Extract version-specific upgrade notice from readme text

I am currently writing a PHP function which should help me to extract an upgrade notice from a given readme text.

This is my source text:

Some stuff before this notice like a changelog with versioning and explanation text.

== Upgrade Notice ==

= 1.3.0 =

When using Master Pro, 1.3.0 is the new minimal required version!

= 1.1.0 =

When using Master Pro, 1.1.0 is the new minimal required version!

= 1.0.0 =

No upgrade - just install :)

[See changelog for all versions](https://plugins.svn.wordpress.org/master-pro/trunk/CHANGELOG.md).

This is the function:

/**
 * Parse update notice from readme file
 *
 * @param string $content
 * @param string $new_version
 *
 * @return void
 */
private function parse_update_notice( string $content, string $new_version ) {
    $regexp  = '~==\s*Upgrade Notice\s*==\s*(.*?=+\s*' . preg_quote( $new_version ) . '\s*=+\s*(.*?)(?=^=+\s*\d+\.\d+\.\d+\s*=+|$))~ms';

    if ( preg_match( $regexp, $content, $matches ) ) {
        $version = trim( $matches[1] );
        $notices = (array) preg_split( '~[\r\n]+~', trim( $matches[2] ) );

        error_log( $version );
        error_log( print_r( $notices, true ) );
    }
}

I am currently stuck at my RegEx. I'm not really getting it to work. This was my initial idea:

Only search after == Upgrade Notice ==
Check if we have a version matching $new_version
Get the matched version between the = x.x.x = as match 1 e.g. 1.1.0
Get the content after the version as match 2 but stopping after an empty new line. The upgrade notice can go over multiple lines but without an empty new line.

Solution

To get the first part after "Upgrade Notice", matching only the first following block with non empty lines, you can omit the s flag to have the dot match a newline and capture matching all following lines that contain at least a single non whitespace character.

^==\h*Upgrade Notice\h*==\R\s*^=\h*(1\.3\.0)\h*=\R\s*^((?:\h*\S.*(?:\R\h*\S.*)*)+)

The line in PHP:

$regexp = '~^==\h*Upgrade Notice\h*==\R\s*^=\h*(' . preg_quote( $new_version ) . ')\h*=\R\s*^((?:\h*\S.*(?:\R\h*\S.*)*)+)~m';

Regex demo

If you want to be able to determine which occurrence after matching "Upgrade Notice", you can use a quantifier to skip the amount of occurrences that start with the version pattern:

^==\h*Upgrade Notice\h*==(?:(?:\R(?!=\h*\d+\.\d+\.\d+\h*=$).*)*\R=\h*(\d+\.\d+\.\d+)\h*=$\s*){2}(^\h*\S.*(?:\R\h*\S.*)+)

^ Start of string
==\h*Upgrade Notice\h*== The starting pattern, where \h* match optional horizontal whitespace characters
(?: Non capture group
- (?:\R(?!=\h*\d+\.\d+\.\d+\h*=$).*)* Match all lines that do not start with a version pattern
- \R=\h* Match a newline and = followed by horizontal whitespace characters
- (\d+\.\d+\.\d+) Capture group 1, match the version
- \h*=$\s* Match horizontal whitespace characters, = and assert the end of the string and match optional whitespace characters
){2} Use a quantifier (in this case {2}) to match n times a version pattern
^ Start of string
( Capture group 2
- (?:\h*\S.*(?:\R\h*\S.*)*)+ Match 1 or more lines that contain at least a single non whitespace character
) Close the group

Regex demo