grep inverted match over string satisfying a pattern

Question

I have strings per line like:

" … “es” completed."
" … “en” completed."
" … “fr” completed."

I am trying to grep inverse match of all this. That is avoid output all the lines that match with the pattern.

echo " … “es” completed." | grep -v " … “(*)” completed."
echo " … “es” completed." | grep -v " … “[*]” completed."
echo " … “en” completed." | grep -v " … “[\w]” completed."
echo " … “fr” completed." | grep -v " … “[\W]” completed."

All this grep ways still output the strings, I don't know if I need use -e as option parameter, but I am not getting the desired result

Answer 1

-e PATTERN, --regexp=PATTERN Use PATTERN as the pattern. This can be used to specify multiple search patterns, or to protect a pattern beginning with a hyphen (-). (-e is specified by POSIX .)

Yes, you should use -e. But also regex up, especially with double quotes in double quotes, dots and asteriks.

Lets say I do not want any line that ends with completed.

| grep -v -e '.*completed\.$'

• $ end of string or line
• * zero or more occurences
• . matches almost everything
• \ escape characters, in this case .