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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".

lgeorget
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Ayoub Hammami
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    Have you tried something yourself already? Could you just use `ls` instead of `ls -l`? – Sjoerd Apr 16 '19 at 09:59
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    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
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    @AyoubHammami Are there spaces etc. in the filenames? – Kusalananda Apr 16 '19 at 11:06
  • @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)?](https://unix.stackexchange.com/questions/128985/why-not-parse-ls-and-what-do-to-instead) – JigglyNaga Apr 18 '19 at 09:34

4 Answers4

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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

dgfjxcv
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    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
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Try this,

awk '$NF~/CSV$/ {print $NF}'  file
  • extracts the lines end with CSV and prints the last field
Siva
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  • thanks but i mean i have file with something like this : -rw-rw-rw 1 root root 6379 May 24 2016 test1.CSV -rw-rw-rw 1 root root 23249 May 25 2016 test2.CSV -rw-rw-rw 1 root root 2995 May 26 2016 test3.CSV and i need to keep just csv file name, i have to use a regex but i dont know which one – Ayoub Hammami Apr 16 '19 at 10:10
  • @AyoubHammami try my update – Siva Apr 16 '19 at 10:55
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Try this: [^ ]* *$

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

fra-san
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Eran Ben-Natan
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You can use ' awk' :

awk '{print $9}' files

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

jcbermu
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