Match all lines with a pattern after a text until pattern matching failure regex

I have a text:


{{Verkleinerungsformen}}
:[1] [[Äpfelchen]], [[Äpfelein]], [[Äpflein]]

{{Oberbegriffe}}
:[1] [[Kernobst]], [[Obst]]; [[Frucht]]
:[4] [[Kot]]
:[7] [[Gut]]

{{Unterbegriffe}}
:[1] [[Augustapfel]], [[Bohnapfel]], [[Bratapfel]], [[Essapfel]], [[Fallapfel]],

I'm interested in extracting all items under {{Oberbegriffe}} that have the pattern [[Text]] including all lines until it reach another line that does not have :[NUMBER-HERE] at the begin

so in the above example it should return an array of these strings:

Kernobst, Obst, Frucht, Kot, Gut

what I have tried is:

re.search(r'{{Oberbegriffe}}\n(?::?\n)?([^\n]+)', text)

But it matches only the full first line. It's ok if there is a way to extract all lines with the pattern and it retruns this string

:[1] [[Kernobst]], [[Obst]]; [[Frucht]]
:[4] [[Kot]]
:[7] [[Gut]]

Solution

You may extract the blocks using

(?m)^{{Oberbegriffe}}(?:\n:\[\d+].*)*

See the regex demo

Then, use \[\[([^][]+)]] pattern to extract the values you need. See this regex demo.

Regex details

(?m) - an inline modifier, same as re.M / re.MULTILINE
^ - start of a line
{{Oberbegriffe}} - literal text
(?:\n:\[\d+].*)* - 0 or more repetitions of a newline followed with :[, then 1+ digits, ], and then any 0 or more characters other than line break chars, as many as possible.

The second regex - \[\[([^][]+)]] - matches [[, then capturing group #1 matching any 1 or more chars other than [ and ], and then ]].

In Python:

with open(filepath, 'r') as fr:
  blocks = re.findall(r'^{{Oberbegriffe}}(?:\n:\[\d+].*)*', fr.read(), flags=re.M)
  print([re.findall(r'\[\[([^][]+)]]', block) for block in blocks])