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Question

我只需要IP地址。如何报废。我的代码正确NOW-

import urllib 
import urllib.request 
from bs4 import BeautifulSoup 

x = urllib.request.urlopen('http://bannedhackersips.blogspot.com/2014_08_04_archive.html') 
soup = BeautifulSoup(x,"html.parser") 
data = soup.find_all("ul", {"class": "posts"}) 

for content in data: 
    print(content.text) 

输出：

[Fail2Ban] SSH: banned 116.10.191.162 
[Fail2Ban] SSH: banned 116.10.191.204 
[Fail2Ban] SSH: banned 61.174.51.232 
[Fail2Ban] SSH: banned 61.174.51.224 
[Fail2Ban] SSH: banned 116.10.191.225 
[Fail2Ban] SSH: banned 200.162.47.130 
[Fail2Ban] SSH: banned 116.10.191.175 
[Fail2Ban] SSH: banned 61.174.51.223 
[Fail2Ban] SSH: banned 61.174.51.234 
[Fail2Ban] SSH: banned 61.174.51.209 
[Fail2Ban] SSH: banned 116.10.191.165 
[Fail2Ban] SSH: banned 106.240.247.220 

Answer 1

您可以从文本与正则表达式中提取：

data = soup.find("ul", {"class": "posts"}) 

import re 

r = re.compile("\d+\.\d+\.\d+\.\d+") 

print(r.findall(data.text)) 
['116.10.191.162', '116.10.191.204', '61.174.51.232', '61.174.51.224', '116.10.191.225', '200.162.47.130', '116.10.191.175', '61.174.51.223', '61.174.51.234', '61.174.51.209', '116.10.191.165', '106.240.247.220'] 

或者作为重复该模式可以分成与splitlines子并从每个子串的末尾分开一次以提取ip：

data = soup.find("ul", {"class": "posts"}) 

ips = [line.rsplit(None, 1)[1] for line in data.text.splitlines() if line] 

print(ips) 
['116.10.191.162', '116.10.191.204', '61.174.51.232', '61.174.51.224', '116.10.191.225', '200.162.47.130', '116.10.191.175', '61.174.51.223', '61.174.51.234', '61.174.51.209', '116.10.191.165', '106.240.247.220'] 

在页面上只有一个posts类，所以查找就足够了，当您遍历find_all时，您实际上正在遍历单个元素列表。