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Kavermoki
Asked:2023-10-20 13:43:18 +0000 UTC2023-10-20 13:43:18 +0000 UTC

如果动态 HTML <li> 元素都相同，如何解析它们？

  • 772
dp.message_handler(text='Столовая')
async def help(message: types.Message, state: FSMContext):
    url = 'https://foodmonitoring.ru/13188/food'
    headers = {'Accept': '*/*',
                   'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/118.0.0.0 Safari/537.36'}
    reg = requests.get(url, headers=headers).text
    soup = BeautifulSoup(reg, 'lxml')
    block = soup.find('div', class_="col-12 col-md align-self-center")
    stolovaya = block.find_all(itemprop="menuListItemLink")
    for item in stolovaya:
        item_text = item.text
        item_href = "https://foodmonitoring.ru" + item.get("href")
        print(f"{item_text}: {item_href}")

大家好。目前，它解析块 (div) 中的整个 xlsx 列表。该块有 html 元素“li”，其中包含我需要的链接，如何才能使其仅解析顶部的第三个“li”元素？请帮帮我..

例子

python

1 个回答

  1. Best Answer

    您存储stolovaya已解析的所有 html 元素的列表。您只需访问索引即可获取所需的元素。在您提供的代码中，li我们可以得到第三个元素，如下所示：

    @dp.message_handler(text='Столовая')
async def help(message: types.Message, state: FSMContext):
    url = 'https://foodmonitoring.ru/13188/food'
    headers = {'Accept': '*/*',
               'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/118.0.0.0 Safari/537.36'}
    reg = requests.get(url, headers=headers).text
    soup = BeautifulSoup(reg, 'lxml')
    block = soup.find('div', class_="col-12 col-md align-self-center")
    stolovaya = block.find_all(itemprop="menuListItemLink")

    item = stolovaya[2]
    item_text = item.text
    item_href = "https://foodmonitoring.ru" + item.get("href")

    print(f"{item_text}: {item_href}")
    • 0

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