This question already has answers here:
Usage of expect command within a heredoc
(1 answer)
Pass commands as input to another command (su, ssh, sh, etc)
(3 answers)
Closed 4 years ago.
I have a script that I need to execute through ssh as another use, is there a way to pass whole script like this:
ssh -t user#server.com sudo -u user2 sh -c << EOF
cd /home
ls
dir=$(pwd)
echo "$dir"
echo "hello"
....
EOF
Returns: sh: -c: option requires an argument
ssh'ing and sudo'ing separately is not an option and putting .sh file directly on the machine is not possible.
sh -c requires a command string as the argument. Since you are reading the commands from standard input (through heredoc), you need to use sh -s option:
ssh -t user#server.com sudo -u user2 sh -s << 'EOF'
cd /home
ls
dir=$(pwd)
echo "$dir"
echo "hello"
...
EOF
From man sh:
-c
string If the -c option is present, then commands are read from string. If there are arguments after the string, they are assigned to the positional parameters, starting with $0.
-s
If the -s option is present, or if no arguments remain after option processing, then commands are read from the standard input. This option allows the positional parameters to be set when invoking an interactive shell.
Need to quote the heredoc marker to prevent the parent shell from interpreting the content.
Related
As the title says, within linux how can I feed input to the bash when I do sudo bash
Lets say I have a bash script that reads the name.
The way I execute the script is through sudo using:
cat read-my-name-script.sh | sudo bash
Lets just say this is how I execute the script throught the network.
Now I want to fill the name automatically, is there a way to feed the input. I tried doing this: cat read-my-name-script.sh < name-input-file | sudo bash where the name-input-file is a file for the input that the user will be using to feed the script.
I am new to linux and learning to automate the input and wanted to create a file for input where the user can fill it and feed it to my script.
This is convoluted, but might do what you want.
sudo bash -c "$(cat read-my-name.sh)" <name-input-file
The -c says the next quoted argument are the commands to run (so, read the script as a string on the command line, instead of from a file), and the calling shell interpolates the contents of the file inside the double quotes before the sudo command gets evaluated. So if read-my-name.sh contains
#!/bin/bash
read -p "I want your name please"
then the command gets expanded into
sudo bash -c '#!/bin/bash
read -p "I want your name please"' <name-input-file
(where of course at this time the shell has actually removed the outer double quotes altogether; I put in single quotes in their place instead to show how this would look as actually executable, syntactically valid code).
I think you need that:
while read -r arg; do sudo bash read-my-name-script.sh "$arg";done <name-input-file
So each line of name-input-file will be passed as argument to sudo bash read-my-name-script.sh
If your argslist located on http server, you can do that:
while read -r arg; do sudo bash read-my-name-script.sh "$arg";done < <(wget -q -O- http://some/address/in/internet/name-input-file)
UPD
add [[ -f name-input-file ]] && readarray -t args <name-input-file
to read-my-name-script.sh
and use "${args[#]}" as arguments of command in the script.
For example echo "${args[#]}" or cmd "${args[0]}" "${args[1]}" ... "${args[100]}" in any order.
In this case you can use
wget -q -O- http://some/address/in/internet/read-my-name-script.sh | bash
for run your script with arguments from name-input-file whitout saving script to the local machine
I download some shell script from example.com with wget and execute it immediately by streaming stdout of wget via a pipe to stdin of the sh command.
wget -O - http://example.com/myscript.sh | sh -
How can I pass arguments to the script?
You need to use -s option while invoking bash for passing an argument to the shell script being downloaded:
wget -O - http://example.com/myscript.sh | bash -s 'arg1' 'arg2'
As per man bash:
-s If the -s option is present, or if no arguments remain after option processing,
then commands are read from the standard input. This option allows
the positional parameters to be set when invoking an interactive shell.
While the accepted answer is correct, it does only work on bash and not on sh as the initial poster requested.
To do this in sh you'll have to add --:
curl https://example.com/script.sh | sh -s -- --my-arg
This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 6 years ago.
I'm trying to write a database call from within a bash script and I'm having problems with a sub-shell stripping my quotes away.
This is the bones of what I am doing.
#---------------------------------------------
#! /bin/bash
export COMMAND='psql ${DB_NAME} -F , -t --no-align -c "${SQL}" -o ${EXPORT_FILE} 2>&1'
PSQL_RETURN=`${COMMAND}`
#---------------------------------------------
If I use an 'echo' to print out the ${COMMAND} variable the output looks fine:
echo ${COMMAND}
screen output:-
#---------------
psql drupal7 -F , -t --no-align -c "SELECT DISTINCT hostname FROM accesslog;" -o /DRUPAL/INTERFACES/EXPORTS/ip_list.dat 2>&1
#---------------
Also if I cut and paste this screen output it executes just fine.
However, when I try to execute the command as a variable within a sub-shell call, it gives an error message.
The error is from the psql client to the effect that the quotes have been removed from around the ${SQL} string.
The error suggests psql is trying to interpret the terms in the sql string as parameters.
So it seems the string and quotes are composed correctly but the quotes around the ${SQL} variable/string are being interpreted by the sub-shell during the execution call from the main script.
I've tried to escape them using various methods: \", \\", \\\", "", \"" '"', \'"\', ... ...
As you can see from my 'try it all' approach I am no expert and it's driving me mad.
Any help would be greatly appreciated.
Charlie101
Instead of storing command in a string var better to use BASH array here:
cmd=(psql ${DB_NAME} -F , -t --no-align -c "${SQL}" -o "${EXPORT_FILE}")
PSQL_RETURN=$( "${cmd[#]}" 2>&1 )
Rather than evaluating the contents of a string, why not use a function?
call_psql() {
# optional, if variables are already defined in global scope
DB_NAME="$1"
SQL="$2"
EXPORT_FILE="$3"
psql "$DB_NAME" -F , -t --no-align -c "$SQL" -o "$EXPORT_FILE" 2>&1
}
then you can just call your function like:
PSQL_RETURN=$(call_psql "$DB_NAME" "$SQL" "$EXPORT_FILE")
It's entirely up to you how elaborate you make the function. You might like to check for the correct number of arguments (using something like (( $# == 3 ))) before calling the psql command.
Alternatively, perhaps you'd prefer just to make it as short as possible:
call_psql() { psql "$1" -F , -t --no-align -c "$2" -o "$3" 2>&1; }
In order to capture the command that is being executed for debugging purposes, you can use set -x in your script. This will the contents of the function including the expanded variables when the function (or any other command) is called. You can switch this behaviour off using set +x, or if you want it on for the whole duration of the script you can change the shebang to #!/bin/bash -x. This saves you explicitly echoing throughout your script to find out what commands are being run; you can just turn on set -x for a section.
A very simple example script using the shebang method:
#!/bin/bash -x
ec() {
echo "$1"
}
var=$(ec 2)
Running this script, either directly after making it executable or calling it with bash -x, gives:
++ ec 2
++ echo 2
+ var=2
Removing the -x from the shebang or the invocation results in the script running silently.
This question already has answers here:
How to run two commands with sudo?
(11 answers)
Closed 7 years ago.
Can we use heredocs to run multiple commands using sudo?
I am facing an issue (need to pass password for every command) while running
multiple commands:
echo 'password'| sudo -S ls
echo 'password'| sudo -S cat /var/log/abc.log
Can anyone help me how to automate this in a script? Like:
echo 'password' | sudo -S << ENDBLOCK
ls
cat
ENDBLOCK
you can run sh -c ..., but remember to quote properly.
sudo sh -c 'id; echo another command ; id'
sudo must see this as a single argument for the sh command.
Of course you can use new line instead of semicolon:
sudo sh -c '
echo "I am root"
id
echo "another command"
id
'
One way is to write a script with all the things you need to do with sudo and then run the script with sudo.
you could put all your commands in a script. Then
sudo ./script.sh
put permissions for script.sh in /etc/sudoers.d; that way you'll never need to type your password again (for that script)
I'm trying to write a shell script that calls another script that then executes a rsync command.
The second script should run in its own terminal, so I use a gnome-terminal -e "..." command. One of the parameters of this script is a string containing the parameters that should be given to rsync. I put those into single quotes.
Up until here, everything worked fine until one of the rsync parameters was a directory path that contained a space. I tried numerous combinations of ',",\",\' but the script either doesn't run at all or only the first part of the path is taken.
Here's a slightly modified version of the code I'm using
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l '\''/media/MyAndroid/Internal storage'\''' "
Within Backup.sh this command is run
rsync $5 "$path"
where the destination $path is calculated from text in Stamp.
How can I achieve these three levels of nested quotations?
These are some question I looked at just now (I've tried other sources earlier as well)
https://unix.stackexchange.com/questions/23347/wrapping-a-command-that-includes-single-and-double-quotes-for-another-command
how to make nested double quotes survive the bash interpreter?
Using multiple layers of quotes in bash
Nested quotes bash
I was unsuccessful in applying the solutions to my problem.
Here is an example. caller.sh uses gnome-terminal to execute foo.sh, which in turn prints all the arguments and then calls rsync with the first argument.
caller.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh 'long path' arg2 arg3"
foo.sh:
#!/bin/bash
echo $# arguments
for i; do # same as: for i in "$#"; do
echo "$i"
done
rsync "$1" "some other path"
Edit: If $1 contains several parameters to rsync, some of which are long paths, the above won't work, since bash either passes "$1" as one parameter, or $1 as multiple parameters, splitting it without regard to contained quotes.
There is (at least) one workaround, you can trick bash as follows:
caller2.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh '--option1 --option2 \"long path\"' arg2 arg3"
foo2.sh:
#!/bin/bash
rsync_command="rsync $1"
eval "$rsync_command"
This will do the equivalent of typing rsync --option1 --option2 "long path" on the command line.
WARNING: This hack introduces a security vulnerability, $1 can be crafted to execute multiple commands if the user has any influence whatsoever over the string content (e.g. '--option1 --option2 \"long path\"; echo YOU HAVE BEEN OWNED' will run rsync and then execute the echo command).
Did you try escaping the space in the path with "\ " (no quotes)?
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l ''/media/MyAndroid/Internal\ storage''' "