Extract csv files from path to multiple files in bash

Question

I have a files multiple file path for backup purpose ,now i want to filter it based on ".csv " fromat. I have tried this.

filesall=(/home/abc/allfiles/*)
files=$((${filesall[@]}|grep -F ".csv"))
files=$(${filesall[@]}|grep -oe "*.csv")

but unable to extract .csv files from $filesall . I know i can achieve it like this

files=(/home/abc/allfiles/*.csv)

any suggestion ,how can i get all .csv files from filesall.Thanks.

Answer 1

This is really not the best way to do this. Why parse the array? A simpler approach would be something like:

path=/home/abc/allfiles
filesall=(${path}/*)
files=(${path}/*.csv)

If you insist on doing it your way, you would have to do something like:

files=($(for file in "${filesall[@]}"; do [[ $file =~ \.csv$ ]] && echo $file; done))

or

files=($(printf "%s\n" "${filesall[@]}" | grep '\.csv$'))

But both of the above break if any of your file names contain whitespace. (They can be made to work with spaces if you precede them with saveIFS="$IFS"; IFS=$'\n' and follow them with IFS="$saveIFS", but they still break on newline.)

Answer 2

Building on terdon’s answer, here’s one that seems to work:

filesall=(all files)
files=(); for file in "${filesall[@]}"; do [[ $file =~ \.csv$ ]] && files+=("$file"); done