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 ]

>>> 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]]

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 ]

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

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