Find the file with higher version-CodePudding

I am working in Linux bash.

I have a directory where I have files like this:

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)?

CodePudding user response：

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"

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"

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

In each file name prefix single digit number with 0

Sort lexicaly all file names

Get the last file name

Remove 0 prefix from 2 digits numbers in filename

CodePudding user response：

awk

awk -F'[-.]' '{int($2"."$3"."$4)>n && n=$2"."$3"."$4}END{print n}' <(ls -1 /path/to/xpto-*.tar.xz)
40.20.30