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I'm trying to run a command remotely with SSH, but my problem is that when I do that, my .basrhc doesn't get executed, and so the PATH is all wrong and my script is breaking.

What I'm doing is

ssh www.remotehost.com 'my_command'

I've also tried things like

ssh www.remotehost.com 'source ~/.bashrc; my_command' and similar variants, but nothing seems to work.

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2 Answers 2

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SSH with a single, simple command argument will not start a shell, which means .bashrc does not get executed.

SSH with a compound command argument such as your second example should start a shell, but it will be a non-interactive one, so that won't run .bashrc by default. See Bash Startup Files in the Bash manual for more information.

However, calling 'source' in your second example command should work. Does your .bashrc have a line like if [ "$PS1" ]; then or if [ -z "$PS1" ]; then at the top? The sample bashrc does have one of these lines, and it prevents the script commands from running unless you are on an interactive shell. As mentioned above, SSH does not start an interactive shell by default when you pass a command argument. Try ssh -t www.remotehost.com 'source ~/.bashrc; my_command', as this will force TTY allocation and act like an interactive shell.

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  • if [ "$PS1" ]; then or if [ -z "$PS1" ]; then did it. In my case it was [ -z "$PS1" ] && return just added source ~/.my_bash_stuff at the top
    – natanavra
    Commented Aug 26, 2019 at 13:06
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Zancheys Post gives you the right solution.

An other aproach could be to provide a simple shell script like following.

#!/bin/sh
# program name: /usr/local/bin/my_command_ssh
#
export PATH=/home/custom/bin:/usr/local/bin:/usr/bin:/bin
my_command

And now you can call it via ssh:

ssh -p22 [email protected] '/usr/local/bin/my_command_ssh'
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