单

soup.find("form",{"class":"c-login__form"})

多个

res=soup.find_all("input")
for each in res:
    print(each)

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'])

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

下面的方法对我很有效

a_tag = soup.find_all("div",class_='full tabpublist')

这应该可以工作:

soup = BeautifulSoup(sdata)
mydivs = soup.findAll('div')
for div in mydivs: 
    if (div.find(class_ == "stylelistrow"):
        print div

或者我们可以使用lxml，它支持xpath和非常快!

from lxml import html, etree 

attr = html.fromstring(html_text)#passing the raw html
handles = attr.xpath('//div[@class="stylelistrow"]')#xpath exresssion to find that specific class

for each in handles:
    print(etree.tostring(each))#printing the html as string