Can I specify a custom command before launching interactive bash session? [duplicate]

Question

Please note: Although this question is essentially the same as this linked question, I did not mark it as "solving my problem" because the answers provided there are not very satisfactory. The answer I accepted below is cleaner and simpler. Take a look!

I know how to pass bash a custom command to execute, e.g. like this:

bash -c "/bin/echo hello"

When I do this, bash executes the argument of -c and then exits. How can I get it to give me an interactive prompt when I'm done? Adding -i has no effect. I don't want to write the custom commands into a file that .bashrc always reads.

I've been making do with bash -c "commands; exec bash -i", but it's not ideal since it starts bash twice. Is there a way to get the same effect while starting bash only once?

Answer 1

You can use the --init-file option like this bash --init-file <(echo "echo hello"). This will not source your .bashrc file at all. If you want the .bashrc file sourced you can add that like this bash --init-file <(echo ". ~/.bashrc; echo hello"). Or you can put all the commands you want to execute in a file and use bash --init-file file_with_commands