正则表达式匹配，直到找到除单词“and”之外的任何字符

Question

I would like to find a regex solution to match a part of the string after Item Number(s) until any characters are found except if it is the word 'and'

s = 'this part 123 should be ignored Item Number(s)92349252 and 30239429434, 124029354,345340332,  234325923 hallo 2121124'

it works if I add specifically hallo

re.match(r'.*?Item Number\(s\)(.*?)hallo.*$', s).group(1)

'92349252 and 30239429434, 124029354,345340332,  234325923 '

however I want it to work for any characters (including hallo ) except if it is the word and .

Answer 1

You dont need regex just use:

a,b,c = s.partition("and")
print(c)

c is the part after and.

Answer 2

We can try using a combination of string split with re.findall . First, split the input on the text Item Number(s) , and retain the second entry in the array. This corresponds to all text to the right of Item Number(s) . Then, use re.split to split on whitespace followed by some content which is not either the word and , a digit, space, or commad. Finally, use re.findall to capture all numbers from the remaining text.

s = 'this part 123 should be ignored Item Number(s)92349252 and 30239429434, 124029354,345340332,  234325923 hallo 2121124'
nums = re.findall(r'\b\d+\b', re.split(r' (?!\band\b|[\d\s,])', s.split('Item Number(s)')[1])[0])
print(nums)

['92349252', '30239429434', '124029354', '345340332', '234325923']

Answer 3

I stated the question incorrectly. The correct question is: find a string containing numbers after Item Number(s) until a word is found except if this word is and .

Rephrasing: find the string after Item Number(s) which have 1 or more digits separated by either zero or more non word character(s) or repeated the word 'and` preceded with a non word character followed by 0 or more non word character(s)

import re
s = '123 ignore Item Number(s)92349252 and,,;^and,and;;;30239429434, 124029354,345340332,  and and 234325923 hallo 2121124'
pattern = r'.*?Item Number\(s\)(((\W*?|(\W+?and)+\W*?)\d+)+)'
m = re.match(pattern, s).group(1)
numbers = re.findall('\d+', m)
print(numbers)

is

['92349252', '30239429434', '124029354', '345340332', '234325923']