Thursday, June 2, 2022

[SOLVED] Find the file with higher version

Issue

I am working in Linux bash.

I have a directory where I have files like this:

  • xpto-10.20.30.tar.xz
  • xpto-10.20.30.tar.xz
  • xpto-20.20.30.tar.xz
  • xpto-30.20.30.tar.xz
  • xpto-10.20.30.tar.xz
  • xpto-40.20.30.tar.xz
  • xpto-40.20.9.tar.xz

I want to get the latest file version name, which should be xpto-40.20.30.tar.xz.

I am using this script:

#!/bin/bash

set_latest () {
  eval "latest=\${$#}"
}
set_latest xpto-*.tar.xz
echo "The highest-numbered file is $latest"

But the result I am getting its xpto-40.20.9.tar.xz

I guess this is because of the number of characters. How can I get the latest file version name (considering this case)? Is there a way as well to get only version "40.30.30"?


Solution

Suggesting:

ls -1 *.tar.xz| sed -E "{s|\.([1-9]\.)|.0\1|;s|\.([1-9]\.)|.0\1|}"  |sort|tail -1|sed -E "s|0([1-9])|\1|g"

In more readable way:

ls -1 *.tar.xz|\
sed -E "{s|\.([1-9]\.)|.0\1|;s|\.([1-9]\.)|.0\1|}"  |\
sort|\
tail -1|\
sed -E "s|0([1-9])|\1|g"

Gradual running

ls -1 *.tar.xz |

ls -1 *.tar.xz |sed -E "{s|\.([1-9]\.)|.0\1|;s|\.([1-9]\.)|.0\1|}" 

ls -1 *.tar.xz |sed -E "{s|\.([1-9]\.)|.0\1|;s|\.([1-9]\.)|.0\1|}"|sort

ls -1 *.tar.xz |sed -E "{s|\.([1-9]\.)|.0\1|;s|\.([1-9]\.)|.0\1|}"|sort|tail -1

ls -1 *.tar.xz |sed -E "{s|\.([1-9]\.)|.0\1|;s|\.([1-9]\.)|.0\1|}"|sort|tail -1|0([1-9])|\1|g"

Explanation

ls -1 *.tar.xz

List all files in current directory with suffix .tar.xz . A file name per line.

sed -E "{s|\.([1-9]\.)|.0\1|;s|\.([1-9]\.)|.0\1|}"

In each file name prefix single digit number with 0

sort

Sort lexicaly all file names

tail -1

Get the last file name

sed -E "s|0([1-9])|\1|g"

Remove 0 prefix from 2 digits numbers in filename



Answered By - Dudi Boy
Answer Checked By - David Goodson (WPSolving Volunteer)
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