How do I run multiple background commands in bash in a single line?

Question

I normally run multiple commands with something like this:

sleep 2 && sleep 3

or

sleep 2 ; sleep 3

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Answer 1

You can use the bash command substitution $(command) like this:

$(command1 ; command2) &

Note that stdin and stdout are still linked to the parent process and redirecting at least the stdout can be tricky. So alternatively you can chain the commands in a single line then pass the string to the bash command to spawn a new process which will handle the execution.

bash -c "command1 ; command2" & 

This is especially useful in a bash script when you need to run multiple commands in background.

This two statements should be equivalent. A bash process is spawn in both cases to handle the command (chain of commands) and the & at the end detaches the execution.

This time you can add &>/dev/null before the & at the end of the command to redirect at least the stdout and avoid the output on the stdout of the parent process. Something like:

bash -c "command1 ; command2" &>/dev/null &