Bash how to include a command that has quotation marks as a variable?

Question

In my bash script, in several places I need to run the command

rsync [stuff] --exclude={"/mnt/*","/proc/*"} [source] [destination]

To avoid typing out the whole list, I want to pack the option --exclude={"/mnt/*","/proc/*"} into a variable called EXCLUDES such that I can type in my script:

rsync [stuff] "$EXCLUDES" [source] [destination]

What is the proper way to achieve this?

Possible duplicate of Bash pass variable as argument with quotes — Jean-François Fabre
– Jean-François Fabre ♦, Commented Nov 5, 2016 at 14:41
There is no problem: $EXCLUDES can expand into something which contains quotation marks; these are not processed in relation to the quotes around "$EXCLUDES". The $VAR syntax cannot perpetrate a "close quote injection attack" against the surrounding syntax. — Kaz
– Kaz, Commented Nov 5, 2016 at 14:46

chepner · Accepted Answer · 2016-11-05 15:38:48Z

2

Use an array:

EXCLUDES=(
  --exclude="/mnt/*"
  --exclude="/proc/*"
)
rsync [stuff] "${EXCLUDES[@]}" [source] [destination]

or a here-doc with the --exclude-from option:

rsync [stuff] --exclude-from - [source] [destination] <<EOF
/mnt/proc/*
/proc/*
EOF

I would recommend against using brace expansion in a script. Your text editor should make it easy to quickly duplicate repeated strings, and the the resulting script will be more readable. Brace expansions are intended for interactive use.

edited Nov 5, 2016 at 15:38

answered Nov 5, 2016 at 15:24

chepner

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