python多行正则表达式findall

Question

I'm trying to find multiple matches across multiple lines of text with a delimiter to stop the search using regex in python... my query works well for what I'm trying to accomplish if what I need is all on the same line: re.findall('([a-zA-Z]{3}\\d-[aAeE][rRsS]\\d.*), output)

the problem is, sometimes the additional data I'm trying to capture doesn't fit on the same line and goes to the next... is there a way to set the pattern match to stop if it either finds the next match or hits a delimiter (= in this case)? Simplified example with two matches below, and I need the ability to capture both...

Example

Port Id Description
3/2/4 Part of aggregate interface lag-4. Next device in path sea1-as2
lag-4, sea1-as2 3/1/2.

Answer 1

It seems that all you have to do is to add [\\s\\S]*? to capture whatever coming in the next line and include the expected stops , | . , | . to stop the match. Note that it is important to make [\\s\\S]*? lazy, otherwise, it will match the whole thing.

print(re.findall(r'([a-zA-Z]{3}\d-[aAeE][rRsS]\d[\s\S]*?\d)(?:,|\.)', output))

output

['sea1-as2 lag-4', 'sea1-as2 3/1/2']

Answer 2

You mentioned [a-zA-Z] and [aAeE][rRsS] . There are several ways to set re.IGNORECASE so that [ae][rs] would suffice.

You didn't make it clear if you're using re.MULTILINE or if you're deleting newlines before evaluating the regex. You end with .* which could trivially become

[^=]*

if you want everything up to the = delimiter.

Alternatively, before evaluating the regex you could split on \\n newline and = equal, so you hand in appropriate size chunks for evaluation.