Extracting names from filename in Bash

Question

I have a directory filled with thousands of files in the format LastnameFirstnameYYYYMMDD.pdf. The last and first name will always been in title case.

I'd like to extract the last name so I can move these files to a directory structure of {first letter of last name}/lastname/full filename. Example: DoeJohn20190327 would be moved to D/Doe/DoeJohn20190327

I'm not sure if the final destination of the file DoeJohn20190327.pdf is like this: D/Doe/DoeJohn20190327/DoeJohn20190327.pdf or D/Doe/DoeJohn20190327/DoeJohn20190327.pdf ? — Philippe Delteil, Mar 29 '19 at 15:37

Philippe Delteil · Answer 1 · 2019-03-29T15:44:22.893

Here you have a solution. I tested it an it creates the folders as you explained.

for filename in *.pdf
do
  echo "Processing file $filename "
  first_letter="${filename:0:1}"
  mkdir -p $first_letter #if already exists won't print error
  last_name=$(echo $filename | sed 's/\([^[:blank:]]\)\([[:upper:]]\)/\1 \2/g'  |awk '{print $1}')
  mkdir -p $first_letter/$last_name
  mv $filename $first_letter/$last_name
done

score 4 · Answer 2 · answered Mar 29 '19 at 07:03

If the lastname is always the shortest trailing string staring with an upper case letter (there are no compound lastnames for example) you could use a shell parameter expansion of the form ${parameter%pattern} in place of a regex solution.

Ex.

for f in [[:upper:]]*[[:upper:]]*; do 
  d="${f:0:1}/${f%[[:upper:]]*}/"
  echo mkdir -p "$d"
  echo mv "$f" "$d"
done
mkdir -p D/Doe/
mv DoeJohn20190327 D/Doe/

Remove the echos when you are satisfied that it is doing the right thing.

See for example Parameter Expansion

Extracting names from filename in Bash

2 Answers2