Extract substring in Bash

Question

Given a filename in the form someletters_12345_moreleters.ext, I want to extract the 5 digits and put them into a variable.

So to emphasize the point, I

Answer 1

similar to substr('abcdefg', 2-1, 3) in php:

echo 'abcdefg'|tail -c +2|head -c 3

Answer 2

shell cut - print specific range of characters or given part from a string

#method1) using bash

 str=2020-08-08T07:40:00.000Z
 echo ${str:11:8}

#method2) using cut

 str=2020-08-08T07:40:00.000Z
 cut -c12-19 <<< $str

#method3) when working with awk

 str=2020-08-08T07:40:00.000Z
 awk '{time=gensub(/.{11}(.{8}).*/,"\\1","g",$1); print time}' <<< $str

Answer 3

Given test.txt is a file containing "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

cut -b19-20 test.txt > test1.txt # This will extract chars 19 & 20 "ST" 
while read -r; do;
> x=$REPLY
> done < test1.txt
echo $x
ST

Answer 4

Generic solution where the number can be anywhere in the filename, using the first of such sequences:

number=$(echo $filename | egrep -o '[[:digit:]]{5}' | head -n1)

Another solution to extract exactly a part of a variable:

number=${filename:offset:length}

If your filename always have the format stuff_digits_... you can use awk:

number=$(echo $filename | awk -F _ '{ print $2 }')

Yet another solution to remove everything except digits, use

number=$(echo $filename | tr -cd '[[:digit:]]')