How to grave the Date from this command in bash script - bash

Question

!/bin/bash
# When a match is not found, just present nothing.
shopt -s nullglob
# Match all .wav files containing the date format.
files=(*[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]*.wav)
if [[ ${#files[#]} -eq 0 ]]; then
echo "No match found."
fi
for file in "${files[#]}"; do
# We get the date part by part
file_date=''
# Sleep it to parts.
IFS="-." read -ra parts <<< "$file"
for t in "${parts[#]}"; do
# Break from the loop if a match is found
if [[ $t == [0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9] ]]; then
file_date=$t
break
fi
done
# If a value was not assigned, then show an error message and continue to the next file.
# Just making sure there is nothing in Array and date before it moves on
if [[ -z $file_date ]]; then
continue
fi
file_year=${file_date:0:4}
file_month=${file_date:4:2}
mkdir -p "$file_year/$file_month"
# -- is just there to not interpret filenames starting with - as options.
echo "Moving: ./"$file "to: " "./"$file_year"/"$file_month
mv "$file" "$file_year/$file_month"
done
Now there are files I would need to do the date to gra the date and then move it like I do now. for example there is a file called meetme.. its a wav file and I have DIR with YYYY/MM and would like to move thoses files without YYYYMMDD in file name already

Answer 1

If you're writing a program to do something with that information, then you might prefer seconds-since-epoch and use date to get the date in the desired format.
$ date -d #$(stat --format='%Y' testdisk.log) +%Y%m%d
20130422
You can also get the ascii representation, and then manipulate the string.
$ stat --format='%y' testdisk.log
2013-04-22 09:11:39.000000000 -0500
$ date_st=$(stat --format='%y' testdisk.log)
$ date_st=${date_st/ */}
$ date_st=${date_st//-/}
$ echo ${date_st}
20130422