Regex match the first line after blank line

Question

I want to check the first line after blank line I tried with ^$.* but not working,

YYYY-MM-DD HH:II:SS Message Log
Message Log

Message Log
not empty line

the desired output: ["YYYY-MM-DD HH:II:SS Message Log","Message Log"]

Answer 1

Not sure why are you using a Regex for this task, I think you can use a simple split .

data = '''YYYY-MM-DD HH:II:SS Message Log
Message Log

Message Log
not empty line'''

relevant_part, _ = data.split('\n\n', 1)
relevant_part.splitlines() # ["YYYY-MM-DD HH:II:SS Message Log","Message Log"]

Answer 2

You could use:

import re


pattern = re.compile("(?:^|\n\n)(.*)")
pattern.findall(data)

OUTPUT

['YYYY-MM-DD HH:II:SS Message Log', 'Message Log']

To test it, I have saved these lines in a file:

YYYY-MM-DD HH:II:SS Message Log
Message Log

Message Log
not empty line

read the content in data and applied the pattern to it.

Answer 3

Your regex fails because you don't match the newline character.

Also .* will match a blank line, so use .+ .

Try

regex = r"(?<=^\n).+"

print(re.search(regex, text, re.MULTILINE).group(0))