sed to remove all after a string?

Question

I'm trying to use sed to remove the rest of a line after: HTTP1.1" 200

I can't figure out how to get sed to understand I want that whole thing as a string to match, including the double quote and the space.

An example for good measure, I want to turn:

"GET /images/loading.gif HTTP/1.1" 200 10819 "https://...

into

"GET /images/loading.gif HTTP/1.1" 200

Answer 1

You have to quote strings with spaces when you use sed (or most other tools) from the commandline. And since you already use the double quote, you have to go for single quotes:

echo '"GET /images/loading.gif HTTP/1.1" 200 10819 "https://...' | \
     sed 's|HTTP/1.1" 200.*|HTTP/1.1" 200|'

gives:

"GET /images/loading.gif HTTP/1.1" 200

Answer 2

With GNU grep:

| grep -o '.*HTTP/1\.1" 200'

With GNU sed:

| sed -r 's/(.*HTTP\/1\.1" 200).*/\1/'

Answer 3

If you just append a newline delimiter after your matched string you can Print only so much of pattern space without having to modify it overmuch. This can usually even work when pattern space contains bytes which are not parts of a character.

sed -n 's|HTTP/1.0” 200|&\n|;P' <in >out

note: portably you'll want to use a literal newline character in place of the n used above, though it does work as written in a GNU sed