打印稿。获取两个字符串之间的字符串。

搜索前缀和后缀之间的最短字符串

前缀-字符串/字符串数组/ null(意味着从开始搜索)。

Postfixes -字符串/字符串数组/ null(意味着搜索直到结束)。

public getStringInBetween(str: string, prefixes: string | string[] | null,
                          postfixes: string | string[] | null): string {

    if (typeof prefixes === 'string') {
        prefixes = [prefixes];
    }

    if (typeof postfixes === 'string') {
        postfixes = [postfixes];
    }

    if (!str || str.length < 1) {
        throw new Error(str + ' should contain ' + prefixes);
    }

    let start = prefixes === null ? { pos: 0, sub: '' } : this.indexOf(str, prefixes);
    const end = postfixes === null ? { pos: str.length, sub: '' } : this.indexOf(str, postfixes, start.pos + start.sub.length);

    let value = str.substring(start.pos + start.sub.length, end.pos);
    if (!value || value.length < 1) {
        throw new Error(str + ' should contain string in between ' + prefixes + ' and ' + postfixes);
    }

    while (true) {
        try {
            start = this.indexOf(value, prefixes);
        } catch (e) {
            break;
        }
        value = value.substring(start.pos + start.sub.length);
        if (!value || value.length < 1) {
            throw new Error(str + ' should contain string in between ' + prefixes + ' and ' + postfixes);
        }
    }

    return value;
}

import re
print re.search('AAA(.*?)ZZZ', 'gfgfdAAA1234ZZZuijjk').group(1)

在python中，可以使用正则表达式(re)模块中的findall方法从字符串中提取子字符串。

>>> import re
>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> ss = re.findall('AAA(.+)ZZZ', s)
>>> print ss
['1234']

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

>>> 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。

使用正则表达式-供进一步参考的文档

import re

text = 'gfgfdAAA1234ZZZuijjk'

m = re.search('AAA(.+?)ZZZ', text)
if m:
    found = m.group(1)

# found: 1234

or:

import re

text = 'gfgfdAAA1234ZZZuijjk'

try:
    found = re.search('AAA(.+?)ZZZ', text).group(1)
except AttributeError:
    # AAA, ZZZ not found in the original string
    found = '' # apply your error handling

# found: 1234