How can I search for a dot an a number in sed or awk and prefix the number with a leading zero? - awk

Question

I am trying to modify the name of a large number of files, all of them with the following structure:
4.A.1 Introduction to foo.txt
2.C.3 Lectures on bar.pdf
3.D.6 Processes on baz.mp4
5.A.8 History of foo.txt
And I want to add a leading zero to the last digit:
4.A.01 Introduction to foo.txt
2.C.03 Lectures on bar.pdf
3.D.06 Processes on baz.mp4
5.A.08 History of foo.txt
At first I am trying to get the new names with sed (FreeBSD implementation):
ls | sed 's/\.[0-9]/0&/'
But I get the zero before the .
Note: replacing the second dot would also work. I am also open to use awk.

Answer 1

While it may have worked for you here, in general slicing and dicing ls output is fragile, whether using sed or awk or anything else. Fortunately one can accomplish this robustly in plain old POSIX sh using globbing and fancy-pants parameter expansions:
for f in [[:digit:]].[[:alpha:]].[[:digit:]]\ ?*; do
# $f = "[[:digit:]].[[:alpha:]].[[:digit:]] ?*" if no files match.
if [ "$f" != '[[:digit:]].[[:alpha:]].[[:digit:]] ?*' ]; then
tail=${f#*.*.} # filename sans "1.A." prefix
head=${f%"$tail"} # the "1.A." prefix
mv "$f" "${head}0${tail}"
fi
done
(EDIT: Filter out filenames that don't match desired format.)

Answer 2

This pipeline should work for you:
ls | sed 's/\.\([0-9]\)/.0\1/'
The sed command here will capture the digit and replace it with a preceding 0.
Here, \1 references the first (and in this case only) capture group - the parenthesized expression.

Answer 3

I am also open to use awk.
Let file.txt content be:
4.A.1 Introduction to foo.txt
2.C.3 Lectures on bar.pdf
3.D.6 Processes on baz.mp4
5.A.8 History of foo.txt
then
awk 'BEGIN{FS=OFS="."}{$3="0" $3;print}' file.txt
outputs
4.A.01 Introduction to foo.txt
2.C.03 Lectures on bar.pdf
3.D.06 Processes on baz.mp4
5.A.08 History of foo.txt
Explanation: I set dot (.) as both field seperator and output field seperator, then for every line I add leading 0 to third column ($3) by concatenating 0 and said column. Finally I print such altered line.
(tested in GNU Awk 5.0.1)

Answer 4

This might work for you (GNU sed):
sed 's/^\S*\./&0/' file
This appends 0 after the last . in the first string of non-empty characters in each line.

Answer 5

In case it helps somebody else, as an alternative to #costaparas answer:
ls | sed -E -e 's/^([0-9][.][A-Z][.])/\10/' > files
To then create the script the files:
cat files | awk '{printf "mv \"%s\" \"%s\"\n", $0, $0}' | sed 's/\.0/\./' > movefiles.sh