How to use negative regex matching with grep -E?
grep, regex
Solution
Use the `-v` option which inverts the matches, selecting non-matching lines
grep -v 'match me'
Another option is to use `-P` which interprets the pattern as a Perl regular expression.
grep -P '^((?!match me).)*$'
Problem
I'm using the following regex via `grep -E` to match a specific string of chars via `|` pipe. ``` $ git log <more switches here> | grep -E "match me" ``` Output: ``` match me once match me twice ``` What I'm really looking for a is a negative match (return all output lines that don't contain the specified string something like the following but `grep` doesn't like it: ``` $ git log <more switches here> | grep -E "^match me" ``` desired output: ``` whatever 1 whatever 2 ``` here is the full output that comes back from the command line: ``` match me once match me twice whatever 1 whatever 2 ``` How to do arrive at the desired output per a negative regex match?