Python regex match across multiple lines

Question

I am trying to match a regex pattern across multiple lines. The pattern begins and ends with a substring, both of which must be at the beginning of a line. I can match across lines, but I can't seem to specify that the end pattern must also be at the beginning of a line.

Example string:

Example=N      ; Comment Line One error=

; Comment Line Two.

Desired=

I am trying to match from Example= up to Desired= . This will work if error= is not in the string. However, when it is present I match Example=N ; Comment Line One error= Example=N ; Comment Line One error=

config_value = 'Example'
pattern = '^{}=(.*?)([A-Za-z]=)'.format(config_value)
match = re.search(pattern, string, re.M | re.DOTALL)

I also tried:

config_value = 'Example'
pattern = '^{}=(.*?)(^[A-Za-z]=)'.format(config_value)
match = re.search(pattern, string, re.M | re.DOTALL)

Answer 1

You may use

config_value = 'Example'
pattern=r'(?sm)^{}=(.*?)(?=[\r\n]+\w+=|\Z)'.format(config_value)
match = re.search(pattern, s)
if match:
    print(match.group(1))

See the Python demo .

Pattern details

• (?sm) - re.DOTALL and re.M are on
• ^ - start of a line
• Example= - a substring
• (.*?) - Group 1: any 0+ chars, as few as possible
• (?=[\\r\\n]+\\w+=|\\Z) - a positive lookahead that requires the presence of 1+ CR or LF symbols followed with 1 or more word chars followed with a = sign, or end of the string ( \\Z ).

See the regex demo .