为什么我不能用 selenium python 抓取特定的亚马逊音乐页面？

Question

https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP

上面提到的纯link，而不是用stack overflow标签点击后直接跳转到的link

这是url。

def get_soup(url):
headers = {'User-Agent':
           'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2228.0 Safari/537.36',
           }
r = requests.get(url, headers=headers)
r.raise_for_status()
return BeautifulSoup(r.text, 'lxml')

url = input("Please enter an Amazon music url:")
soup = get_soup(url)

我通过它请求时出错，为什么会这样？

Please enter an Amazon music url:https://www.amazon.com/Prettymuch-EP- 
PRETTYMUCH/dp/B07CF6YXDP 
Traceback (most recent call last):
  File "D:/Pycharm (4)/selemin.py", line 4, in <module>
    import amazon
  File "D:\Pycharm (4)\amazon.py", line 63, in <module>
    soup = get_soup(url)
  File "D:\Pycharm (4)\amazon.py", line 12, in get_soup
    r.raise_for_status()
  File "C:\Users\HP\AppData\Local\Programs\Python\Python37-32\lib\site- 
    packages\requests\models.py", line 940, in raise_for_status
    raise HTTPError(http_error_msg, response=self)
 requests.exceptions.HTTPError: 404 Client Error: Not Found for url: 
 https://www.amazon.com/Prettymuch-EP-PRETTYMUCH/dp/B07CF6YXDP%20

Answer 1

查看错误告诉您的内容 - 这是不同的 url。特别是，它会抛出一个错误，因为它以 %20 结尾。这意味着在您输入的末尾有一个 space。我建议处理您的输入以避免这种情况，例如

new_url = url.strip()

为什么我不能用 selenium python 抓取特定的亚马逊音乐页面？

Why can't i scrape the particular Amazon music page with selenium python?

python

beautifulsoup

amazon