如何在Shell脚本中同时运行两个命令？

Question

I am trying to make a shell script run jeykll -w serve and sass --watch style.scss:style.css so I don't have to worry each time I want to develop locally. So I made a shell script, but obviously, if I put the jekyll command first, it won't run the sass one till jekyll is done. So how can run two commands at once? Do I have to make a command to open another tab in the terminal and then run sass? there must be a better way of doing that.

Answer 1

You are looking to fork the process so the current shell is not waiting on the command I assume.

jeykll -w serve && sass --watch style.scss:style.css

The above will wait for the first command to complete with an non-errored status

jeykll -w serve ; sass --watch style.scss:style.css

The above will wait for the first command to complete and disregard the exit status.

So if I understand correctly you want multiple commands to run near simultaneously. For this you use the & operator at the end of a command reference.

The Bash & (ampersand) is a builtin control operator used to fork processes. From the Bash man page, "If a command is terminated by the control operator &, the shell executes the command in the background in a subshell".

jeykll -w serve &
sass --watch style.scss:style.css &