擅长:python、mysql、java
<p>代码中的问题是您使用了错误的方法名findALL。。
soup对象中没有findALL方法,因此不会为此返回任何方法。
要解决新代码使用find\u all的问题,findAll也应该起作用(小写双l)。希望这件事能让你明白。你知道吗</p>
<pre><code>import requests
from bs4 import BeautifulSoup
link = "https://www.amazon.in/Power-Banks/b/ref=nav_shopall_sbc_mobcomp_powerbank?ie=UTF8&node=6612025031"
def amazon(url):
sourcecode = requests.get(url)
sourcecode_text = sourcecode.text
soup = BeautifulSoup(sourcecode_text, "html.parser")
# add "html.parser" as second arg , so you not get a warning .
# use soup.find_all for new code , also soup.findAll should work
for link in soup.find_all('a', {'class': 'a-link-normal aok-block a-text-normal'}):
href = link.get('href')
print(href)
amazon(link)
</code></pre>