how many files find found?

Question

I'm writing a script where I want to error out if the file I'm searching for exists in multiple locations, and tell the user the locations (the find results). So I've got a find like:

file_location=$(find $dir -name $file -print)

I'm thinking it should be simple to see if the file is found in multiple places, but I must not be matching what find uses to separate results with (seems like space sometimes, and a newline others). As such, rather than matching on that, I want to see if there are any characters after $file in $file_location.

I'm checking for

echo "$file_location" | grep -q "${file}."; then

and this still doesn't work. So I guess I don't care what I use, except I want to capture $file_location as a result of the find, and then check that. Can you suggest a good way?

Answer 1

Something like below if you want to avoid errors on eols and such

files=() 
while IFS= read -d $'\0' -r match; do 
  files+=("$match") 
done < <(find "$dir" -name "$file" -print0) 
(${#files[@]} > 1) && printf '%s\n' "${files[@]}"

Or in bash 4+

shopt -s globstar dotglob 
files=("$dir"/**/"$file") 
((${#files[@]} > 1)) && printf '%s\n' "${files[@]}"

Answer 2

If you specify the full name in the find command, the matches on name will be unique. That is, if you say find -name "hello.txt", just files named hello.txt will be found.

What you can do is something like

find $dir -name $file -printf '.'
                      ^^^^^^^^^^^

this will print as many . as matches are found. Then, to see how many files are found with this name it is just a matter of counting the number of dots you got as output.

Answer 3

found=$(find "$dir" -name "$file" -ls)
count=$(wc -l <<< "$found")
if [ "$count" -gt 1 ]
then
  echo "I found more than one:"
  echo "$found"
fi

For zero matches found you will still receive a 1 because of the intransparent way the shell strips a trailing newline with the $() operator, so effectively one line output and zero lines output are both one line in the end. See xxd <<< "" for demonstration of the automatic appending of a newline when used as input again. A simple way to circumvent this is to add a fake newline in the beginning of the string, so no empty string can happen: found=$(echo; find …), and then subtract one from the number of lines.

EDIT: I changed the usage of -printf "%p\n" in my answer to -ls which performs a proper quoting of newlines. Otherwise file names with newlines in them would mess up the counting.

Answer 4

No need for find here if you're running a new (4.0+) bash which can do recursive globbing itself; just load glob results directly into a shell array, and check its length:

shopt -s nullglob globstar # enable recursive globbing, and null results
file_locations=( "$dir"/**/"$file" )
echo "${#file_locations[@]} files named $file found under $dir; they are:"
printf '  %q\n' "${file_locations[@]}"

If you don't want to mess with nullglob, then:

shopt -s globstar # enable recursive globbing
file_locations=( "$dir"/**/"$file" )

# without nullglob, a failed match will return the glob expression itself
# to test for this, see if our first entry exists
if [[ -e ${file_locations[0]} ]]; then
  echo "No instances of $file found under $dir"
else
  echo "${#file_locations[@]} files named $file found under $dir; they are:"
  printf '  %q\n' "${file_locations[@]}"
fi

---

You can still use an array to unambiguously read find results on old versions of bash; unlike more naive approaches, this will work even when file or directory names contain literal newlines:

file_locations=( )
while IFS= read -r -d '' filename; do
  file_locations+=( "$filename" )
done < <(find "$dir" -type f -name "$file" -print0)
echo "${#file_locations[@]} files named $file found under $dir; they are:"
printf '  %q\n' "${file_locations[@]}"

Answer 5

I recommend using:

find . -name blong.txt -print0

Which tells find to join its output together with null \0 characters. Makes it easier to use awk with the -F flag or xargs with the -0 flag.

Answer 6

Try:

N=0

for i in `find $dir -name $file -printf '. '`
do
    N=$((N+1))
done

echo $N