Using grep to display second character in string?

Question

I have a string that is always 3 characters... the first one and the last one are always the same.

example:

▁▅█

Is there a simple way to only display the one in the middle? (which is the only one that changes)

Answer 1

Yes, try doing this, and pick your preferred method =) :

With grep:

echo "ixi" | grep -oP "^.\K."

With cut:

echo "ixi" | cut -c2

With bash parameter expansion :

x='ixi'; echo ${x:1:1}

With sed:

echo "ixi" | sed 's/.\(.\)./\1/'

or

echo "ixi" | sed 's/\(^.\|.$\)//g'

With perl:

echo "ixi" | perl -lne 'print $& if /^.\K./'

With ruby:

echo "ixi" | ruby -ne 'print $_.split(//)[1]'

With awk:

echo 'ixi' | awk '{split($0, a, ""); print a[2]}'

With python:

echo "ixi" | python -c 'print list("'$(cat)'")[1]'

or

python -c 'import sys; print list(sys.argv[1])[1]' ixi

NOTE

• \K restart the match to zero (see pcre doc)
• $(cat) in python is a shell hack to get STDIN

Answer 2

You could also use sed instead of grep:

sed 's/.\(.\)./\1/'

This says:

1. s/ ... match the expression up to the next / character.
2. . ... match the first character (any).
3. \(.\) ... match the next character and remember it.
4. . ... match the third character (any).
5. / ... denotes the end of the expression.
6. \1 ... replace the entire string that matched with the character that was remembered.
7. / ... end of replacement text.

Thus:

$ echo "abc" | sed 's/.\(.\)./\1/'

Will print:

b

Answer 3

        ︙
or

echo "ixi" | awk '{print substr($0,2,1)}'

or

expr "ixi" : '.\(.\).'

Answer 4

For variety, here's one way with dd:

echo ixi | dd bs=1 skip=1 count=1 2>/dev/null

If your grep doesn't support perl (-P), you could do it like this:

echo ixi | grep -o . | sed -n 2p

These alternatives only work for one-line input.