BeautifulSoup 不能使用 if is None then continue 语句来避免

Question

代码如下：import requestsfrom bs4 import BeautifulSoupticker='FAST'url = "https://www.google.com/search?q=nasdaq+%s+earnings+reaction+history&tbs=qdr:m"%(ticker)response = requests.get(url)soup = BeautifulSoup(response.text, "html.parser")# for searchs in soup.find_all('div', {'class':'BNeawe s3v9rd AP7Wnd'}):cache = []for searchs in soup.find_all('div', {'class':'kCrYT'}):    if searchs.find('a')['href'] is None: continue    cache.append(searchs.find('a')['href'])    print(''.join(cache))当 .find('a')['href'] 没有返回结果时，我希望使用 if continue 语句来避免类型错误。然而，它并没有完成这项工作。有人可以指出克服它的方法吗？

Answer 1

取决于你想用块做什么，如果只有在href之后使用并使用后代组合select器指定具有子元素的父类和属性href

import requests

from bs4 import BeautifulSoup

ticker='FAST'

url = "https://www.google.com/search?q=nasdaq+%s+earnings+reaction+history&tbs=qdr:m"%(ticker)

response = requests.get(url)

soup = BeautifulSoup(response.text, "html.parser")

cache = [searchs['href'] for searchs in soup.select('div.kCrYT [href]')]

print(''.join(cache))

Answer 2

使用下面的代码从TypeError.

try:

    cache.append(searchs.find('a')['href'])

except TypeError:

    continue

Answer 3

不要试图在一行中放这么多：

...

for searchs in soup.find_all('div'):

    tag = searchs.find('a')

    #print(searchs)

    if not tag:

        continue

    try:

        h = tag['href']

    except KeyError as e:

        continue

    print(h)