如何使用 python findall 提取公共部分？

Question

I have an issue with re.findall

eg.

text = '[1]xxxxxxxx[2]xxxxxxxx[3]xxxxxx[4]xxxxxxxxxend'
re.findall('(\[\d{1,2}\].*?)(?:\[\d{1,2}\]|end)',text)

what I want is to extract ["[1]xxxxxxxx","[2]xxxxxxxx","[3]xxxxxx","[4]xxxxxxxxx"] .

However when I did re.findall('(\[\d{1,2}\].*?)(?:\[\d{1,2}\]|end)',text)

I got ['[1]xxxxxxxx', '[3]xxxxxx']

Any luck by this question

Answer 1

The non-capturing group, (?:...) , does not create a separate memory buffer with the text matched, but it still consumes the text matched, ie it is added to the match value and the regex index is advanced.

You need a non-consuming pattern here, a positive lookahead:

re.findall(r'\[\d{1,2}\].*?(?=\[\d{1,2}\]|end)', text)

See the regex demo .

The (?=\[\d{1,2}\]|end) pattern matches a ocation that is immediately followed with [ , one or two digits and then ] , or end char sequence.