How to get a part of a file name in shell script?

Question

I have a file that has the format nome1-nome2-0.0.0.0.war, and through the shell wanted to take the version (0.0.0.0) and save it in a variable and then use it to create a directory.

Obs.: The versions change, but the names and extension are the same. Thank you!

Answer 1

You can do it in some ways, one of them is:

ls nome1-nome2-0.0.0.0.war| egrep  -wo '[\.0-9]+'

The output of this will be: 0.0.0.0

Explanation:

I created a file called name1-Nome2-0.0.0.0.War, executed a grep using the following regex '[\.0-9]+', the parameter o and w, were necessary to match only what the regex contemplated.

A script that will scan a folder and create other folders based on your app versions, could be:

#!/bin/bash

for i in $(ls *.war|egrep  -wo '[\.0-9]+')
do
    mkdir $i
done

Answer 2

ls *.war | awk -F '-' {'print $3'} | awk -F '.war' {'print $1'}
while read versao
do
    mkdir $versao 
done