I'm creating a bash script which will wrap a docker run command, passing in all arguments. The docker cmd being run has a -e parameter like so:
docker run --rm -it --name some_name -v $(pwd):/some_dir some_image some_command -e -r -t
However, the bash script for some reason is not passing in the -e parameter.
For example I have the following test.sh script.
#!/bin/bash
echo "Echoing the command arguments."
echo $#
The -e parameter is not passed through to $# when in the first position.
$ ./test.sh -e -r -t
Echoing the command arguments.
-r -t
$ ./test.sh -r -e -t
Echoing the command arguments.
-r -e -t
Ultimately I would like to be able to call the docker command as follows, simply passing in all given parameters from the bash script to the docker command.
docker run --rm -it --name some_name -v $(pwd):/some_dir some_image some_command $#
This may be confusing to users if they are expecting the -e parameter to be passed in and the associate activity does not happen.
Is there a way to allow the -e parameter to pass through?
Change your test to:
#!/bin/bash
echo "Echoing the command arguments:"
printf ' - %q\n' "$#"
...and you'll see -e present.
A POSIX-conforming implementation of echo will print -e on output when given that string as its first argument. bash's implementation does not comply unless both set -o posix and shopt -s xpg_echo runtime options are set.
Related
Why does the last example throw an error but the others work? Bash is invoked in any case.
#!/bin/bash
function hello {
echo "Hello! user=$USER, uid=$UID, home=$HOME";
}
# Test that it works.
hello
# ok
bash -c "$(declare -f hello); hello"
# ok
sudo su $USER bash -c "$(declare -f hello); hello"
# error: bash: -c: line 1: syntax error: unexpected end of file
sudo -i -u $USER bash -c "$(declare -f hello); hello"
It fail because of the -i or --login switch:
It seems like when debugging with -x
$ set -x
$ sudo -i -u $USER bash -c "$(declare -f hello); hello"
++ declare -f hello
+ sudo -i -u lea bash -c 'hello ()
{
echo "Hello! user=$USER, uid=$UID, home=$HOME"
}; hello'
Now if doing it manually it cause the same error:
sudo -i -u lea bash -c 'hello ()
{
echo "Hello! user=$USER, uid=$UID, home=$HOME"
}; hello'
Now lets just do a simple tiny change that make it work:
sudo -i -u lea bash -c 'hello ()
{
echo "Hello! user=$USER, uid=$UID, home=$HOME";}; hello'
The reason is that sudo -i runs everything like an interactive shell. And when doing so, every newline character from the declare -f hello is internally turned into space. The curly-brace code block need a semi-colon before the closing curly-brace when on the same line, which declare -f funcname does not provide since it expands the function source with closing curly brace at a new line.
Now lets make this behaviour very straightforward:
$ sudo bash -c 'echo hello
echo world'
hello
world
It executes both echo statements because they are separated by a newline.
but:
$ sudo -i bash -c 'echo hello
echo world'
hello echo world
It executes the first echo statement that prints everything as arguments because the newline has been replaced by a space.
It is the same code in all examples, so it should be ok.
Yes, "$(declare -f hello); hello" is always the same string. But it is processed differently by sudo su and sudo -i as found out by Lea Gris .
sudo -i quotes its arguments before passing them to bash. This quoting process seems to be undocumented and very poor. To see what was actually executed, you can print the argument passed to bash -c inside your ~/.bash_profile/:
Content of ~/.bash_profile
cat <<EOF
# is executed as
$BASH_EXECUTION_STRING
# resulting in output
EOF
Some examples of sudo -i's terrible and inconsistent quoting
Linebreaks are replaced by line continuations
$ sudo -u $USER -i echo '1
2'
# is executed as
echo 1\
2
# resulting in output
12
Quotes are escaped as literals
$ sudo -u $USER -i echo \'single\' \"double\"
# is executed as
echo \'single\' \"double\"
# resulting in output
'single' "double"
But $ is not quoted
$ sudo -u $USER -i echo \$var
# is executed as
echo $var
# resulting in output
Some side notes:
There might be a misunderstanding in your usage of su.
sudo su $USER bash -c "some command"
does not execute bash -c "echo 1; echo 2". The -c ... is interpreted by su and passed as -c ... to $USER's default shell. Afterwards, the remaining arguments are passed to that shell as well. The executed command is
defaultShellOfUSER -c "some command" bash
You probably wanted to write
sudo su -s bash -c "some command" "$USER"
Interactive shells behave differently
su just executes the command specified by -c. But sudo -i starts a login shell, in your case that login shell seems to be bash (this is not necessarily the case, see section above).
An interactive bash session behaves different from bash -c "..." or bash script.sh. An interactive bash sources files like .profile, .bash_profile, and enables history expansion, aliases, and so on. For a full list see the section Interactive Shell Behavior in bash's manual.
I was trying to run some docker-compose command over ssh using bash script like below. I mean I have an executable shell script deploy.sh which contains below code snippets
ssh -tt -o StrictHostKeyChecking=no root#142.32.45.2 << EOF
DIR=test
echo \${DIR}
docker-compose --project-name \${DIR} up -d
EOF
But the DIR variable doesn't get expanded while passing as a parameter to docker-compose. It executes like below. While echo \${DIR} gives correct output i.e test.
docker-compose --project-name ${DIR} up -d
ssh -tt -o StrictHostKeyChecking=no root#142.32.45.2 <<'EOF'
DIR=test
echo ${DIR}
docker-compose --project-name ${DIR} up -d
EOF
Get rid of the \$ - it is preventing your variable expansion. But on second review, I see that's your intention. If you want to prevent all variable expansion until your code gets executed on the remote host, try putting the heredoc word in quotes. That way, the $ gets passed to the script being passed to ssh.
As a second suggestion ( as per my comment below ), I would consider just sending a parameterized script to the remote host and then executing it ( after changing its permissions ).
# Make script
cat >compose.sh <<'EOF'
#!/bin/bash
DIR=$1
docker-compose --project-name $DIR
EOF
scp -o StrictHostKeyChecking=no compose.sh root#142.32.45.2:
ssh -o StrictHostKeyChecking=no root#142.32.45.2 chmod +x ./compose.sh \; ./compose.sh test
I want to echo a string into the /etc/hosts file. The string is stored in a variable called $myString.
When I run the following code the echo is empty:
finalString="Hello\nWorld"
sudo bash -c 'echo -e "$finalString"'
What am I doing wrong?
You're not exporting the variable into the environment so that it can be picked up by subprocesses.
You haven't told sudo to preserve the environment.
\
finalString="Hello\nWorld"
export finalString
sudo -E bash -c 'echo -e "$finalString"'
Alternatively, you can have the current shell substitute instead:
finalString="Hello\nWorld"
sudo bash -c 'echo -e "'"$finalString"'"'
You can do this:
bash -c "echo -e '$finalString'"
i.e using double quote to pass argument to the subshell, thus the variable ($finalString) is expanded (by the current shell) as expected.
Though I would recommend not using the -e flag with echo. Instead you can just do:
finalString="Hello
World"
I am trying to use an environment variable in a bash script that needs to run as sudo with source.
I have the following file (my_file.sh)
echo "this is DOMAIN = $DOMAIN"
I have the DOMAIN environment variable in my session..
and now I need to run
sudo -E bash -c "source ./my_file.sh"
but the output does not display the value for $DOMAIN. instead it is empty.
if I change the command to be
sudo -E bash -c "echo $DOMAIN"
I see the correct value..
what am I doing wrong?
With the command line:
sudo -E bash -c "source ./my_file.sh"
you are running a script that may refer to environment variables that would need to be exported from a parent shell to be visible.
On the other hand:
sudo -E bash -c "echo $DOMAIN"
expands the value of $DOMAIN in the parent shell, not inside your sudo line.
To demonstrate this, try your "working" solution with single quotes:
sudo -E bash -c 'echo $DOMAIN'
And to make things go, try exporting the variable:
export DOMAIN
sudo -E bash -c "source ./my_file.sh"
Or alternately, pass $DOMAIN on the command line:
sudo -E bash -c "source ./my_file.sh $DOMAIN"
And have your script refer to $1.
What is a procedure to decorate an arbitrary bash command to execute it in a subshell? I cannot change the command, I have to decorate it on the outside.
the best I can think of is
>bash -c '<command>'
works on these:
>bash -c 'echo'
>bash -c 'echo foobar'
>bash -c 'echo \"'
but what about the commands such as
echo \'
and especially
echo \'\"
The decoration has to be always the same for all commands. It has to always work.
You say "subshell" - you can get one of those by just putting parentheses around the command:
x=outer
(x=inner; echo "x=$x"; exit)
echo "x=$x"
produces this:
x=inner
x=outer
You could (ab)use heredocs:
bash -c "$(cat <<-EOF
echo \'\"
EOF
)"
This is one way without using -c option:
bash <<EOF
echo \'\"
EOF
What you want to do is exactly the same as escapeshellcmd() in PHP (http://php.net/manual/fr/function.escapeshellcmd.php)
You just need to escape #&;`|*?~<>^()[]{}$\, \x0A and \xFF. ' and " are escaped only if they are not paired.
But beware of security issues...
Let bash take care of it this way:
1) prepare the command as an array:
astrCmd=(echo \'\");
2) export the array as a simple string:
export EXPORTEDastrCmd="`declare -p astrCmd| sed -r "s,[^=]*='(.*)',\1,"`";
3) restore the array and run it as a full command:
bash -c "declare -a astrCmd='$EXPORTEDastrCmd';\${astrCmd[#]}"
Create a function to make these steps more easy like:
FUNCbash(){
astrCmd=("$#");
export EXPORTEDastrCmd="`declare -p astrCmd| sed -r "s,[^=]*='(.*)',\1,"`";
bash -c "declare -a astrCmd='$EXPORTEDastrCmd';\${astrCmd[#]}";
}
FUNCbash echo \'\"