Question

import bs4
import requests
import re

r = requests.get('https://www.the961.com/latest-news/lebanon-news/').text

soup = bs4.BeautifulSoup(r, 'lxml')

for article in soup.find_all('article'):

    title = article.h3.text
    print(title)

    date = article.find('span', class_='byline-part date')
    if date: print('Date:', date.text)

    author = article.find('span', class_="byline-part author")
    if author: print('Author:', author.text)

    link = article.find('h3', class_='title').a['href']
    link_r = requests.get(link).text
    
    soup_link = bs4.BeautifulSoup(link_r, 'lxml')

// 从标题中抓取链接，然后打开该链接并尝试抓取整篇文章，对此很陌生，所以我不知道该怎么做！

    for article in soup_link.find_all('article'):
        paragraph = article.find('p')
        print(paragraph)


    print()

Answer 1

在某些页面上，<p> 标记不在 <article> 下，因此返回 None。相反，要抓取所有段落（以及 <li> 标记，如果存在）使用以下 CSS 选择器：.entry-content > p, .entry-content li。

要使用 CSS 选择器，请使用 .select() 方法而不是 .find_all()。

在您的代码示例中：

import bs4
import requests


r = requests.get("https://www.the961.com/latest-news/lebanon-news/").text
soup = bs4.BeautifulSoup(r, "lxml")

for article in soup.find_all("article"):
    title = article.h3.text
    print(title)

    date = article.find("span", class_="byline-part date")
    if date:
        print("Date:", date.text)

    author = article.find("span", class_="byline-part author")
    if author:
        print("Author:", author.text, "\n")

    link = article.find("h3", class_="title").a["href"]
    link_r = requests.get(link).text
    soup_link = bs4.BeautifulSoup(link_r, "lxml")
    
    # Select all `p` tags (and `li`) under the class `entry-content`
    for page in soup_link.select(".entry-content > p, .entry-content li"):
        print(page.get_text(strip=True))
    print("-" * 80)
    print()

bs4 p 标签返回 None

1 个答案: