Replace multiple underscores in filename using bash script

Question

Doing some reading here and here I found this solution to replace two underscores in filenames with only one using bash:

for file in *; do
  f=${file//__/_}
  echo $f
done;

However how do I most easily expand this expression to replace an arbitrary number of underscores with only one?

Answer 1

Typically, it's going to be faster to just put your original code in a loop than to do anything else.

for file in *; do
  f=$file
  while [[ $f = *__* ]]; do
    f=${f//__/_}
  done
  echo "$f"
done

Even better, if you're on a modern shell release, you can enable extended globs, which provide regex-like functionality:

shopt -s extglob
for file in *; do
  f=${file//+(_)/_}
  echo "$f"
done

Answer 2

You could use a simple regex using sed

for file in *; do
  f=$(echo "$file" | sed -e 's/_\+/_/')
  echo "$f"
done;

This regex matches one or more underscores (_\+) and substitutes them with only one (_)

Answer 3

GNU tr has --squeeze-repeats:

$ echo foo_______bar | tr --squeeze-repeats _
foo_bar

If you're using BSD tr you can use -s instead:

$ echo foo_______bar | tr -s _
foo_bar

Answer 4

This Shellcheck-clean pure shell code should work with any POSIX-compliant shell, including bash and dash:

for file in *; do
    while :; do
        case $file in
            *__*)   file=${file%%__*}_${file#*__};;
            *)      break;;
        esac
    done
    printf '%s\n' "$file"
done

• ${file%%__*} expands to $file with the first __ and all characters after it removed (e.g. a__b__c produces a).
• ${file#*__} expands to $file with all characters up to and including the first __ removed (e.g. a__b__c produces b__c).
• See the accepted, and excellent, answer to Why is printf better than echo? for an explanation of why printf '%s\n' "$file" is used instead of echo "$file".