How to execute a remote command over ssh with arguments?
In my .bashrc I define a function which I can use on the command line later:
function mycommand() {
ssh [email protected] cd testdir;./test.sh "$1"
}
When using this command, just the cd command is executed on the remote host; the test.sh command is executed on the local host. This is because the semicolon separates two different commands: the ssh command and the test.sh command.
I tried defining the function as follows (note the single quotes):
function mycommand() {
ssh [email protected] 'cd testdir;./test.sh "$1"'
}
I tried to keep the cd command and the test.sh command together, but the argument $1 is not resolved, independent of what I give to the function. It is always tried to execute a command
./test.sh $1
on the remote host.
How do I properly define mycommand, so the script test.sh is executed on the remote host after changing into the directory testdir, with the ability to pass on the argument given to mycommand to test.sh?
Answer
Do it this way instead:
function mycommand {
ssh [email protected] "cd testdir;./test.sh \"$1\""
}
You still have to pass the whole command as a single string, yet in that single string you need to have $1 expanded before it is sent to ssh so you need to use "" for it.
Update
Another proper way to do this actually is to use printf %q to properly quote the argument. This would make the argument safe to parse even if it has spaces, single quotes, double quotes, or any other character that may have a special meaning to the shell:
function mycommand {
printf -v __ %q "$1"
ssh [email protected] "cd testdir;./test.sh $__"
}
- When declaring a function with
function,()is not necessary. - Don't comment back about it just because you're a POSIXist.