使用sed，可以对字符串执行如下操作:

echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"

结果是1234。

你可以使用相同的正则表达式对re.sub函数做同样的事情。

>>> re.sub(r'.*AAA(.*)ZZZ.*', r'\1', 'gfgfdAAA1234ZZZuijjk')
'1234'

在基本sed中，捕获组由\(..\)表示，但在python中由(..)表示。

>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> start = s.find('AAA') + 3
>>> end = s.find('ZZZ', start)
>>> s[start:end]
'1234'

然后，如果您愿意，也可以在re模块中使用regexp，但在您的情况下这不是必要的。

另一种方法是使用列表(假设你正在寻找的子字符串是由数字组成的，只是):

string = 'gfgfdAAA1234ZZZuijjk'
numbersList = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
output = []

for char in string:
    if char in numbersList: output.append(char)

print(f"output: {''.join(output)}")
### output: 1234

此外，您可以在波纹函数中找到所有的组合

s = 'Part 1. Part 2. Part 3 then more text'
def find_all_places(text,word):
    word_places = []
    i=0
    while True:
        word_place = text.find(word,i)
        i+=len(word)+word_place
        if i>=len(text):
            break
        if word_place<0:
            break
        word_places.append(word_place)
    return word_places
def find_all_combination(text,start,end):
    start_places = find_all_places(text,start)
    end_places = find_all_places(text,end)
    combination_list = []
    for start_place in start_places:
        for end_place in end_places:
            print(start_place)
            print(end_place)
            if start_place>=end_place:
                continue
            combination_list.append(text[start_place:end_place])
    return combination_list
find_all_combination(s,"Part","Part")

结果:

['Part 1. ', 'Part 1. Part 2. ', 'Part 2. ']

使用PyParsing

import pyparsing as pp

word = pp.Word(pp.alphanums)

s = 'gfgfdAAA1234ZZZuijjk'
rule = pp.nestedExpr('AAA', 'ZZZ')
for match in rule.searchString(s):
    print(match)

收益率:

[[1234]]

只需一行代码就可以做到

>>> import re

>>> re.findall(r'\d{1,5}','gfgfdAAA1234ZZZuijjk')

>>> ['1234']

结果将收到列表…