在文本之后匹配所有带有模式的行，直到模式匹配失败正则表达式

Question

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]]

Answer 1

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])