Extract part of string using grep

Question

I have 3 files in a directory

MYO144064T
MYO144064TA
MYO144064TX

Where the digits and 11th character will change. In my csh script, I want to extract the letters 'A' and 'X' probably using grep

The variable $study equals MYO144064

What I have managed so far is very cumbersome

ls | grep $study | cut -c 11 | sed 's/\///'

Which gives me

A
X

How can I do this with minimal processes and without the additional blank line?

Solution

ls | grep -oP "(?<=$study.)[A-Z]$"

will return any uppercase letter that is preceded by the contents of $study plus one arbitrary character (the T in your example) and followed by the end of the line.

_{The -P option (Perl regular expressions) is needed to be able to use the positive lookbehind expression (?<=...), but might not be available on every system and platform.}

Comments

[moadeep]

The 11th character will change, it may be A or X but could easily be M, Y, O or T

[rob]

then ".$" will always get you the last character