bash event not found trying to match and exclude parenthesis in grep

Question

In a very long line I'll summarize with:

(foo),(bar,baz(word,right),(end)

I want to print only:

      (bar,baz(word,right

To match the second parenthesis, I exclude the word that follows the third one:

$ grep -oP "\\(.*(?!word).*right"

but Bash interprets the exclamation mark:

-bash: !word: event not found

Protecting the exclamation mark with single quote fails with grep: missing )

$ grep -oP '\\(.*(?!word).*right'
$ grep -oP '\\((?!word)(.*right)'

Protecting the exclamation mark with backslash fails with grep: unrecognized character after (? or (?-

Any idea?

Note: -P is for Perl regex and -o is to print only the matching part of a line

grep -o '(b[^)]*' works in that case. Your attempted usage of the negative look ahead operator doesn't make sense. What exactly do you want to match? — Stéphane Chazelas, Mar 31 '14 at 14:03
use set +H to turn off history expansion, set -H to turn it back on. — glenn jackman, Mar 31 '14 at 14:15
@glennjackman Or just use proper quoting: single quotes in this case. (Which gets to the next step that the regex isn't syntactically correct.) — Gilles 'SO- stop being evil', Mar 31 '14 at 20:06

score 11 · Answer 1 · answered Mar 31 '14 at 15:09

The rules are different for single quotes versus double quotes.

For the reason you show, double quotes can't be used reliably in bash, because there's no sane way to escape an exclamation mark.

$ grep -oP "\\(.*(?!word).*right"
bash: !word: event not found

$ grep -oP "\\(.*(?\!word).*right"
grep: unrecognized character after (? or (?-

The second is because bash passes through \! rather than ! to grep. Showing this:

$ printf '%s' "\!"
\!

When you tried single quotes, the double backslash doesn't mean an escaped backslash, it means two backslashes.

$ printf '%s' '\\(.*(?!word).*right'
\\(.*(?!word).*right

Inside single quotes, everything is literal, and there are no escapes, so the way to write the regular expression you're trying is:

$ grep -oP '\(.*(?!word).*right'

Stéphane Chazelas · Answer 2 · 2014-03-31T14:39:30.813

3

If you want to match from the second open parentheses up until (but not including) the next closing parentheses:

grep -Po '\(.*?\K\([^)]*'

Or portably with sed:

sed -n 's/^[^(]*([^(]*\(([^)]*\).*/\1/p'

To match the right most ( that is not followed by word up to the rightmost right after that:

grep -Po '.*\K\((?!word).*right'

edited Mar 31 '14 at 14:39

answered Mar 31 '14 at 14:19

Stéphane Chazelas

544,893

score 2 · Answer 3 · answered Mar 31 '14 at 13:57

2

You can do it with simple awk:

$ echo '(foo),(bar,baz(word,right),(end)' | awk -F'),' '{print $2}'
(bar,baz(word,right

answered Mar 31 '14 at 13:57

cuonglm

153,898

score 0 · Answer 4 · answered Mar 31 '14 at 14:05

Look for the comma parens before and aft instead:

grep -oP '(?<=,)\(.*(?=\),)'

Example

$ echo '(foo),(bar,baz(word,right),(end)' | grep -oP '(?<=,)\(.*(?=\),)'
(bar,baz(word,right

The lookahead and behind can only look for explicit strings, it cannot find things such as .*.

References

Lookahead and Lookbehind Zero-Length Assertions

score 0 · Answer 5 · answered Oct 02 '14 at 09:39

0

If for some reason you cannot use single quotes as suggested in Mikel's answer, you can temporarily turn off history expansion using set +H (turn it back on with set -H), as suggested by glenn jackman in comments.

answered Oct 02 '14 at 09:39

Skippy le Grand Gourou

3,293

bash event not found trying to match and exclude parenthesis in grep

5 Answers5

Example

References