Escaping problem with find command

I need to find everything in a directory, excluding certain subdirectories and files. My script needs to call this as a function:

function findStuff()
# define exclusions
ignore_dirs=("$1" "*foo*") # exclude base dir
ignore_files=("one.txt" "two.txt" "*three*.txt")
# build patterns for find command
dir_pattern=""
file_pattern=""
for i in "$ignore_dirs[@]"; do dir_pattern=$dir_pattern" ! -path "$i""; done
for i in "$ignore_files[@]"; do file_pattern=$file_pattern" ! -name "$i""; done
# find
find "$1 $dir_pattern $file_pattern"
# now do other stuff with the results...


findStuff /some/base/dir


But this gives me a No such file or directory error.

No such file or directory

So I wanted to see what the command actually was and tried echo find "$1 $dir_pattern $file_pattern" and pasted that on the command line and it worked. Then I pasted that into the script and run it, and it also worked!

echo find "$1 $dir_pattern $file_pattern"

So i think it's failing because of some escaping problem. What have I done wrong?

1 Answer
1

find will use the first arguments (up to the first argument that starts with - or that is ! or () that it gets as the top-level paths to search. You are giving find a single argument when you call it in your function, the string $1 $dir_pattern $file_pattern (with the variables expanded). This path is not found.

find

-

!

(

find

$1 $dir_pattern $file_pattern

You also include literal double quotes in the arguments that you intend do give to find. Double quoting is done to prevent the shell from expanding glob patterns and from splitting on whitespaces (or whatever the IFS variable contains), but if you use e.g. ! -name "thing" then the double quotes would be part of the pattern that find uses to compare against the filenames.

find

IFS

! -name "thing"

find

Use arrays, and quote the separate arguments to find properly:

find

myfind ()
local ignore_paths=( "$1" "*foo*" )
local ignore_names=( "one.txt" "two.txt" "*three*.txt" )

local path_args=()
for string in "$ignore_paths[@]"; do
path_args+=( ! -path "$string" )
done

local name_args=()
for string in "$ignore_names[@]"; do
name_args+=( ! -name "$string" )
done

find "$1" "$path_args[@]" "$name_args[@]"



Each time we append to path_args and name_args above, we add three elements to the list, !, -path or -name, and "$string". When expanding "$path_args[@]" and "$name_args[@]" (note the double quotes), the elements will be individually quoted.

path_args

name_args

!

-path

-name

"$string"

"$path_args[@]"

"$name_args[@]"

Equivalent implementation suitable for /bin/sh:

/bin/sh

myfind () (
topdir=$1

set --

# paths to ignore
for string in "$topdir" "*foo*"; do
set -- "$@" ! -path "$string"
done

# names to ignore
for string in "one.txt" "two.txt" "*three*.txt"; do
set -- "$@" ! -name "$string"
done

find "$topdir" "$@"
)


In the sh shell we only have a single array available to us, which is the list of positional parameters, $@, so we collect our find options in that. The bash-specific solution could also be written to use a single array, obviously, and the sh variation would run in bash too.

sh

$@

find

bash

sh

bash

And lastly, the output of your echo test is not an accurate representation of the command that would have been executed by your function.

echo

Consider this:

cat "my file name"


which runs cat on something called my file name, and

cat

my file name

echo cat "my file name"


which outputs the string cat my file name. This is due to the fact that the shell removes quotes around strings before executing the command. Running that command, cat would look for three files, not one.

cat my file name

cat

Your command worked well when you copy-pasted it into the shell, because you included the literal double quotes in the string that was outputted by echo (by escaping them), but that was not the actual command executed by your function.

echo

-name "somefile.txt"

-name 'somefile.txt'

-name

-path

"$string"

find

find

$string

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