Ripgrep: Output only matched patterns (data extraction)

[Hrxn]

Hi,

just been trying ripgrep for the first time, but I couldn't find an easy and straightforward way to just output the matching pattern.

For example, trying to search in a file like this:
rg "(\{.*\}\})" example.txt

The regex matching works as intended, correct result is highlighted in bold red text.

But when I tried to redirect ("pipe") the output to another program I realized that the actual output is still the unchanged input.

Am I missing something here? I couldn't find an option just for this output, seemed pretty straightforward to me, but neither rg --help or the front page here seem to mention something..

[BurntSushi]

It's a missing feature: #34. It's on my list of things to do soon, but it's part of a larger refactoring effort, so it might take a little longer.

There is a work around today though. You can use the -r/--replace flag:

rg "(\{.*\}\})" example.txt -r '$1'

I think that should do it. (On mobile so i haven't checked.)

[Hrxn]

rg "(\{.*\}\})" example.txt -r '$1'

Not really. Replacing works as intended apparently, in this case:

• match and take capture group 1
• replace with reference 1, i.e. capture group 1 again

But the thing is, the actual output is still unchanged, because the program always returns the whole line where the match was found.

To help illustrate things better, I made an example file: https://fnpaste.com/YZvd

But I found a way around this: Changing the rx syntax to match the whole line, and then replace the whole line with capture group 1.

For example:
rg ".*40 (\{.*\}\}).*" example.txt -r "$1"

This gives me the desired part. But there is a big caveat: I use the 40 as an anchor to make the group match the correct part, but this is specific for this example only, I don't actually know the value.

I tried ".*(\{.*\}\}).*" in my text editor and it actually works there, whereas the same pattern only matches the last {...}} part of the target inside the line with rg.

I'm not really familiar with the intricacies of regular expressions in Rust, so this needs a bit of tweaking, I guess, although it should definitely be possible.

[BurntSushi]

You're right. I forgot to add that you need to make your regex match the entire line in order for -r/--replace to work like -o/--only-matching.

I think this gives the desired output for you though:

$ rg '^.*?(\{.*\}\}).*?$' example.txt -r '$1'

The trick is to make the .* lazy by using .*?. When it's lazy, it will try to match the least possible. Otherwise, it tries to match the most possible.

[Hrxn]

Thanks!

That did the trick just fine, greedy vs. lazy, obviously..

I'll close this issue for now.
We have a good workaround, and you already said that this is on your list of things to do, so I guess all is fine..