Removing substrings from a shell variable

Question

I have some string that read like this one here

Dio - We Rock-Greatest Hits (2CD) (2004)

I use this

NewDirectoryName=${NewDirectoryName%'(" ")'* }
NewDirectoryName=${NewDirectoryName//[^A-Za-z ]/ }

and it gives me this left over :

NewDirectoryName is -> Dio We Rock Greatest Hits Cd

it removes the (2CD) and (2004) alltogether leaving me with Cd.

What puzzles me is if it removes the (2004), then why does it turn (2CD) into Cd. Why only remove (2 and the ) and why is it changing D to d?

How do I get rid of the entire thing to get this:

Dio We Rock Greatest Hits

Answer 1

As I commented, your first expansion does nothing. The literal string (" ") is not in the string so nothing is removed. Then then 2nd expansion removes all characters that are not a letter or a space. That's why "CD" stays. There's nothing in your code that would lower-case the "d" in "CD".

If you want to remove all bits in parentheses:

shopt -s extglob
NewDirectoryName='Dio - We Rock-Greatest Hits (2CD) (2004)'

NewDirectoryName=${NewDirectoryName//\(*([^\)])\)}
# ...................................aa........cc
# .....................................bbbbbbbb
# (a) a literal open parenthesis
# (b) zero or more non-close-parenthesis characters
# (c) a literal close parenthesis

echo ">$NewDirectoryName<"

>Dio - We Rock-Greatest Hits  <

To remove trailing whitespace:

NewDirectoryName=${NewDirectoryName%%+([[:blank:]])}

This uses "extended pattern matching" patterns, documented at the end of this section of the manual: https://www.gnu.org/software/bash/manual/bashref.html#Pattern-Matching