Match a line break on the line before another match

Question

I have a file with the following pattern:

A = 1
B = 2
C = 3
A = 10
B = 20
C = 30

I would like match only the line break before A = *

My attempt would also matches A = * as well as the line break

Code:

[rn]+.*A.*

But I only need the line break alone. I also understand the first A would get sacrificed as there is no like above it.

Is it possible to use lookbehind?

EDIT:
My attempt works but leave 2 groups which I can just access group 2. I was hoping to get this done in only 1 group

([.*rn])(.*A.*)

Asked By: Kevin_ALA

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Source

Answer 1

$[s]+^ works in Adobe brackets, every RegExp editor is different, but I would generally try using the $ (end of line) and ^ (beginning of line) combined with s which is any whitespace including new lines to match what you want.

Edit:

Try this for the lookahead:

($s*)?^(?=A)

Answered By: richlowe

Answer 2

I presume you are trying to split the string into ABC groups. Look for a n with a lookahead = ‘A’

text = """
A = 1
B = 2
C = 3
A = 10
B = 20
C = 30
"""

re.split(r"n(?=A)", text)

Returns:

['A = 1nB = 2nC = 3', 'A = 10nB = 20nC = 30n']

If you want the actual newline, use the same pattern with re.search() or re.finditer() to get the match object(s).

Answered By: RootTwo

Match a line break on the line before another match

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