Question

How can I make sed filter matching lines according to some expression, but ignore non-matching lines, instead of letting them print?

As a real example, I want to run scalac (the Scala compiler) on a set of files, and read from its -verbose output the .class files created. scalac -verbose outputs a bunch of messages, but we’re only interested in those of the form [wrote some-class-name.class].
What I’m currently doing is this (|& is bash 4.0’s way to pipe stderr to the next program):

$ scalac -verbose some-file.scala ... |& sed 's/^\[wrote \(.*\.class\)\]$/\1/'

This will extract the file names from the messages we’re interested in, but will also let all other messages pass through unchanged! Of course we could do instead this:

$ scalac -verbose some-file.scala ... |& grep '^\[wrote .*\.class\]$' |
  sed 's/^\[wrote \(.*\.class\)\]$/\1/'

which works but looks very much like going around the real problem, which is how to instruct sed to ignore non-matching lines from the input. So how do we do that?

Answer 1

Another way with plain sed:

sed -e 's/.../.../;t;d'

s/// is a substituion, t without any label conditionally skips all following commands, d deletes line.

No need for perl or grep.

(edited after Nicholas Riley’s suggestion)