BASH: Print Nth Letter From Beginning

Question

I want to read all file and print Nth letter (just letter) from beginning in all lines.

I read file like this:

while read line

do

echo $line >> text.txt

done < input.txt

Then i do this:

cut -c3 text.txt

But i realize that it doesn't working as i want in these inputs:

He llo

or

re+set

I want to get 'l' for first input and 's' for second input.

But this code found ' ' for first input and '+' for second input

And anubhava wrote this code:

awk -v n=3 '{gsub(/ +/, ""); print substr($0, n, 1)}' file

It was working for Nth letter except space

But now i want print Nth letter (just letter)

What should i do ?

Answer 1

You can use this awk:

awk -v n=3 '{gsub(/ +/, ""); print substr($0, n, 1)}' file

Testing:

s='re set'
awk -v n=3 '{gsub(/ +/, ""); print substr($0, n, 1)}' <<< "$s"
s

s='He llo'
awk -v n=3 '{gsub(/ +/, ""); print substr($0, n, 1)}' <<< "$s"
l

s='1 2 345'
awk -v n=3 '{gsub(/ +/, ""); print substr($0, n, 1)}' <<< "$s"
3