How to get the last 2 characters of a string with last character being A or B and second to the last character being 1-360? (REGEX GREP)

Question

I'm not really using regex in a daily basis and I'm still new to this.

For example, I have these strings and this is the format of the strings:

APPLE20B 50A
APPLE30A 60B
APPLE12B 5B
APPLE360A 360B
APPLE 56B

Basically, I want to get the last letter (A or B) and the digit before the last letter (or a digit after the letter/before the digit which is also A or B too). There are also a format like APPLE56B that doesn't have digit+letter in the middle.

Expected Output :

50A
60B
5B
360B
56B

I tried grep -o '.\{2\}$' but it only outputs the last 2 characters:

0A
0B
5B
0B
6B

and obviously, it's not dynamic for the digits. Any help would be appreciated.

Answer 1

grep -o would indeed work with the correct pattern

grep -oP '[0-9]+[AB]$'

With Perl,

perl -nle'print $& if /[0-9]+[AB]$/'

perl -nle'print for /([0-9]+[AB])$/'

In all cases, you can provide the input via STDIN or by passing a file name to read as an argument.

Answer 2

Try this:

cat input-file | perl -ne 'print "$1\n" if (m/([0-9]+[AB])$/)'

Answer 3

This might work for you (GNU grep):

grep -o '\(360\|3[0-5][0-9]\|[1-2][0-9][0-9]\|[1-9][0-9]\|[1-9]\)[AB]\>' file

This will print each value on a separate line from 1A/1B to 360A/360B .

To space separate these values use:

grep -o '\(360\|3[0-5][0-9]\|[1-2][0-9][0-9]\|[1-9][0-9]\|[1-9]\)[AB]\>' file |
paste -sd' '