Use generated argument array/string in find

Question

I want to pass a generated argument array (or string) to a find command in bash, but everything I tried doesn't work. It seems to be something with the quoting, but I don't know how to solve this.

Here is what I tried.

#!/bin/bash
path="/path/to/folders/"
excludes=()
excludes+=(" -not -path \"./cache/*\"")
excludes+=(" -not -path \"./tmp/*\"")
find $path -type f \( "${excludes[@]}" \) >test.txt

I just get this message find: Der Pfad muß vor dem Suchkriterium stehen: -not -path "./cache/*"

anubhava · Accepted Answer · 2016-01-04 19:36:42Z

3

You don't need to quote multiple arguments, try:

path="/path/to/folders/"
excludes=( -not -path "./cache/*" -not -path "./tmp/*" )
find "$path" -type f \( "${excludes[@]}" \) >test.txt

answered Jan 4, 2016 at 19:36

anubhava

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2 Comments

John Bollinger Over a year ago

More precisely, he must not quote multiple arguments in a manner that combines them into fewer arguments. Doing so is the essential problem of his attempt. He can still build his excludes array in multiple steps, though.

anubhava Over a year ago

That is exactly what I meant by multiple arguments, doing that in multiple steps using += is not a problem.

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