Remove middle of name in bash

Question

I have 300+ files named:

T1_0000106_FS1_MAX_5743.nii.gz  T1_0000214_FS1_MAX_5475.nii.gz
T1_0000107_FS1_MAX_5477.nii.gz  T1_0000215_FS1_MAX_6162.nii.gz

I would like to remove everything between T1 and _5/6*.nii.gz so:

T1_5743.nii.gz  T1_5475.nii.gz
T1_5477.nii.gz  T1_6162.nii.gz 

I can't figure out why it isn't working; I tried (from another post):

 for file in *.gz;
 do new_name=$(sed 's/_[^.]*_/_g' <<< "new_name");
 mv "$file" "$new_name"; done

and variations of rename/sed but nothing changes.

Answer 1

Problems with your script include, at least,

• s/_[^.]*_/_g is not a valid sed command. You appear to want s/_[^.]*_/_/g, or in this case, s/_[^.]*_/_/ would do fine as well.

• <<< "new_name" redirects the literal string new_name into sed. Possibly you mean <<< "$new_name"

Personally, though, I would not bother with sed for this job, especially if you have a large number of files. Bash is perfectly capable of doing fairly complex string manipulation itself, and your needs don't push it too hard. Consider:

for f in *.gz; do
    # The value of $f with the longest trailing match to _* removed
    head=${f%%_*}

    # The value of $f with the longest leading match to *_ removed
    tail=${f##*_}

    new_name="${head}_${tail}"

    # Sanity check and avoid needless errors
    if [ "$f" != "$new_name" ]; then
        mv "$f" "$new_name"
    fi
done