grep 拆分匹配

Question

I need to parse a CHANGELOG in Keep a changelog format with grep (or awk, etc in shell/bash) and get the last version (the first one after [Unreleased] tag).

It means, split this file with block '\n## ', ignore the first ([Unreleased]) and get the second (if exists).

With nodeJS, it's very easy and readable CHANGELOG.split(/\n## /)[2]; But I can't make it work with grep... grep -zoP -m 1 "(\n##.*)(\n##.*)?(\n## )?" CHANGELOG.md grep -zoP -m 1 "(\n##.*)(\n##.*)?(\n## )?" CHANGELOG.md

I can't make the regex match group with multiline even using (.|\n)+ Since I'm on it since few days and trying again and again, the Machine Learning found this ##(?:[^be]+[^#]*###)+[^#]* but, it looks like too heavy for just "block split with \n## ".

# Changelog
All notable changes to this project will be documented in this file.

The format is based on [Keep a Changelog](https://keepachangelog.com/en/1.0.0/),
and this project adheres to [Semantic Versioning](https://semver.org/spec/v2.0.0.html).

## [Unreleased]

## [1.0.0] - 2017-06-20
### Added
{...}

### Changed
{...}

### Removed
{...}

## [0.3.0] - 2015-12-03
{...}

I need to capture the block:

## [1.0.0] - 2017-06-20
### Added
{...}

### Changed
{...}

### Removed
{...}

UPDATE #1

I found one working (see in regex101.com ) with (?=\n##.*?)(\n##.*?)(?=\n## |$) and now, just need to print Match 2

Any help? Thank you!

Answer 1

This perl one-liner does the job, it reads the file in “slurp” mode and prints the data you're looking for:

perl -0777 -ane '/## \[Unreleased]\R\R\K##[\s\S]+(?=## \[\d)/ && print$&' logfile
## [1.0.0] - 2017-06-20
### Added
{...}

### Changed
{...}

### Removed
{...}

Explanation:

/                       # regex delimiter
    ## \[Unreleased]        # literally
    \R\R                    # 2 linebreak
    \K                      # forget all we have seen until this position
    ##[\s\S]+               # 2 # followed by 1 or more any character including newline
    (?=## \[\d)             # positive lookahead, make sure we have ## [digit after (previous relaese)
/                       # regex delimiter

If this regex matches, then print what is matched print$&

Answer 2

Thanks to @Toto answer that help me go near the solution.

Here's how I end up:

perl -0777 -ane '/## \[Unreleased][\s\S]+?\K(\n## [\s\S]+?)(?=\n## |$)/ && print$&' CHANGELOG.md

Answer 3

Ed can do this.

#!/bin/sh

cp CHANGELOG.md stack

cat >> extract.ed << EOF
/2017/
.,\$w new-changelog.txt
EOF

cat >> ex2.ed << EOF
/2015/
-1
.,\$d
wq
EOF

ed -s stack < extract.ed
ed -s new-changelog.txt < ex2.ed
rm -v ./extract.ed ./ex2.ed