CODEWITHSUNDEEP
bashshellawkgrep
Yash Yadav
Yash Yadav

Reputation: 21

List numeric files which are higher than particular number

I have list of files in directory,ending with numeric. As below:

play_football_3
play_football_4
play_football_5
play_football_15
play_football_59

I'm able to extract the last numeric digit of above files. as

echo "play_football_5" | cut -f3 -d"_"

Now Here i'm trying to list all files which have higher version then play_football_5.

Expected Output:

play_football_15
play_football_59

Upvotes: 2

Views: 89

Answers (5)

stack0114106
stack0114106

Reputation: 8711

Using Perl one-liner

> ll
total 0
-rw-r--r-- 1 xxxxxx devlgrp 0 Oct 30 14:41 play_football_5
-rw-r--r-- 1 xxxxxx devlgrp 0 Oct 30 14:41 play_football_4
-rw-r--r-- 1 xxxxxx devlgrp 0 Oct 30 14:41 play_football_3
-rw-r--r-- 1 xxxxxx devlgrp 0 Oct 30 14:41 play_football_15
-rw-r--r-- 1 xxxxxx devlgrp 0 Oct 30 14:41 play_football_59
> perl -ne ' BEGIN { @files=glob("play*");foreach(@files){ ($file=$_)=~s/.*_(\d+)/\1/g; print "$_\n" if $file > 5 } exit }'
play_football_15
play_football_59
>

Upvotes: 0

nsa
nsa

Reputation: 565

You can also keep your original idea doing something like this:

for file in $(ls | cut -d_ -f3)
do
  if [[ "$file" -gt 5 ]]
  then
    echo "play_football_$file"
  fi
done

If you do that while in the directory that has your files you will get:

play_football_15
play_football_59

Upvotes: 0

Mark Setchell
Mark Setchell

Reputation: 207445

If you have bash 2.02-alpha1, or newer, you can turn on "extended globbing" and look for files starting play_football_ and not ending with the digits 0-5 like this:

shopt -s extglob

ls play_football_!([0-5])

Here is a reference to start learning more about it.

Upvotes: 3

oliv
oliv

Reputation: 13249

You can use awk for this:

printf "%s\n" play_football_* | awk -F_ '$3>5'

printf will list all files in the current directory start with play_football_ and awk filters the files with the number greater than 5

Upvotes: 3

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626758

You may use

awk -F'_' '$3 > 5'

See the online awk demo

With -F'_', you set the delimiter to an underscore and with '$3 > 5', only those lines (records) are printed where the third field is greater than 5.

Upvotes: 0

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