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我经常需要供应商安全公告页面上列出的 CVE 列表。有时这很容易复制，但通常它们与一堆文本混合在一起。

我有一段时间没有接触 Python，所以我认为这将是一个很好的练习来弄清楚如何提取这些信息——特别是因为我一直发现自己是手动操作的。

这是我当前的代码：

#!/usr/bin/env python3

# REQUIREMENTS
#   python3
#   BeautifulSoup (pip3 install beautifulsoup)
#   python 3 certificates (Applications/Python 3.x/ Install Certificates.command) <-- this one took me forever to figure out!

import sys
if sys.version_info[0] < 3:
    raise Exception("Use Python 3:  python3 " + sys.argv[0])
from urllib.request import urlopen
from bs4 import BeautifulSoup
import re

#specify/get the url to scrape
#url ='https://chromereleases.googleblog.com/2020/02/stable-channel-update-for-desktop.html'
#url = 'https://source.android.com/security/bulletin/2020-02-01.html'
url = input("What is the URL?  ") or 'https://chromereleases.googleblog.com/2020/02/stable-channel-update-for-desktop.html'
print("Checking URL: " + url)

# CVE regular expression
cve_pattern = 'CVE-\d{4}-\d{4,7}'

# query the website and return the html
page = urlopen(url).read()

# parse the html returned using beautiful soup
soup = BeautifulSoup(page, 'html.parser')

count = 0

############################################################
# ANDROID === search for CVE references within <td> tags ===

# find all <td> tags
all_tds = soup.find_all("td")

#print all_tds

for td in all_tds:
    if "cve" in td.text.lower():
        print(td.text)


############################################################
# CHROME === search for CVE reference within <span> tags ===

# find all <span> tags
all_spans = soup.find_all("span")

for span in all_spans:
    # this code returns results in triplicate
    for i in re.finditer(cve_pattern, span.text):
        count += 1
        print(count, i.group())


    # this code works, but only returns the first match
#   match = re.search(cve_pattern,span.text)
#   if match:
#       print(match.group(0))

我为 Android URL 所做的工作正常；我遇到的问题是 Chrome URL。他们在<span>标签内有 CVE 信息，我正在尝试利用正则表达式将其提取出来。

使用这种re.finditer方法，我最终得到一式三份的结果。使用re.search它错过了 CVE-2019-19925 的方法——他们在同一行列出了两个 CVE。

您能否就实现此功能的最佳方式提供任何建议？

4

1 回答 1

3

我终于自己解决了。不需要 BeautifulSoup；现在一切都是正则表达式。为了解决我之前看到的重复/三次结果，我将 re.findall 列表结果转换为字典（保留唯一值的顺序）并返回到列表。

import sys
if sys.version_info[0] < 3:
    raise Exception("Use Python 3:  python3 " + sys.argv[0])
import requests
import re

# Specify/get the url to scrape (included a default for easier testing)
### there is no input validation taking place here ###
url = input("What is the URL?  ") #or 'https://chromereleases.googleblog.com/2020/02/stable-channel-update-for-desktop.html'
print()

# CVE regular expression
cve_pattern = r'CVE-\d{4}-\d{4,7}'

# query the website and return the html
page = requests.get(url)

# initialize count to 0
count = 0

#search for CVE references using RegEx
cves = re.findall(cve_pattern, page.text)

# after several days of fiddling, I was still getting double and sometimes triple results on certain pages.  This next line
# converts the list of objects returned from re.findall to a dictionary (which retains order) to get unique values, then back to a list.
# (thanks to https://stackoverflow.com/a/48028065/9205677)
# I found order to be important sometimes, as the most severely rated CVEs are often listed first on the page
cves = list(dict.fromkeys(cves))

# print the results to the screen
for cve in cves:
    print(cve)
    count += 1

print()
print(str(count) + " CVEs found at " + url)
print()
于 2020-02-13T22:06:11.687 回答