在shell脚本中包含一大块代码

Question

I have a number of shell scripts that all look like this:

#!/bin/bash

cd ~/Dropbox/cms_sites/examplesite/media
sass -C --style compressed --update css:css

cd ~/Dropbox/cms_sites/examplesite

rm -f ./cache/*.html
rm -fr ./media/.sass-cache/
rm -fr ./admin/media/.sass-cache/


rsync -auvzhL . username@host:/home/username/remote_folder

(I know the use of cd seems weird, but they have evolved!)

Now, all these scripts have a few differences, in that they have different usernames, hosts, local folder and remote folder names, and I want an inexperienced user to be able to run them without arguments (so he can drag and drop them into a terminal without issue).

What I'd like to do is something like:

#!/bin/bash

cd ~/Dropbox/cms_sites/examplesite/media
sass -C --style compressed --update css:css

cd ~/Dropbox/cms_sites/examplesite

include ~/scripts/common.sh

rsync -auvzhL . username@host:/home/username/remote_folder

then have a file in common.sh that looks like:

rm -f ./cache/*.html
rm -fr ./media/.sass-cache/
rm -fr ./admin/media/.sass-cache/

so that I can easily change sections of the code in lots of scripts at once.

Is this possible, or is there a better way to do this without using arguments and having one script?

Answer 1

Use the source command. It's bash's version of 'include'

Answer 2

No need for "include" if the script is executable:

~/scripts/common.sh

If the script is not executable or does not have an appropriate shebang line then you'll need to specify the interpreter:

bash ~/scripts/common.sh