Bash Square Bracket Expansion-CodePudding

So the official bash manual states that "For example, the regular expression ‘[0123456789]’ matches any single digit, whereas ‘[^()]’ matches any single character that is not an opening or closing parenthesis,", copied a link at the bottom of this question, for context.

So I tested it every which way I could think of, to try and do the "negate" part of this, but I could not get it to work:

$ cat test
a line
b line
c line
d line
$ grep [^abc] test
a line
b line
c line
d line
$ grep '[^abc]' test
a line
b line
c line
d line
$ grep '[^(abc)]' test
a line
b line
c line
d line
[$ grep [^(abc)] test
bash: syntax error near unexpected token `('

https://www.gnu.org/software/grep/manual/html_node/Character-Classes-and-Bracket-Expressions.html

I was expecting just line D to be shown

CodePudding user response：

[^abc] matches any character which is not a, b, or c.

It matches a line because it includes , l, i, n, and e, none of which are the excluded characters.

To ensure that no character in the string matches any in the list you would need grep '^[^abc]*$' test

CodePudding user response：

To stick with the bash tag on the question, here’s a pure Bash option that takes advantage of extglob (which is mostly enabled by default).

while IFS= read -r line; do
  [[ "$line" = *([^abc]) ]] && printf '%s\n' "$line"
done < test