Python 如何抓取网站内的特定链接？

我成功地浏览了标题和链接

我想用链接中的主要文章替换摘要选项卡（因为标题和摘要是一样的）

link=”https://www.vanglaini.org“+article.a['href']

（例如）

请帮我修改代码

下面是我的代码

import pandas as pd
import requests
from bs4 import BeautifulSoup

source = requests.get('https://www.vanglaini.org/').text
soup = BeautifulSoup(source, 'lxml')

list_with_headlines = []
list_with_summaries = []
list_with_links = []

for article in soup.find_all('article'):
    if article.a is None:
        continue
    headline = article.a.text.strip()
    summary = article.p.text.strip()
    link = "https://www.vanglaini.org" + article.a['href']
    list_with_headlines.append(headline)
    list_with_summaries.append(summary)
    list_with_links.append(link)

news_csv = pd.DataFrame({
    'Headline': list_with_headlines,
    'Summary': list_with_summaries,
    'Link' : list_with_links,
})

print(news_csv)
news_csv.to_csv('test.csv')

---

只需在for循环内部再次执行请求并获取标记文本

import pandas as pd
import requests
from bs4 import BeautifulSoup

source = requests.get('https://www.vanglaini.org/').text
soup = BeautifulSoup(source, 'lxml')

list_with_headlines = []
list_with_summaries = []
list_with_links = []

for article in soup.find_all('article'):
    if article.a is None:
        continue
    headline = article.a.text.strip()
    link = "https://www.vanglaini.org" + article.a['href']
    list_with_headlines.append(headline)
    list_with_links.append(link)
    soup = BeautifulSoup(requests.get(link).text, 'lxml')
    list_with_summaries.append(soup.select_one(".pagesContent").text)

news_csv = pd.DataFrame({
    'Headline': list_with_headlines,
    'Summary': list_with_summaries,
    'Link' : list_with_links,
})

print(news_csv)
news_csv.to_csv('test.csv')

Csv将是这样的

---

@anddrew这就是你想要的？是的，非常感谢。我会问你是否还有其他问题，希望你能回答。（我是python新手）如何在一个文件中向csv添加新新闻？我的意思是继续每天爬行，并在相同的csv文件中添加新信息。我有这个错误