BeautifulSoup中是否有任何严格的findAll函数？

Question

I am using Python- 2.7 and BeautifulSoup

Apologies if I am unable to explain what exactly I want

There is this html page in which data is embedded in specific structure I want to pull the data ignoring the first block

But the problem is when I do-

self.tab = soup.findAll("div","listing-row") 

It also gives me the first block which is actually (unwanted html block)-

("div","listing-row wide-featured-listing")

I am not using

soup.find("div","listing-row")

since I want all the classes named "listing-row " only in that entire page.

How can I ignore the class named "listing-row wide-featured-listing" ?

Help/Guidance in any form is appreciated. Thanks a lot !

Answer 1

Or, you may make a CSS selector to match the class exactly to listing-row :

soup.select("div[class=listing-row]")

Demo:

>>> from bs4 import BeautifulSoup
>>> 
>>> data = """
... <div>
...     <div class="listing-row">result1</div>
...     <div class="listing-row wide-featured-listing">result2</div>
...     <div class="listing-row">result3</div>
... </div>
... """
>>> 
>>> soup = BeautifulSoup(data, "html.parser")
>>> print [row.text for row in soup.select("div[class=listing-row]")]
[u'result1', u'result3']

Answer 2

You could just filter out that element:

self.tab = [el for el in soup.find_all('div', class_='listing-row')
            if 'wide-featured-listing' not in el.attr['class']]

You could use a custom function:

self.tab = soup.find_all(lambda e: e.name == 'div' and
                                   'listing-row' in e['class'] and
                                   'wide-featured-listing' not in el.attr['class'])