你可以在你的代码中找到这个函数的第一个子字符串(通过字符索引)。此外，还可以找到子字符串后面的内容。

def FindSubString(strText, strSubString, Offset=None):
    try:
        Start = strText.find(strSubString)
        if Start == -1:
            return -1 # Not Found
        else:
            if Offset == None:
                Result = strText[Start+len(strSubString):]
            elif Offset == 0:
                return Start
            else:
                AfterSubString = Start+len(strSubString)
                Result = strText[AfterSubString:AfterSubString + int(Offset)]
            return Result
    except:
        return -1

# Example:

Text = "Thanks for contributing an answer to Stack Overflow!"
subText = "to"

print("Start of first substring in a text:")
start = FindSubString(Text, subText, 0)
print(start); print("")

print("Exact substring in a text:")
print(Text[start:start+len(subText)]); print("")

print("What is after substring \"%s\"?" %(subText))
print(FindSubString(Text, subText))

# Your answer:

Text = "gfgfdAAA1234ZZZuijjk"
subText1 = "AAA"
subText2 = "ZZZ"

AfterText1 = FindSubString(Text, subText1, 0) + len(subText1)
BeforText2 = FindSubString(Text, subText2, 0) 

print("\nYour answer:\n%s" %(Text[AfterText1:BeforText2]))

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']

Python 3.8中保证文本包含子字符串的一行代码:

text[text.find(start:='AAA')+len(start):text.find('ZZZ')]

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;
}

令人惊讶的是，没有人提到这是我的一次性脚本的快速版本:

>>> x = 'gfgfdAAA1234ZZZuijjk'
>>> x.split('AAA')[1].split('ZZZ')[0]
'1234'