这招对我很管用:

for div in mydivs:
    try:
        clazz = div["class"]
    except KeyError:
        clazz = ""
    if (clazz == "stylelistrow"):
        print div

其他答案对我不起作用。

在其他回答中，findAll被用于soup对象本身，但我需要一种方法来对从findAll之后获得的对象中提取的特定元素中的对象执行类名查找。

如果您试图在嵌套的HTML元素中进行搜索，以按类名获取对象，请尝试下面的-

# parse html
page_soup = soup(web_page.read(), "html.parser")

# filter out items matching class name
all_songs = page_soup.findAll("li", "song_item")

# traverse through all_songs
for song in all_songs:

    # get text out of span element matching class 'song_name'
    # doing a 'find' by class name within a specific song element taken out of 'all_songs' collection
    song.find("span", "song_name").text

注意事项:

I'm not explicitly defining the search to be on 'class' attribute findAll("li", {"class": "song_item"}), since it's the only attribute I'm searching on and it will by default search for class attribute if you don't exclusively tell which attribute you want to find on. When you do a findAll or find, the resulting object is of class bs4.element.ResultSet which is a subclass of list. You can utilize all methods of ResultSet, inside any number of nested elements (as long as they are of type ResultSet) to do a find or find all. My BS4 version - 4.9.1, Python version - 3.8.1

关于@Wernight对上面关于部分匹配的答案的评论…

你可以部分匹配:

<div class="stylelistrow"> and <div class="stylelistrow button">

西班牙凉菜汤:

from gazpacho import Soup

my_divs = soup.find("div", {"class": "stylelistrow"}, partial=True)

两者都将被捕获并作为Soup对象列表返回。

截至BeautifulSoup 4+，

如果你只有一个类名，你可以像这样把类名作为参数传递:

mydivs = soup.find_all('div', 'class_name')

或者如果你有多个类名，只需将类名列表作为参数传递:

mydivs = soup.find_all('div', ['class1', 'class2'])

直接的方法是:

soup = BeautifulSoup(sdata)
for each_div in soup.findAll('div',{'class':'stylelist'}):
    print each_div

确保你使用了findAll的外壳，它不是findAll

以下操作应该可以工作

soup.find('span', attrs={'class':'totalcount'})

用你的类名替换'totalcount'，用你正在寻找的标签替换'span'。此外，如果类包含多个带空格的名称，只需选择一个并使用即可。

附注:这个函数用给定的条件找到第一个元素。如果你想找到所有的元素，那么将'find'替换为'find_all'。