How to name a file in the deepest level of a directory tree

Question

How do I name one random file in the deepest level of a directory tree using basic Bash commands and script?

I was searching for a way to do it as an one-liner and without functions. Assuming also that I'm starting in the current directory and then work towards the depth.

find, grep, sed and awk are acceptable commands.

My current attempt looks like this and does not work:

find -type d | declare COUNT=-1; declare P=""; while read LINE ; do echo $LINE; declare C; C=$(echo $LINE | cut -c3- | sed "s/\//\n\//g" | grep '/' -c); echo $C; if [ $COUNT -gt $C ]; then let COUNT=$C; let P=$LINE; echo "Done"; fi; done

This would only find the directory.

How could this be solved in the most simple way?

Answer 1

That's an odd request!

I'd use find + awk to grab a file in the deepest directory:

bash-3.2$ deepest=$(find / -type f | awk -F'/' 'NF > depth {
>     depth = NF;
>     deepest = $0;
> }
>
> END {
>     print deepest;
> }')

Using ${deepest} in your mv command is left as an exercise but the following five lines may help you further:

bash-3.2$ echo "${deepest}"
/Developer/SDKs/MacOSX10.6.sdk/System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/gems/1.8/gems/activesupport-2.3.5/lib/active_support/vendor/tzinfo-0.3.12/tzinfo/definitions/America/Argentina/Buenos_Aires.rb

bash-3.2$ echo "${deepest%.*}"
/Developer/SDKs/MacOSX10.6.sdk/System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/gems/1.8/gems/activesupport-2.3.5/lib/active_support/vendor/tzinfo-0.3.12/tzinfo/definitions/America/Argentina/Buenos_Aires

bash-3.2$ echo "${deepest%/*}"
/Developer/SDKs/MacOSX10.6.sdk/System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/gems/1.8/gems/activesupport-2.3.5/lib/active_support/vendor/tzinfo-0.3.12/tzinfo/definitions/America/Argentina

bash-3.2$ echo "${deepest##*/}"
Buenos_Aires.rb

bash-3.2$ echo "${deepest##*.}"
rb

---

Following update to question:

find -type d [...] "This would only find the directory. [...] How could this be solved in the most simple way?".

By supplying -type f to find to find all files (f), not all directories (d).

Answer 2

You are missing some curly-bracketing, you inverted the comparison, and you need to print the result:

find -type d | {
  declare COUNT=-1
  declare P=""
  while read LINE
    do echo $LINE
    declare C=$(echo $LINE | cut -c3- | sed "s/\//\n\//g" | grep '/' -c)
    echo $C
    if [ $C -gt $COUNT ]; then let COUNT=$C; let P=$LINE; echo "Done"; fi
  done
  echo deepest: "$P"
}

Slightly improved version, with debugging stuff thrown away:

find -type d -links 2 | (
  declare COUNT=-1
  P=""
  while IFS= read -r LINE; do
    declare C=$(echo $LINE | tr -cd / | wc -c)
    if [ $C -gt $COUNT ]; then let COUNT=$C; P=$LINE; fi
  done
  echo deepest: "$P"
)

Answer 3

find -type f | awk -F/ 'NF > maxdepth { maxdepth = NF; file = $0 }; END {print file}'

And it seems this is essentially the same as this answer on the other question you posted, what's wrong with that/this one?

Answer 4

I would go with these, although they will fail on file names containing leading spaces. The first outputs just the file name, the second includes the path to that file too:

find -type f | sed 's:[^/]*/: :g' | LC_ALL=C sort | head -1 | sed 's/^ *//'

find -type f | sed 'h;s:[^/]*/: :g;G;s/\n/\t/' | LC_ALL=C sort | head -1 | sed 's/.*\t//'

Answer 5

function step () {
    d=$1
    d=$((d+1))
    res=$(find -mindepth $d -type d ! -empty)
    test -n "$res" && step $d || echo $((d-1))
}

find -mindepth $(step 0) -type d ! -empty | head -n 1 

Recursively step deeper and deeper into directories, as long as they contain a file (! -empty). From the last step, we have to subtract 1, and we can then use the result in a second find command.