Python - 如何使用正则表达式查找文本？

Question

Please bear with me, I'm new in Python. I have a text, and I want to get the value after ^s until the next ^ so for example there's ^s100^ then the value is 100 . This is what I've tried so far:

#!/usr/bin/python

import re

text="^request^ #13#10#13#10^s100^GET http://facebook.com #13#10Host: http://facebook.com #13#10X-Online-Host: http://facebook.com #13#10X-Forward-Host: http://facebook.com #13#10Connection: Keep-Alive#13#10#13#10"
if re.split(r'\^s',text):
    print "found it"

The problem is that it always returns found it even if I change the regex to re.split(r'\\^bla',text) and basically any text, it will always return found it Please help me to fix it.

Answer 1

What you probably want is re.search :

import re

text="^request^ #13#10#13#10^s100^GET http://facebook.com #13#10Host: http://facebook.com #13#10X-Online-Host: http://facebook.com #13#10X-Forward-Host: http://facebook.com #13#10Connection: Keep-Alive#13#10#13#10"
m = re.search(r'\^s(.*)\^',text)
print m.group(1)  # 100

Answer 2

You don't need as much code, try the following:

import re
text = "^request^ #13#10#13#10^s100^GET http://facebook.com #13#10Host: http://facebook.com #13#10X-Online-Host: http://facebook.com #13#10X-Forward-Host: http://facebook.com #13#10Connection: Keep-Alive#13#10#13#10"
match = re.search(r"\^s(\d+)\^", text)
if match:
    print match.group(1)

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Regex Explanation:

\^s(\d+)\^

Match the character “^” literally «\^»
Match the character “s” literally (case sensitive) «s»
Match the regex below and capture its match into backreference number 1 «(\d+)»
   Match a single character that is a “digit” (ASCII 0–9 only) «\d+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the character “^” literally «\^»

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