Have sed ignore non-matching lines

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?

user_4375327 · Answer

Rapsey raised a relevant point about multiple substitutions expressions.

First, quoting an Unix SE answer, you can "prefix most sed commands with an address to limit the lines to which they apply".
Second, you can group commands within curly braces {} (separated with a semi-colon ; or a new line)
Third, add the print flag p on the last substitution

Syntax:

sed -n -e '/^given_regexp/ {s/regexp1/replacement1/flags1;[...];s/regexp1/replacement1/flagsnp}'

Example (see Here document for more details):

Code:
```
sed -n -e '/^ha/ {s/h/k/g;s/a/e/gp}' <
```


Result:
keke

user_12233858 · Answer

sed -n '/.../!p'

There is no need for a substitution.

Have sed ignore non-matching lines

5 Answers