Parse a changelog and extract changes for a version

Question

I have a changelog file in markdown which contains all changes between each version of my app like that :

## Version 1.0.6

* first change
* second change
* third change

## Version 1.0.5

* first foo change
* second foo change

## Version 1.0.4

* and so on...

What I want is to extract in a script the changes content for a version. For example I would to extract the changes for the version 1.0.5, so it should print :

* first foo change
* second foo change

The ideal way would be ./getVersionChanges version filename which would those 2 params :

version : the version to extract changes

filename : the filename to parse

How can I achieve this with sed, awk, grep, or whatever ?

Answer 1

A slightly more elaborate awk solution, which

• exits once the block of interest has been printed,
• ignores blank lines,
• doesn't include the header line.

awk -v ver=1.0.5 '
 /^## Version / { if (p) { exit }; if ($3 == ver) { p=1; next } } p && NF
' file

---

As script getVersionChanges:

#!/usr/bin/env bash

awk -v ver="$1" '
 /^## Version / { if (p) { exit }; if ($3 == ver) { p=1; next } } p && NF
' "$2"

---

Explanation:

• Regex condition /^## Version / matches the header line of a block of lines with version-specific information, by looking for substring ## Version at the start (^) of the line and, if found, executing the associated code block ({ ... }):

  • if (p) { exit } exits (stops processing), if the p (print) flag has previously been set, because that implies that the block after the one of interest has been reached, i.e. that the block of interest has now been fully processed.

  • if ($3 == ver) { p=1; next } checks if the 3rd whitespace-separated field ($3) on the header line matches the given version number (passed via option -v ver=1.0.5 and therefore stored in variable ver) and, if so, sets custom variable p, which serves as a flag indicating whether to print a line, to 1 and moves on to the next line (next), so as not to print the header line itself.
    In other words: p containing 1 indicates for subsequent lines that the version-specific block of interest has been entered, and that its lines should (potentially) be printed.

• Condition p && NF implicitly prints the line at hand if the condition matches, which is the case if the print flag p is set and (&&) the line at hand has at least one field (based on the number of fields being reflected in built-in variable NF), i.e. if the line is non-blank, thereby effectively skipping empty and all-whitespace lines in the block of interest.

  • Note that both operands of && use implicit Boolean logic: a value of 0 (which a non-initialized custom variable such as p defaults to) is implicitly false, whereas any nonzero value implicitly true.

Answer 2

A rather short awk script will extract the block you want.

#!/bin/sh

awk -v version="$1" '/## Version / {printit = $3 == version}; printit;' "$2"

A sample:

$ ./getVersionChanges 1.0.5 filename
## Version 1.0.5

* first foo change
* second foo change

$

Answer 3

Try this. You can replace /tmp/data with your file name and 'Version 1.0.5' with your search pattern. Note that this does not strip any empty lines.

sed  '1,/Version 1.0.5/d;/Version/Q' /tmp/data

Output:

* first foo change
* second foo change  

Explanation

By default sed will print the lines. So we just change the logic to delete the lines which we do not need.

Select everything between line 1 and pattern and delete it

 1,/Version 1.0.5/d

Quit when you find the pattern

 /Version/Q