Python 如何使用BeautifulSoup查找所有下一个链接

我目前正在通过预设一个名为number_of_pages的变量来删除特定网站的所有页面。在添加我不知道的新页面之前，预设此变量一直有效。例如，下面的代码是3页，但网站现在有4页

base_url = 'https://securityadvisories.paloaltonetworks.com/Home/Index/?page='
number_of_pages = 3
for i in range(1, number_of_pages, 1):
   url_to_scrape = (base_url + str(i))

我想使用BeautifulSoup查找网站上的所有下一个链接。下面的代码查找第二个URL，但不查找第三个或第四个URL。如何在删除所有页面之前建立一个列表

base_url = 'https://securityadvisories.paloaltonetworks.com/Home/Index/?page='
CrawlRequest = requests.get(base_url)
raw_html = CrawlRequest.text
linkSoupParser = BeautifulSoup(raw_html, 'html.parser')
page = linkSoupParser.find('div', {'class': 'pagination'})
for list_of_links in page.find('a', href=True, text='next'):
  nextURL = 'https://securityadvisories.paloaltonetworks.com' + list_of_links.parent['href']
print (nextURL)

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有几种不同的方法来实现分页。这是其中之一

其思想是初始化一个无止境循环，并在没有“下一个”链接时将其中断：

如果执行，您将看到打印的以下消息：

Processing page: #1; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=
Processing page: #2; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=2
Processing page: #3; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=3
Processing page: #4; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=4
Done.

请注意，为了提高性能并在请求之间持久保存cookie，我们正在使用维护一个web抓取会话

Processing page: #1; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=
Processing page: #2; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=2
Processing page: #3; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=3
Processing page: #4; url: https://securityadvisories.paloaltonetworks.com/Home/Index/?page=4
Done.