How to extract the CSV file name from each line of this file

Question

I have a file which contains lines like those:

-rw-rw-rw 1 root root 6379 May 24 2016 test1.CSV 
-rw-rw-rw 1 root root 23249 May 25 2016 test2.CSV 
-rw-rw-rw 1 root root 2995 May 26 2016 test3.CSV

and I need to keep just the CSV file names. I have to use a regex but I don't know which one.

The expected result is "test1.CSV \n test2.CSV \n test3.CSV".

Have you tried something yourself already? Could you just use ls instead of ls -l? — Sjoerd, Apr 16 '19 at 09:59
Where do you keep this string, how do you produce it, and what would you want to do with it? If you are looping over all CSV files in a directory, consider using for file in *.csv; do ...; done, for example, instead of storing the ls -l or ls output. — Kusalananda, Apr 16 '19 at 10:03
the file is produce by a system and i have to format it to keep just csv file names. The file contains a lot of line looking like what i have said, so i do keep csv file name ==> test1.csv \n test2.csv \n test3.csv — Ayoub Hammami, Apr 16 '19 at 10:14
@AyoubHammami: Does "CSV" in the file have to be transformed to "csv" in the output? (Any more changes?) — DonHolgo, Apr 16 '19 at 12:00
Not a duplicate, but relevant reading: Why not parse ls (and what do to instead)? — JigglyNaga, Apr 18 '19 at 09:34

score 2 · Answer 1 · answered Apr 16 '19 at 10:15

You can use grep

grep -io '[^ ]*\.csv' filename

-i, --ignore-case Ignore case distinctions, so that characters that differ only in case match each other.

-o, --only-matching Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.

[^ ]*\.csv matches any character but space ending with .csv

I am assuming filename with no space

or you can use awk

awk '{print $(NF)}' filename | grep -i '\.csv'

variable $(NF) is the last field in every line

you can accept it as answer, if it solves your problem, as it will help other persons in future if in case — dgfjxcv, Apr 16 '19 at 10:34

Siva · Answer 2 · 2019-04-16T12:22:43.967

1

Try this,

awk '$NF~/CSV$/ {print $NF}'  file

extracts the lines end with CSV and prints the last field

edited Apr 16 '19 at 12:22

answered Apr 16 '19 at 10:03

Siva

9,077

thanks but i mean i have file with something like this : -rw-rw-rw 1 root root 6379 May 24 2016 test1.CSV -rw-rw-rw 1 root root 23249 May 25 2016 test2.CSV -rw-rw-rw 1 root root 2995 May 26 2016 test3.CSV and i need to keep just csv file name, i have to use a regex but i dont know which one – Ayoub Hammami Apr 16 '19 at 10:10
@AyoubHammami try my update – Siva Apr 16 '19 at 10:55

score 1 · Answer 3 · edited Apr 16 '19 at 12:07

1

Try this: [^ ]* *$

Or simply, if relevant, gawk '{print $NF}'

edited Apr 16 '19 at 12:07

fra-san

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answered Apr 16 '19 at 10:19

Eran Ben-Natan

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score 0 · Answer 4 · answered Apr 16 '19 at 10:16

0

You can use ' awk' :

awk '{print $9}' files

It will work whenever the filename is the 9th field in the line.

answered Apr 16 '19 at 10:16

jcbermu

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How to extract the CSV file name from each line of this file

4 Answers4