command for "strip out every line that doesn't contain xxxxx"
 

Ok, so I'm stringing together some line commands here. Is there a line command that will do the following:

Given some newline seperated data (such as ls -l ) that's piped in ( with | ), do the following:

Remove every line that does NOT contain "abcdefg" on it. Or even better, could it remove every line that does not contain a regexp like /a*bcd(e|g)/;

Yes. It's called grep. :)

Specifically, it PRINTS every line that DOES contain the expression. Use 'egrep' for the more extended regexps, like above.

the -v option to grep will remove any line that contains the following exp. such as:

Every line which does *not* include 'whatever' will be returned.

--Shade

I think the OP wanted to remove those that do NOT contain the expression. Which would be equivalent to printing those that DO contain the expression. Hence, no -v.

Wow. I think you're right. These double negatives confuse me :)

--Shade

Thanks. Duuh. I'm embarassed. (I think my double negatives confused me too.)

No problem. :) Don't be embarassed, I've asked plenty of obvious questions.