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]))

如果你想寻找多次出现的情况。

content ="Prefix_helloworld_Suffix_stuff_Prefix_42_Suffix_andsoon"
strings = []
for c in content.split('Prefix_'):
    spos = c.find('_Suffix')
    if spos!=-1:
        strings.append( c[:spos])
print( strings )

或者更快:

strings = [ c[:c.find('_Suffix')] for c in content.split('Prefix_') if c.find('_Suffix')!=-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']

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'

>>> s = '/tmp/10508.constantstring'
>>> s.split('/tmp/')[1].split('constantstring')[0].strip('.')

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]]