正则表达式查找两个单词之间的多个值

Question

I'm trying to retrieve multiple values between two words, delimiting a specific part of the text. The pattern is this:

(^\d\d\d\d)\D+

I've tried a lot of regex, but I was not successful.

Below is an example of an attempt that is not working, because it is only returning the first value.

Livro[\s\S]*?(^\d\d\d\d)\D+[\s\S]*(?=em moeda corrente)

The text where I am trying to apply the regex is as follows. In bold the values I want to retrieve.

UPDATE I changed the example because sometimes the first four digits don't have the '/ dd' after them.

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CERTIDAO DE DIVIDA ATIVA Nr:XXXXXXXXX 6A

Inscrigao Pessoa Receita

5588 39783 03 -1SS VARIAVEL

Dispositivo Legal do Principal 03 - Artigos 55, 57, 58, 59, 63, 64, 151 e 153, no subiteém 14.01 da Lista de Servicos e na Tabela 03, inciso Ill, da Lei Complementar n° 12/1994, com alteragé6es dadas pelas Leis Complementares Municipais n° 56/1997, 116/2000, 196/2002, 217/2003, 270/2006, 314/2008, 320/2008, 399/2011 e 502/2015, 538/2017 e artigo 4° da Lei Complementar n° 4124/2000.

Livro: 14 _ Folha: 17583 a Data: 18/04/201 9 - a Doc.--Receita -Origam do Débite Principal. Corregao. “AcréscimoD.A Multa == dures. Total

2016 03 = =ISS VARIAVEL 36,80 6,47 2,16 4,33 20,33 70,09

2016 03 ISS VARIAVEL 116,00 20,38 6,82 13,64 62,74 219,58

2016 03 ISS VARIAVEL 340,00 59,74 19,99 39,97 179,88 639,58

2016 G3 ISS VARIAVEL 246,40 43,29 14,48 28,97 127,46 466,60

2016 /10 O03 ISS VARIAVEL 56,00 9,84 3,29 6,59 28,31 104,03

2016 /11 03 ISS VARIAVEL 623,84 109,61 36,67 73,35 308,05 1.161 52

2016 /12 03 ISS VARIAVEL 20,40 3,58 4,20 2,40 * 9,83 37,41

TOTAL, em moeda corrente, atualizado até: 23/06/2020 2.682,81 A) Atualizag&o Monetaria: artigos 153, paragrafo 1°, 200, | e 209, todos da Lei Complementar Municipal n° 12/94; artigo 4°, da Lei Complementar Municipal n° 1124/2000.

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I'm testing here https://regex101.com/r/tzgGVT/2 ( updated )

Thanks in advance;-)

Answer 1

As an alternative, you could make use of the regex PyPi module using the \G anchor to get the values in bold in capture group 1.

(?:^Livro.*(?:\r?\n(?!\d{4}/\d).*)*\r?\n|\G)(\d{4})/\d+.*\r?\n(?=(?:\d{4}/\d.*\r?\n)*.*?\bem moeda corrente\b)

In parts

• (?: Non capture group
  • ^Livro.*(?:\r?\n(?.\d{4}/\d)?*)*\r?\n Match start and lines that don't start with 4 digits / and digit
  • | Or
  • \G Assert the position at the end of the previous match
• ) Close non capture group
• (\d{4}) Capture group 1 , match 4 digits
• /\d+.*\r?\n Match / and 1+ digits followed by the rest of the line
• (?= Positive lookahead, assert what is on the right is
  • (?:\d{4}/\d.*\r?\n)* Repeat 0+ times matching a line the starts with 4 digits / and digit
  • .*?\bem moeda corrente\b Match a line that contains em moeda corrente
• ) Close positive lookahead

Regex demo | Python demo

Example code

import regex

pattern = r"(?:^Livro.*(?:\r?\n(?!\d{4}/\d).*)*\r?\n|\G)(\d{4})/\d+.*\r?\n(?=(?:\d{4}/\d.*\r?\n)*.*?\bem moeda corrente\b)"

print(regex.findall(pattern, s, regex.MULTILINE))

Output

['2016', '2016', '2016', '2016', '2016', '2016', '2016']

Answer 2

I think the complete pattern you are looking for is a year/month where year is four digits and month is 1 or 2 digits, not more , so followed by a [space]. In regular expression form:

import re

found = re.findall(r'(\d\d\d\d)/\d\d? ', text)
print(found)

Outputs:

['2016', '2016', '2016', '2016', '2016', '2016', '2016']

Or if you want the entire line for which the expression matches, just leave out the parentheses:

found = re.findall(r'\d\d\d\d/\d\d? .*', text)
for line in found: print(line)

Outputs:

2016/6 03 = =ISS VARIAVEL 36,80 6,47 2,16 4,33 20,33 70,09
2016/7 03 ISS VARIAVEL 116,00 20,38 6,82 13,64 62,74 219,58
2016/8 03 ISS VARIAVEL 340,00 59,74 19,99 39,97 179,88 639,58
2016/9 G3 ISS VARIAVEL 246,40 43,29 14,48 28,97 127,46 466,60
2016/10 O03 ISS VARIAVEL 56,00 9,84 3,29 6,59 28,31 104,03
2016/11 03 ISS VARIAVEL 623,84 109,61 36,67 73,35 308,05 1.161 52
2016/12 03 ISS VARIAVEL 20,40 3,58 4,20 2,40 * 9,83 37,41

Or split the text in two by a marker string (eg 'Livro') and search in the second part for 4 digits immediately after a newline character (is beginning of a line)

parts = text.split('Livro')
found = re.findall(r'\n(\d\d\d\d)', parts[1])

Answer 3

You need to make use of the findall() method of the re module. Moreover, the following path I think is the one that you're looking for: '\n(\d+)\/(?=[\s\S]+ em moeda corrente)'

>>> re.findall('\n(\d+)\/(?=[\s\S]+ em moeda corrente)', text)
['2016', '2016', '2016', '2016', '2016', '2016', '2016']

Try it out at: https://regex101.com/r/RoOB0t/2