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bashquoting

How do I use a variable to specify a path to exclude in `find`?

I'm trying to use a variable to specify an exclusion path in the find command.

This snippet works:

x="*test/.util/*"
y="*test/sh/*"

find test -type f -name "*.sh" -not -path "$x" -not -path "$y"

But I'd like to move -not -path into the variables, as in:

x="-not -path *test/.util/*"
y="-not -path *test/sh/*"

# error: find: -not -path *test/.util/*: unknown primary or operator
find test -type f -name "*.sh" "$x" "$y"

# Tried removing quotes
# error: find: test/.util/foo.sh: unknown primary or operator
find test -type f -name "*.sh" $x $y

I also tried adding quotes to the paths in the variables, but that results in no path filtering.

# no syntax error but does not exclude the paths
x="-not -path '*test/.util/*'"
y="-not -path '*test/sh/*'"

I'm using OSX Mavericks; GNU bash, version 3.2.51(1)-release (x86_64-apple-darwin13).

What am I doing wrong? Thanks

Solution

  • find is receiving -not -path *test/.util/* as a single argument, not as 3 separate arguments that it requires. You can use an array instead.

    x=(-not -path "*test/.util/*")
y=(-not -path "*test/sh/*")

find test -type f -name "*.sh" "${x[@]}" "${y[@]}"

    When quoted, ${x[@]} expands into a series of separate words, one per array element, and each word remains properly quoted, so the patterns are passed literally, as intended, to find.