有人可以幫助我更正此代碼，以便它可以獲取如下鏈接串列嗎？
我將不勝感激任何幫助。
我的代碼：
import requests
from bs4 import BeautifulSoup
header = {
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/98.0.4758.102 Safari/537.36',
}
restaurantLinks = open('pages.csv')
print(restaurantLinks)
urls = [url.strip() for url in restaurantLinks.readlines()]
restlist = []
for link in urls:
print("Opening link:" str(link))
response=requests.get(link, headers=header)
soup = BeautifulSoup(response.text, 'html.parser')
productlist = soup.find_all('div', class_='cNjlV')
print(productlist)
productlinks =[]
for item in productlist:
for link in item.find_all('a', href=True):
productlinks.append('https://www.tripadvisor.com' link['href'])
print(productlinks)
restlist.append(productlinks)
print(restlist)
df = pd.DataFrame(restlist)
df.to_csv('links.csv')
uj5u.com熱心網友回復：
而不是append()你串列中的元素嘗試extend()它：
restlist.extend(productlinks)
例子
import requests
from bs4 import BeautifulSoup
header = {
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/98.0.4758.102 Safari/537.36',
}
urls = ['https://www.tripadvisor.com/Attractions-g187427-Activities-oa60-Spain.html']
restlist = []
for link in urls:
print("Opening link:" str(link))
response=requests.get(link, headers=header)
soup = BeautifulSoup(response.text, 'html.parser')
restlist.extend(['https://www.tripadvisor.com' a['href'] for a in soup.select('a:has(h3)')])
df = pd.DataFrame(restlist)
df.to_csv('links.csv', index=False)
轉載請註明出處，本文鏈接：https://www.uj5u.com/houduan/478742.html