list files with the same first 3 letter followed by a sequence # with shell script

Question

First time to post here. How do I group files with the same first 3 letters followed by a sequence number in a directory? e.g.

VHS-01-001.avi
VHS-01-002.avi
-------------
VHS-02-001.avi
VHS-02-002.avi
VHS-02-003.avi
---------
Hi8-01-001.avi
Hi8-01-002.avi
Hi8-01-003.avi

so that I can pass each group of video files in a function as the following:

encode(){
  for avi in "$@" 
  do..
}

Answer 1

Maybe not very smart solution:

• sort files by name
• loop through names

• compare characters of last loop:

  last=""
  ls -1 $1 | sort | while read file; do
      sub=${file:0:3}
      [ "$last" != "$sub" ] && { echo "NEW GROUP"; last="$sub"; }
      echo "[$sub] $file"
  done
  

Instead of echo-ing collect filenames inside an array ...

Just an idea ... example output:

NEW GROUP                                                                                                                                                                                                                                    
[Hi8] Hi8-01-002.avi                                                                                                                                                                                                                         
NEW GROUP                                                                                                                                                                                                                                    
[VHS] VHS-01-001.avi                                                                                                                                                                                                                         
[VHS] VHS-01-002.avi
[VHS] VHS-02-002.avi
NEW GROUP
[XZU] XZU

Edit 1: based on Anthony Geoghegan 's answer avoid the pipes at the beginning of the loop and use bash globbing. Take a look at his comment.

Improved script:

last=""
for file in *avi; do
    sub=${file:0:3}
    [ "$last" != "$sub" ] && { echo "NEW GROUP"; last="$sub"; }
    echo "[$sub] $file"
done

Edit 2:

as asked by @ Tony Tan in his third comment: here you find a straigt forward solution to parse the collected file names to a function. There are many ways to do so. And I don't have much experience in bash scripting ... ;)

#!/bin/bash

SOURCE_DIR="$1"
cd "$SOURCE_DIR" || { echo "could not read dir '$SOURCE_DIR'"; exit 1; }

function parseFiles() {
  echo "parsing files:"
  echo "$1"
}

last=""
declare -a fileGroup

for file in *avi; do
  # first 3 chars of filename
  sub=${file:0:3}

  if test -z "$last"; then
    # last is empty. first loop
    last="$sub"
  elif test "$last" != "$sub"; then
    # new file group detected, parse collected
    parseFiles "${fileGroup[*]}"
    # reset array
    fileGroup=()
    last="$sub"
  fi

  # append name to array
  fileGroup[${#fileGroup[@]}]=$file
done

parseFiles "${fileGroup[*]}"

Answer 2

With zsh:

files=(???-??-*.avi)
for prefix (${(Mu)files#???-??-}) encode $prefix*.avi

(or encode ${(M)files:#$prefix*})

The equivalent with the GNU shell (bash) and tools would be:

while IFS= read -u3 -rd '' prefix; do
  encode "$prefix-"*.avi 3<&-
done 3< <(printf '%s\0' ???-??-*.avi | grep -oz '^...-..-' | sort -zu)

Same principle. We get the list of files matching the ???-??-*.avi pattern in the current directory, extract the part that matches ((M)/grep -o) ???-??- (regexp ...-..-), unique them ((u)/sort -u), and then loop over that list of prefixes.