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

• steeldriver: You need to use a regular expression that matches the strings, for example, " ... ".." completed\." - not a "wildcard"

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 .

• shakaran: I gonna try, the problem is that there is more "completed" strings, so I really need the full regex. Also notice that the double quotes are not normal like ", are " (probably difficult to notice)
• shakaran: It seems that grep -v -e " ... ".*" completed\.$" work perfect ;)