I need to gather some information which is not provided by Facebook Analytics. For example, the original url and headline of an article promoted on Facebook as a link post. This info is buried in the html code of a Facebook post but I struggle to dig it out. Will appreciate your help.
Let's take this example: https://www.facebook.com/bbcnews/posts/10156428513547217
I identified classes for a link (bbc.in...): "_6ks"
and headline: 'mbs _6m6 _2cnj _5s6c'
The code below doesn't return anything:
from bs4 import BeautifulSoup
import requests
link = 'https://www.facebook.com/bbcnews/posts/10156428513547217'
r = requests.get(link)
soup = BeautifulSoup(r.content, "lxml")
for paragraph in soup.find_all("div", class_="_6ks"):
for a in paragraph("a"):
print(a.get('href'))
for paragraph in soup.find_all("div", class_='mbs _6m6 _2cnj _5s6c'):
for a in paragraph("a"):
print(a.get('hover'))
Another way to achieve the same would be something like below:
from bs4 import BeautifulSoup
import requests
link = 'https://www.facebook.com/bbcnews/posts/10156428513547217'
res = requests.get(link,headers={'User-Agent':'Mozilla/5.0'})
comment = res.text.replace("-->", "").replace("<!--", "")
soup = BeautifulSoup(comment, "lxml")
items = soup.select_one('.mbs a')
print(items.get("href")+"\n",items.text)
The reason you are not able to getting any output is b'coz both of those divs are placed cleverly placed within comment tags <!-- --> . Comments are ignored by the parsers. If you print the soup, both of the divs are present but within the comment tags.
We can get the comments and then make a new soup using that to bypass this.
from bs4 import BeautifulSoup
from bs4 import Comment
import requests
link = 'https://www.facebook.com/bbcnews/posts/10156428513547217'
headers={'User-Agent':'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:64.0) Gecko/20100101 Firefox/64.0'}
r = requests.get(link,headers=headers)
soup = BeautifulSoup(r.content, "lxml")
comments=soup.find_all(string=lambda text:isinstance(text,Comment))
soup=BeautifulSoup(comments[0], "lxml")
for paragraph in soup.find_all("div", class_="_6ks"):
for a in paragraph("a"):
print(a.get('href'))
print('-------------------------------------------------------------------')
for paragraph in soup.find_all("div", class_='mbs _6m6 _2cnj _5s6c'):
for a in paragraph("a"):
print(a.text)
Output
https://l.facebook.com/l.php?u=https%3A%2F%2Fbbc.in%2F2FP4EgR&h=AT3jWrl9cgJEY-8NBLgbvOEtDSZ8dBABo4TJaVJ66QBbWdCsBypvAkN6MD7VhJoOgy_LGJeomQAlcwtex_Ab-7TvWXhKkLB1m_TjzxOSk3R2uP8qTUL3aTTj4Pcz2ZSZunWxZsPtOlJSpay_AtQfNTuLTUQ80OrtvRiDMs8duN3b27IH2UPnGThQ_YGJAcYJdPE3R9JbyxSQNhJ8yTmaRJe8pMNbgVkentXU4p3liys2IQvphwRd0V8ANmo-4xvKj1dRADHy3hOyUkcv_L2u8Z4WpLx1AZQCTitvfSLvhQRMZ0cK1vIjkuv3gfurRf250p3D54GxQZIsVLymDzNtLbOnigIuFRHfQFAUSBDzJGTqQB3hs4lilYyFXIqaC2cdXwDp8GDrmYbgRWmEMmN6A5fHDdRlF4m7MXJO0vJ_7uqkh0TAdcvTSc0dqt5Wv3wOoEN5S1b2ddLZOp3DFwApAGkSHsOtW7Pjc-STFljuV045ERsUWUbmnALSl9vxB6tiZ0poa3aGxZqnlFqsaTB-A8plwCWp5ed9JALlurBco447aELbpuRexqoOajxTvS_yW9BdSXaufzpbPFKaNt5go7uf4GjdekpITCApJo2JoAOzzsfKHdg1MXasOCw
-------------------------------------------------------------------
MPs put forward rival Brexit plans
Related
I'm having trouble extracting a particular link from each of the web pages I'm considering.
In particular, considering for example the following websites:
https://lefooding.com/en/restaurants/ezkia
https://lefooding.com/en/restaurants/tekes
I would like to know if there is a unique way to extract the field WEBSITE (above the map) shown in the table on the left of the page.
For the reported cases, I would like to extract the links:
https://www.ezkia-restaurant.fr/
https://www.tekesrestaurant.com/
There are no unique tags to refer to and this makes extraction difficult.
I've thought of a solution using the selector, but it doesn't seem to work. For the first link I have:
from bs4 import BeautifulSoup
import requests
url = "https://lefooding.com/en/restaurants/ezkia"
res = requests.get(url)
soup = BeautifulSoup(res.text, 'html.parser')
data = soup.find("div", {"class": "e-rowContent"})
print(data)
but there is no trace of the link I need here. Does anyone know of a possible solution?
Try this:
import requests
from bs4 import BeautifulSoup
urls = [
"https://lefooding.com/en/restaurants/ezkia",
"https://lefooding.com/en/restaurants/tekes",
]
with requests.Session() as s:
for url in urls:
soup = [
link.strip() for link
in BeautifulSoup(
s.get(url).text, "lxml"
).select(".pageGuide__infos a")[-1]
]
print(soup)
Output:
['https://www.ezkia-restaurant.fr']
['https://www.tekesrestaurant.com/']
I am currently coding a price tracker for different websites, but I have run into an issue.
I'm trying to scrape the contents of a h1 tag using BeautifulSoup4, but I don't know how. I've tried to use a dictionary, as suggested in
https://stackoverflow.com/a/40716482/14003061, but it returned None.
Can someone please help? It would be appreciated!
Here's the code:
from termcolor import colored
import requests
from bs4 import BeautifulSoup
import smtplib
def choice_bwfo():
print(colored("You have selected Buy Whole Foods Online [BWFO]", "blue"))
url = input(colored("\n[ 2 ] Paste a product link from BWFO.\n", "magenta"))
url_verify = requests.get(url, headers=headers)
soup = BeautifulSoup(url_verify.content, 'html5lib')
item_block = BeautifulSoup.find('h1', {'itemprop' : 'name'})
print(item_block)
choice_bwfo()
here's an example URL you can use:
https://www.buywholefoodsonline.co.uk/organic-spanish-bee-pollen-250g.html
Thanks :)
This script will print content of <h1> tag:
import requests
from bs4 import BeautifulSoup
url = 'https://www.buywholefoodsonline.co.uk/organic-spanish-bee-pollen-250g.html'
# create `soup` variable from the URL:
soup = BeautifulSoup(requests.get(url).content, 'html.parser')
# print text of first `<h1>` tag:
print(soup.h1.get_text())
Prints:
Organic Spanish Bee Pollen 250g
Or you can do:
print(soup.find('h1', {'itemprop' : 'name'}).get_text())
When I try to get the link on a web page, bs4 doesn't catch the entire link, it stops before the **?ref**.....
I'll explain the question through the code:
imdb_link = "https://www.imdb.com/chart/top?ref_=nv_mv_250"
site = requests.get(imdb_link)
soup = BeautifulSoup(site.text,'lxml')
for items in soup.find("table",class_="chart").find_all(class_="titleColumn"):
link = items.find("a").get('href')
print(link)
The output is:
/title/tt0111161/
/title/tt0068646/
/title/tt0071562/
/title/tt0468569/
/title/tt0050083/
/title/tt0108052/
/title/tt0167260/
...and so on..
But it's wrong, as you can see by seeing the web page, because it might be:
/title/tt0111161/?ref_=adv_li_tt
/title/tt0068646/?ref_=adv_li_tt
...and so on...
How can I get the entire link? I mean the ?ref_=adv_li_tt too?
I use Python 3.7.4
Overall it might be interesting to try and work out how to get the full link - which I think you will need selenium for to allow javascript to run on page, you don't need the full link as seen on rendered page. What you have, with addition of prefix https://www.imdb.com, is perfectly serviceable.
import requests
from bs4 import BeautifulSoup as bs
with requests.Session() as s:
r = s.get('https://www.imdb.com/chart/top?ref_=nv_mv_25')
soup = bs(r.content, 'lxml')
links = ['https://www.imdb.com' + i['href'] for i in soup.select('.titleColumn a')]
for link in links:
r = s.get(link)
soup = bs(r.content, 'lxml')
print(soup.select_one('title').text)
You could let selenium load page so content renders then pass over to bs4 to get links as on page:
from selenium import webdriver
from bs4 import BeautifulSoup as bs
d = webdriver.Chrome()
d.get('https://www.imdb.com/chart/top?ref_=nv_mv_25')
soup = bs(d.page_source, 'lxml')
d.quit()
links = ['https://www.imdb.com' + i['href'] for i in soup.select('.titleColumn a')]
I want to crawl the homepage of youtube to pull out all the links of videos. Following is the code
from bs4 import BeautifulSoup
import requests
s='https://www.youtube.com/'
html=requests.get(s)
html=html.text
s=BeautifulSoup(html,features="html.parser")
for e in s.find_all('a',{'id':'video-title'}):
link=e.get('href')
text=e.string
print(text)
print(link)
print()
Nothing is happenning when I run the above code. It seems like the id is not getting discovered. What am I doing wrong
It is because you are not getting the same HTML as your browser have.
import requests
from bs4 import BeautifulSoup
s = requests.get("https://youtube.com").text
soup = BeautifulSoup(s,'lxml')
print(soup)
Save this code's output to a file named test.html and run. You will see that it is not the same as the browser's, as it looks corrupted.
See these questions below.
HTML in browser doesn't correspond to scraped data in python
Python requests not giving me the same HTML as my browser is
Basically, I recommend you to use Selenium Webdriver as it reacts as a browser.
Yes, this is a strange scrape, but if you scrape at the 'div id="content"' level, you are able to get the data you are requesting. I was able to get the titles of each video, but it appears youtube has some rate limiting or throttling, so I do not think you will be able to get ALL of the titles and links. At any rate, below is what I got working for the titles:
import requests
from bs4 import BeautifulSoup
url = 'https://www.youtube.com/'
response = requests.get(url)
page = response.text
soup = BeautifulSoup(page, 'html.parser')
links = soup.find_all('div', id='content')
for each in links:
print(each.text)
May be this could help for scraping all videos from youtube home page,
from bs4 import BeautifulSoup
import requests
r = 'https://www.youtube.com/'
html = requests.get(r)
all_videos = []
soup = BeautifulSoup(html.text, 'html.parser')
for i in soup.find_all('a'):
if i.has_attr('href'):
text = i.attrs.get('href')
if text.startswith('/watch?'):
urls = r+text
all_videos.append(urls)
print('Total Videos', len(all_videos))
print('LIST OF VIDEOS', all_videos)
This code snippet will selects all links from youtube.com homepage that contains /watch? in their href attribute (links to videos):
from bs4 import BeautifulSoup
import requests
soup = BeautifulSoup(requests.get('https://www.youtube.com/').text, 'lxml')
for a in soup.select('a[href*="/watch?"]'):
print('https://www.youtube.com{}'.format(a['href']))
Prints:
https://www.youtube.com/watch?v=pBhkG2Zwf-c
https://www.youtube.com/watch?v=pBhkG2Zwf-c
https://www.youtube.com/watch?v=gnn7GwqXek4
https://www.youtube.com/watch?v=gnn7GwqXek4
https://www.youtube.com/watch?v=AMKDVfucPfA
https://www.youtube.com/watch?v=AMKDVfucPfA
https://www.youtube.com/watch?v=daQcqPHx9uw
https://www.youtube.com/watch?v=daQcqPHx9uw
https://www.youtube.com/watch?v=V_MXGdSBbAI
https://www.youtube.com/watch?v=V_MXGdSBbAI
https://www.youtube.com/watch?v=KEW9U7s_zks
https://www.youtube.com/watch?v=KEW9U7s_zks
https://www.youtube.com/watch?v=EM7ZR5z3kCo
https://www.youtube.com/watch?v=EM7ZR5z3kCo
https://www.youtube.com/watch?v=6NPHk-Yd4VU
https://www.youtube.com/watch?v=6NPHk-Yd4VU
https://www.youtube.com/watch?v=dHiAls8loz4
https://www.youtube.com/watch?v=dHiAls8loz4
https://www.youtube.com/watch?v=2_mDOWLhkVU
https://www.youtube.com/watch?v=2_mDOWLhkVU
...and so on
I am trying to scrape this site . I managed to do it by using urllib and beautifulsoup. But urllib is too slow. I want to have asynchronous requests because the urls are thousands. I found that a nice package is grequests.
example:
import grequests
from bs4 import BeautifulSoup
pages = []
page="https://www.spitogatos.gr/search/results/residential/sale/r100/m100m101m102m103m104m105m106m107m108m109m110m150m151m152m153m154m155m156m157m158m159m160m161m162m163m164m165m166m167m168m169m170m171m172m173m174m175m176m177m178m179m180m181m182m183m184m185m186m187m188m189m190m191m192m193m194m195m196m197m198m106001m125000m"
for i in range(1,1000):
pages.append(page)
page="https://www.spitogatos.gr/search/results/residential/sale/r100/m100m101m102m103m104m105m106m107m108m109m110m150m151m152m153m154m155m156m157m158m159m160m161m162m163m164m165m166m167m168m169m170m171m172m173m174m175m176m177m178m179m180m181m182m183m184m185m186m187m188m189m190m191m192m193m194m195m196m197m198m106001m125000m"
page = page + "/offset_{}".format(i*10)
rs = (grequests.get(item) for item in pages)
a=grequests.map(rs)
The problem is that I don't know how to continue and use beautifulsoup. So as to get the html code of every page.
It would be nice to hear your ideas. Thank you!
Refer to the script below, also check the link of the source. It will help.
reqs = (grequests.get(link) for link in links)
resp=grequests.imap(reqs, grequests.Pool(10))
for r in resp:
soup = BeautifulSoup(r.text, 'lxml')
results = soup.find_all('a', attrs={"class":'product__list-name'})
print(results[0].text)
prices = soup.find_all('span', attrs={'class':"pdpPriceMrp"})
print(prices[0].text)
discount = soup.find_all("div", attrs={"class":"listingDiscnt"})
print(discount[0].text)
Source: https://blog.datahut.co/asynchronous-web-scraping-using-python/