Using grep with regular expression to filter out matches

Question

I'm trying to use grep with -v for invert-match along with -e for regular expression. I'm having trouble getting the syntax right.

I'm trying something like

tail -f logFile | grep -ve "string one|string two"

If I do it this way it doesn't filter If I change it to

tail -f logFile | grep -ev "string one|string two"

I get

grep: string one|string two: No such file or directory

I have tried using () or quotes, but haven't been able to find anything that works.

How can I do this?

Answer 1

The problem is that by default, you need to escape your |'s to get proper alternation. That is, grep interprets "foo|bar" as matching the literal string "foo|bar" only, whereas the pattern "foo\|bar" (with an escaped |) matches either "foo" or "bar".

To change this behavior, use the -E flag:

tail -f logFile | grep -vE 'string one|string two' 

Alternatively, use egrep, which is equivalent to grep -E:

tail -f logFile | egrep -v 'string one|string two' 

Also, the -e is optional, unless your pattern begins with a literal hyphen. grep automatically takes the first non-option argument as the pattern.

Answer 2

You need to escape the pipe symbol when -e is used:

tail -f logFile | grep -ve "string one\|string two" 

EDIT: or, as @Adam pointed out, you can use the -E flag:

tail -f logFile | grep -vE "string one|string two"