Regex with negative lookahead across multiple lines

Question

For the past few hours I\'ve been trying to match address(es) from the following sample data and I can\'t get it to work:

medicalHistory      None
address            


        

Answer 1

The problem with your regex is that + is greedy and goes until it finds a character out of that group, the @ in the first case and - in the second.

Another approach is to use a non-greedy quantifier and a positive look-ahead for a newline followed by a word-character, like (python version):

re.findall(r'address\s+.*?(?=\n\w)', s, re.DOTALL)

It yields:

['address             24 Lewin Street, KUBURA, \n                NSW, Australia',
 'address             16 Yarra Street, \n                                     LAWRENCE, VIC, Australia']

Answer 2

I would do it this way:

address\s+((?![\r\n]+\w)[0-9a-zA-Z, \r\n\t])+

See it here on Regexr.

This ((?![\r\n]+\w)[0-9a-zA-Z, \r\n\t])+ is the important part, where I say, match the next character from [0-9a-zA-Z, \r\n\t], if (?![\r\n]+\w) is not following. This is matching what you expect.

In both your cases the regex stopped matching because of a character that is not included in your character class. If you want to go that way than you would need to combine a lazy quantifier and a positive lookahead:

address\s+([0-9a-zA-Z, \n\r\t]+?)(?=\r\w)

[0-9a-zA-Z, \n\r\t]+? is matching as less as possible till the condition (?=\r\w) is true.

See it here at Regexr