Styjun

I am scraping a few URLs in batch using BeautifulSoap.

Here is my script (only relevant stuff):

import urllib2
from bs4 import BeautifulSoup
quote_page = 'https://example.com/foo/bar'
page = urllib2.urlopen(quote_page)
soup = BeautifulSoup(page, 'html.parser')
url_box = soup.find('div', attrs='class': 'player')
print url_box

This gives 2 different kinds of print depending on the HTML of URL (about half pages gives first print and rest give the second print).

Here's first kind of print:

<div class="player">
<video class="video-js vjs-fluid video-player" height="100%" id="some-player" poster="https://example.com/path/to/jpg/random.jpg" width="100%"></video>
<span data-type="trailer-src" data-url="https://example.com/path/to/mp4/random.mp4"></span>
</div>

And here's the other:

<div class="player">
<img alt="Image description here" src="https://example.com/path/to/jpg/random.jpg"/>
</div>

I want to extract the image URL which is poster in first and src in second.

Any ideas how I can do that so same script extracts that URL from either kind of print?

P.S The first print also has a mp4 link which I do not need.

You can use the get() method to get the value of attrs from the targeted tag.

You should be able to do something like this:

if url_box.find('video'):
 url = url_box.find('video').get('poster')
 mp4 = ulr_box.find('span').get('data-url')
if url_box.find('img'):
 url = url_box.find('img').get('src')

Decide which version you are dealing with and split accordingly:

firstVersion = '''<div class="player">
<video class="video-js vjs-fluid video-player" height="100%" id="some-player" poster="https://example.com/path/to/jpg/random.jpg" width="100%"></video>
<span data-type="trailer-src" data-url="https://example.com/path/to/mp4/random.mp4"></span>
</div>'''

secondVersion = '''<div class="player">
<img alt="Image description here" src="https://example.com/path/to/jpg/random.jpg"/>
</div>'''

def extractImageUrl(htmlInput):
 imageUrl = ""
 if "poster" in htmlInput:
 imageUrl = htmlInput.split('poster="')[1].split('"')[0]
 elif "src" in htmlInput:
 imageUrl = htmlInput.split('src="')[1].split('"')[0]
 return imageUrl