Question

我想从下面的多行字符串中检索Auth=值。我试过Python re.match没有成功。如果我能得到任何帮助，我将不胜感激？

SID = DQAAALsAAABCeyCMlOaYMHkv55TUQFxA71fxE1LpgpmL1G_o8YennFwBhar2I_LNmJjGjvLHVQy8tSRfYdLnUIHhKyD0FTZBzXyG_s8U4Pt97n9hPz68ZFSM42Qv6Qxuk74TQygHJXhjLWXNuD5mMsh8_MAs-nmhSToNFIyWoP-uTZ_LN2yQS1o9MB43fzuIIxp-1euXGxMceVVrjyidrYeEB13HS5kMHH-HGjiZhoIJBmu5es7pLPj9Ie8NJZ1K3kFhdVEJa4sLSID = DQAAAL4AAACypRIVyVXcs5zYIeUEt9v-wEwPKgQ8Oe23_URsDeHCg-rR2qQK4dTxPV1J6BPTO-6Zly2H9t4sVhm0vHe8IT6sKLdX2IQ8PgGMtSHQNkpQ8zEan0CyFyUetbSW4af6mlk2pksDpvXNm5GtNTj5eTwkCQUmgGep42u5iuCGFy-o9a1cQWz45NO_J8zIYnBdOqlheNTqaMWpi4hpr-_u8Muzs4RjlEbkuYfDu7MrdsJAFwxf0BVW2cGBtB-K2jwaK7wAuth = 873hdyjsbcvuei73hckwoxnaodbc8dnskc8HU1mKRqxh6yEU-9tqx148GqC7h90_190ZzxpEZOHAH5HTptliylRXvMPyqPyijMNu21bOA6ZhvZFuL8YNB3KF63YuV0n5TFJd1-rMI2LQIdPMVBnsxnEGrLIeFOugAFCZ_3OelAc4XjeKdDvIowxkNnvaooXT4kxtkQWzieA3JRKy3Y-Lbi7E0qiXC99GtHVDh5VWvdTs2LCv3wnRULtLp6ZCoToZ8qoUWMzU9PZldsKSnE

Answer 1

import re

text = "SID=DQAAALsAAABCeyCMlOaYMHkv55TUQFxA71fxE1LpgpmL1G_o8YennFwBhar2I_LNmJjGjvLHVQy8tSRfYdLnUIHhKyD0FTZBzXyG_s8U4Pt97n9hPz68ZFSM42Qv6Qxuk74TQygHJXhjLWXNuD5mMsh8_MAs-nmhSToNFIyWoP-uTZ_LN2yQS1o9MB43fzuIIxp-1euXGxMceVVrjyidrYeEB13HS5kMHH-HGjiZhoIJBmu5es7pLPj9Ie8NJZ1K3kFhdVEJa4sLSID=DQAAAL4AAACypRIVyVXcs5zYIeUEt9v-wEwPKgQ8Oe23_URsDeHCg-rR2qQK4dTxPV1J6BPTO-6Zly2H9t4sVhm0vHe8IT6sKLdX2IQ8PgGMtSHQNkpQ8zEan0CyFyUetbSW4af6mlk2pksDpvXNm5GtNTj5eTwkCQUmgGep42u5iuCGFy-o9a1cQWz45NO_J8zIYnBdOqlheNTqaMWpi4hpr-_u8Muzs4RjlEbkuYfDu7MrdsJAFwxf0BVW2cGBtB-K2jwaK7w*Auth=*873hdyjsbcvuei73hckwoxnaodbc8dnskc8HU1mKRqxh6yEU-9tqx148GqC7h90_190ZzxpEZOHAH5HTptliylRXvMPyqPyijMNu21bOA6ZhvZFuL8YNB3KF63YuV0n5TFJd1-rMI2LQIdPMVBnsxnEGrLIeFOugAFCZ_3OelAc4XjeKdDvIowxkNnvaooXT4kxtkQWzieA3JRKy3Y-Lbi7E0qiXC99GtHVDh5VWvdTs2LCv3wnRULtLp6ZCoToZ8qoUWMzU9PZldsKSnE"
m = re.search('Auth=(.+)',text).groups()[0]
print m

结果： '* 873hdyjsbcvuei73hckwoxnaodbc8dnskc8HU1mKRqxh6yEU-9tqx148GqC7h90_190ZzxpEZOHAH5HTptliylRXvMPyqPyijMNu21bOA6ZhvZFuL8YNB3KF63YuV0n5TFJd1-rMI2LQIdPMVBnsxnEGrLIeFOugAFCZ_3OelAc4XjeKdDvIowxkNnvaooXT4kxtkQWzieA3JRKy3Y-Lbi7E0qiXC99GtHVDh5VWvdTs2LCv3wnRULtLp6ZCoToZ8qoUWMzU9PZldsKSnE'

Answer 2

尝试使用re.search代替re.match，match checks for a match only at the beginning of the string, while search checks for a match anywhere in the string。

在创建正则表达式时，您可能还需要re.MULTILINE标志，这样的事情可以满足您的需求：

re.search("Auth=(.*)", data, re.MULTILINE)

如果验证不是最后一个值，请将.*替换为匹配的正则表达式，直到下一个值。

Answer 3

安德鲁的回答是正确的。另外，我会提到命名组功能。您可以轻松地从正则表达式中提取部件：

import re

text = "SID=whateverAuth=myauthvalue"
m = re.search('Auth=(?P<auth>.+)', text, re.MULTILINE)
print m.group('auth') #should print myauthvalue

Python官方文档： http://docs.python.org/library/re.html

Answer 4

不需要regexp，而是在行上循环并在=上拆分。或许有点啰嗦......

#!/usr/bin/python

with open("auth") as fd:
    for line in fd:
        if 'Auth' in line:
            auth=line.split('=')[-1].strip()
            found=True
        if found:
            auth.join(line.strip())

*873hdyjsbcvuei73hckwoxnaodbc8dnskc8HU1mKRqxh6yEU-9tqx148GqC7h90_190ZzxpEZOHAH5HTptliylRXvMPyqPyijMNu21bOA6ZhvZFuL8YNB3KF63YuV0n5TFJd1-rMI2LQIdPMVBnsxnEGrLIeFOugAFCZ_3OelAc4XjeKdDvIowxkNnvaooXT4kxtkQWzieA3JRKy3Y-Lbi7E0qiXC99GtHVDh5VWvdTs2LCv3wnRULtLp6ZCoToZ8qoUWMzU9PZldsKSnE

Answer 5

如果字符串是多行并且总是以相同的顺序（SID，LSID，Auth），则可以使用：

wantedresult = originalstring.strip().split('\n')[-1].split('=')[-1]

strip是必要的，以删除大多数服务器包含在原始字符串末尾的额外换行符。

使用Python Re模块从多行提取文本

5 个答案: