by the spaces around the equals signs. A token with whitespace after it is interpreted as a command name; so what you intended as variable names causes the Command not found errors.
You may need to set your PATH and you really should read the advanced bash scripting guide. See also this answer.
I also suggest to debug your script by starting it with #!/bin/bash -vx as its first line. And you should make it executable with chmod u+x at least.
Perhaps groupadd might not be available on your system.
best thing to do is add the full path before your executables:
change useradd to /usr/sbin/useradd
change groupadd to /usr/sbin/groupadd
will cure the command not found.
remember this programs will probably need to run as root to work.
I have a shell script which needs non-root user account to run certain commands and then change the user to root to run the rest of the script. I am using SUSE11.
I have used expect to automate the password prompt. But when I use
spawn su -
and the command gets executed, the prompt comes back with root and the rest of the script does not execute.
Eg.
< non-root commands>
spawn su -
<root commands>
But after su - the prompt returns back with user as root.
How to execute the remaining of the script.
The sudo -S option does not help as it does not run sudo -S ifconfig command which I need to find the IP address of the machine.
I have already gone through these links but could not find a solution:
Change script directory to user's homedir in a shell script
Changing unix user in a shell script
sudo will work here but you need to change your script a little bit:
$ cat 1.sh
id
sudo -s <<EOF
echo Now i am root
id
echo "yes!"
EOF
$ bash 1.sh
uid=1000(igor) gid=1000(igor) groups=1000(igor),29(audio),44(video),124(fuse)
Now i am root
uid=0(root) gid=0(root) groups=0(root)
yes!
You need to run your command in <<EOF block and give the block to sudo.
If you want, you can use su, of course. But you need to run it using expect/pexpect that will enter password for you.
But even in case you could manage to enter the password automatically (or switch it off) this construction would not work:
user-command
su
root-command
In this case root-command will be executed with user, not with root privileges, because it will be executed after su will be finished (su opens a new shell, not changes uid of the current shell). You can use the same trick here of course:
su -c 'sh -s' <<EOF
# list of root commands
EOF
But now you have the same as with sudo.
There is an easy way to do it without a second script. Just put this at the start of your file:
if [ "$(whoami)" != "root" ]
then
sudo su -s "$0"
exit
fi
Then it will automatically run itself as root. Of course, this assumes that you can sudo su without having to provide a password - but that's out of scope of this answer; see one of the other questions about using sudo in shell scripts for how to do that.
Short version: create a block to enclose all commands to be run as root.
For example, I created a script to run a command from a root subdirectory, the segment goes like this:
sudo su - <<EOF
cd rootSubFolder/subfolder
./commandtoRun
EOF
Also, note that if you are changing to "root" user inside a shell script like below one, few Linux utilities like awk for data extraction or defining even a simple shell variable etc will behave weirdly.
To resolve this simply quote the whole document by using <<'EOF' in place of EOF.
sudo -i <<'EOF'
ls
echo "I am root now"
EOF
The easiest way to do that would be to create a least two scripts.
The first one should call the second one with root privileges. So every command you execute in the second script would be executed as root.
For example:
runasroot.sh
sudo su-c'./scriptname.sh'
scriptname.sh
apt-get install mysql-server-5.5
or whatever you need.