Delete files with regular expression

You can use the following command to delete all files matching your criteria:

ls | grep -P "^A.*[0-9]{2}$" | xargs -d"\n" rm

How it works:

1. ls lists all files (one by line since the result is piped).

2. grep -P "^A.*[0-9]{2}$" filters the list of files and leaves only those that match the regular expression ^A.*[0-9]{2}$

   • .* indicates any number of occurrences of ., where . is a wildcard matching any character.

   • [0-9]{2} indicates exactly two occurrences of [0-9], that is, any digit.

3. xargs -d"\n" rm executes rm line once for every line that is piped to it.

Where am I wrong?

For starters, rm doesn't accept a regular expression as an argument. Besides the wildcard *, every other character is treated literally.

Also, your regular expression is slightly off. For example, * means any occurrences of ... in a regular expression, so A* matches A, AA, etc. and even an empty string.

For more information, visit Regular-Expressions.info.

Or using find:

find your-directory/ -name 'A*[0-9][0-9]' -delete

This solution will deal with weird file names.

See the filename expansion section of the bash man page:

rm A*[0-9][0-9]