ls - how to output only filename, date (ISO 8601) to CSV?

Question

I have a number of files in a directory:

img1_1.jpg, img1_2.jpg ... img1_199.jpg, img1_2000.jpg

I would like to output a listing of these files to a csv containing only: filename, datetime

For example: img1_116.jpg,2011-05-25 22:00:49.000000000 +0000

I am trying:

ls -l --time-style="full-iso" img1* | awk '/^-/ && $1=$1' OFS=","

But the date/time keeps getting split up due to the space in the date/time format:

img1_116.jpg,2011-05-25,22:00:49.000000000,+0000

(This is just an example, my actual output still contains the rights, user, group etc)

Is there a way to:

1. Export the CSV without splitting the date/time format?
2. Have ls and select only filename and date/time?

Answer 1

Does something like ls -l --time-style=full-iso | awk '/^-/ {printf "%s,%s %s\n",$NF,$6,$7}' do what you want?