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I need to get the CNPJ, but, ignore the match if string contains INTERESSADO. I need to do this using only regex, can't use if condition.

I thought that this patterns will works: (?!interessado: ).* CNPJ\W+(?P<cnpj>\d+\.\d+\.\d+\s?\/\d+-\s?\d+)

RECLAMADO: FOO LTDA - CNPJ: 99.999.999/9999-99
RECLAMADO: FOO FOO LTDA - CNPJ: 99.999.999/9999-99
TERCEIRO INTERESSADO: FOO FOO
TERCEIRO INTERESSADO: FOO FOO FOO FOO - CNPJ: 99.999.999/9999-99
TERCEIRO INTERESSADO: FOO FOO IT'S A TEST - CNPJ: 99.999.999/9999-99
TERCEIRO INTERESSADO: TEST
INTERESSADO: TEST - CNPJ: 99.999.999/9999-99

In this case, my pattern match all lines, but i need only the RECLAMADO and `INTERSSADO CNPJ's.

I'm using regex101 to test the patterns.

obs: i'm using regex .i and .s flags.

Daniel Bailo
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2 Answers2

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get the CNPJ, but, ignore the match if string contains INTERESSADO

That is, accepting the lines with CNPJ, and rejecting the lines with INTERESSADO.

There might be several ways to accomplish it. One of them is employ the notion of lookahead and lookbehind. (More Details)

In short, what we need to do is rejecting INTERESSADO with (?!...) while processing each line. Here is a simple demo for your case, shown below.

Codes:

import re

str = """RECLAMADO: FOO LTDA - CNPJ: 99.999.999/9999-99
RECLAMADO: FOO FOO LTDA - CNPJ: 99.999.999/9999-99
TERCEIRO INTERESSADO: FOO FOO
TERCEIRO INTERESSADO: FOO FOO FOO FOO - CNPJ: 99.999.999/9999-99
TERCEIRO INTERESSADO: FOO FOO IT'S A TEST - CNPJ: 99.999.999/9999-99
TERCEIRO INTERESSADO: TEST
INTERESSADO: TEST - CNPJ: 99.999.999/9999-99"""

lines = str.split("\n")
regex = "^((?<!interessado).(?!interessado))*(cnpj)[^0-9./-]*(?P<cnpjvalue>[0-9./-]*)$"
ptn = re.compile(regex,re.I|re.S)   # re.I for .i flag  ;  re.S for .s flag
for l in lines:
    m = ptn.match(l)
    if m:
        print("(Matched) cnpjvalue is " + m.group("cnpjvalue"))
    else:
        print("(Ignored)   ... ")

Output:

(Matched) cnpjvalue is 99.999.999/9999-99
(Matched) cnpjvalue is 99.999.999/9999-99
(Ignored)   ... 
(Ignored)   ... 
(Ignored)   ... 
(Ignored)   ... 
(Ignored)   ...
Michael Lee
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  • Instead of `^((?<!interessado).(?!interessado))*` that is redundant, you can use `^(?:(?!interessado).)*` or (why not but IMO less efficient), `^(?:.(?<!interessado))*`. It's useless to test backward and forward for each position. – Casimir et Hippolyte Nov 24 '21 at 00:51
  • @CasimiretHippolyte Thank you for clarifying that. In fact, I've already known that it's redundant before I posted it. Because I have emphased that this is a rather **simple demo**, since it just works fine, I would leave the further implementation to the asker. – Michael Lee Nov 24 '21 at 06:17
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You can use

import re
text = "RECLAMADO: FOO LTDA - CNPJ: 99.999.999/9999-99\nRECLAMADO: FOO FOO LTDA - CNPJ: 99.999.999/9999-99\nTERCEIRO INTERESSADO: FOO FOO\nTERCEIRO INTERESSADO: FOO FOO FOO FOO - CNPJ: 99.999.999/9999-99\nTERCEIRO INTERESSADO: FOO FOO IT'S A TEST - CNPJ: 99.999.999/9999-99\nTERCEIRO INTERESSADO: TEST\nINTERESSADO: TEST - CNPJ: 99.999.999/9999-99"
print( re.findall(r'^(?!.*interessado: ).* CNPJ\W+(\d+\.\d+\.\d+\s?\/\d+-\s?\d+)', text, re.M | re.I) )

See the Python demo. Output:

['99.999.999/9999-99', '99.999.999/9999-99']

See the regex demo. Details:

  • ^ - start of a line (due to re.M)
  • (?!.*interessado: ) - only go on matching if there is no interessado: + space on the line
  • .* - any zero or more chars other than line break chars as many as possible
  • - a space
  • CNPJ - a fixed string
  • \W+ - one or more non-word chars (may match across lines! If you do not need it, use [^\w\r\n]+)
  • (\d+\.\d+\.\d+\s?\/\d+-\s?\d+) - Group 1: your return value, one or more digits and . twice, then one or more digits, an optional whitespace, /, 1 or more digits, -, an optional whitespace, one or more digits.

Note that \s matches line breaks, so use [^\S\n] / [^\S\r\n] to match horizontal whitespace only.

Wiktor Stribiżew
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