I'm trying to extract the url for a "background-image" on a Soundcloud page (ex. https://soundcloud.com/ohwondermusic/drive). I'm not quite sure why I'm finding this so difficult compared to extracting urls from other webpages that I've found guides for online.
From the example webpage I linked, I would like this url: https://i1.sndcdn.com/artworks-000125017075-di2n0i-t500x500.jpg that can be found by right clicking the album artwork and choosing 'inspect element' when in the Chrome browser.
I would like some way to consistently do this for other Soundcloud pages too (ie get the URL that would be found by inspecting the album artwork, the URL that ends in 500x500.jpg).
Does anyone know how to do this?
Edit: I've used various codes to attempt to solve, along the lines of:
def pull2(url):
html = urllib2.urlopen(url)
soup = BeautifulSoup(html)
readOnly = soup.body.find_all('div', attrs={'class': 'image readOnly customImage'})
print readOnly.attrs['style']
or
def test(url):
html = urllib2.urlopen(url)
soup = BeautifulSoup(html)
imgs = soup.findAll("div", {"class":"thumb-pic"})
for img in imgs:
print img.a['href'].split("imgurl=")[1]
It looks like you should just be able to grab the style from the correct span on each page with something like:
soup.find("span", class_="sc-artwork")['style']
Then, write a regex to extract the url from that or split it on "url"
Related
I am a newbie to web scraping. I am trying to get FASTA file from here, but somehow I cannot. First of all the problem starting for me span tag, I tried some couple of suggestions but not working for me I am suspecting that maybe there is a privacy problem
The FASTA file in this class, but when I run this code, I just can see FASTA title:
url = "https://www.ncbi.nlm.nih.gov/nuccore/193211599?report=fasta"
res = requests.get(url)
soup = BeautifulSoup(res.text, "html.parser")
fasta_data = soup.find_all("div")
for link in soup.find_all("div", {"class": "seqrprt seqviewer"}):
print link.text
url = "https://www.ncbi.nlm.nih.gov/nuccore/193211599?report=fasta"
res = requests.get(url)
soup = BeautifulSoup(res.text, "html.parser")
fasta_data = soup.find_all("div")
for link in soup.find_all("div", {"class": "seqrprt seqviewer"}):
print link.text
##When I try to reach directly via span, output is empty.
div = soup.find("div", {'id':'viewercontent1'})
spans = div.find_all('span')
for span in spans:
print span.string
Every scraping job involves two phases:
Understand the page that you want to scrape. (How it works? content loaded from Ajax? redirections? POST? GET? iframes? antiscraping stuff?...)
Emulate the webpage using your favourite framework
Do not write a single line of code before to work on point 1. Google network inspector is your friend, use it!
Regarding your webpage, it seems that the report is loaded into a viewer getting data from this url:
https://www.ncbi.nlm.nih.gov/sviewer/viewer.fcgi?id=193211599&db=nuccore&report=fasta&extrafeat=0&fmt_mask=0&retmode=html&withmarkup=on&tool=portal&log$=seqview&maxdownloadsize=1000000
Use that url and you will get your report.
I am trying to scrape website, but I encountered a problem. When I try to scrape data, it looks like the html differs from what I see on google inspect and from what I get from python. I get this with http://edition.cnn.com/election/results/states/arizona/house/01 I tried to scrape election results. I used this script to check HTML part of the webpage, and I noticed that they different. There is no classes that I need, like section-wrapper.
page =requests.get('http://edition.cnn.com/election/results/states/arizona/house/01')
soup = BeautifulSoup(page.content, "lxml")
print(soup)
Anyone knows what is the problem ?
http://data.cnn.com/ELECTION/2016/AZ/county/H_d1_county.json
This site use JavaScript fetch data, you can check the url above.
You can find this url in chrome dev-tools, there are many links, check it out
Chrome >>F12>> network tab>>F5(refresh page)>>double click the .josn url>> open new tab
import requests
from bs4 import BeautifulSoup
page=requests.get('http://edition.cnn.com/election/results/states/arizona/house/01')
soup = BeautifulSoup(page.content)
#you can try all sorts of tags here I used class: "ad" and class:"ec-placeholder"
g_data = soup.find_all("div", {"class":"ec-placeholder"})
h_data = soup.find_all("div"),{"class":"ad"}
for item in g_data:print item
#print '\n'
#for item in h_data:print item
Is there a "general" way to scrape link titles from any website in Python? For example, if I use the following code:
from urllib.request import url open
from bs4 import BeautifulSoup
site = "https://news.google.com"
html = urlopen(site)
soup = BeautifulSoup(html.read(), 'lxml');
titles = soup.findAll('span', attrs = { 'class' : 'titletext' })
for title in titles:
print(title.contents)
I am able to extract nearly every headline title from news.google.com. However, if I use the same code at www.yahoo.com, I am unable to due to a different HTML formatting.
Is there a more general way to do this so that it works for most sites?
No, each site is different and if you make a more general scraper, it will get more data that isn't as specific as every headline title.
For instance the following would get every headline title from google and would also probably get them from yahoo also.
titles = soup.find_all('a')
for title in titles:
print(title.get_text())
However it would also get you all of the headers and other links which would muddy up your results. (there are approximately 150 links on that google page that aren't headlines)
Not, that's why we need CSS selector and XPath, but if there are small number of page, there is a convenient way to do that:
site = "https://news.google.com"
if 'google' in site:
filters = {'name':'span', "class" : 'titletext' }
elif 'yahoo' in site:
filters = {'name':'blala', "class" : 'blala' }
titles = soup.findAll(**filters)
for title in titles:
print(title.contents)
For an extra curricular school project, I'm learning how to scrape a website. As you can see by the code below, I am able to scrape a form called, 'elqFormRow' off of one page.
How would one go about scraping all occurrences of the 'elqFormRow' on the whole website? I'd like to return the URL of where that form was located into a list, but am running into trouble while doing so because I don't know how lol.
import bs4 as bs
import urllib.request
sauce = urllib.request.urlopen('http://engage.hpe.com/Template_NGN_Convert_EG-SW_Combined_TEALIUM-RegPage').read()
soup = bs.BeautifulSoup(sauce, 'lxml')
for div in soup.find_all('div', class_='elqFormRow'):
print(div.text.strip())
You can grab the URLs from a page and follow them to (presumably) scrape the whole site. Something like this, which will require a little massaging depending on where you want to start and what pages you want:
import bs4 as bs
import requests
domain = "engage.hpe.com"
initial_url = 'http://engage.hpe.com/Template_NGN_Convert_EG-SW_Combined_TEALIUM-RegPage'
# get urls to scrape
text = requests.get(initial_url).text
initial_soup = bs.BeautifulSoup(text, 'lxml')
tags = initial_soup.findAll('a', href=True)
urls = []
for tag in tags:
if domain in tag:
urls.append(tag['href'])
urls.append(initial_url)
print(urls)
# function to grab your info
def scrape_desired_info(url):
out = []
text = requests.get(url).text
soup = bs.BeautifulSoup(text, 'lxml')
for div in soup.find_all('div', class_='elqFormRow'):
out.append(div.text.strip())
return out
info = [scrape_desired_info(url) for url in urls if domain in url]
URLlib stinks, use requests. If you need to go multiple levels down in the site put the URL finding section in a function and call it X number of times, where X is the number of levels of links you want to traverse.
Scrape responsibly. Try not to get into a sorcerer's apprentice situation where you're hitting the site over and over in a loop, or following links external to the site. In general, I'd also not put in the question the page you want to scrape.
Does anyone know how to use beautifulsoup in python.
I have this search engine with a list of different urls.
I want to get only the html tag containing a video embed url. and get the link.
example
import BeautifulSoup
html = '''https://archive.org/details/20070519_detroit2'''
#or this.. html = '''http://www.kumby.com/avatar-the-last-airbender-book-3-chapter-5/'''
#or this... html = '''https://www.youtube.com/watch?v=fI3zBtE_S_k'''
soup = BeautifulSoup.BeautifulSoup(html)
what should I do next . to get the html tag of video, or object or the exact link of the video..
I need it to put it on my iframe. i will integrate the python to my php. so getting the link of the video and outputting it using the python then i will echo it on my iframe.
You need to get the html of the page not just the url
use the built-in lib urllib like this:
import urllib
from bs4 import BeautifulSoup as BS
url = '''https://archive.org/details/20070519_detroit2'''
#open and read page
page = urllib.urlopen(url)
html = page.read()
#create BeautifulSoup parse-able "soup"
soup = BS(html)
#get the src attribute from the video tag
video = soup.find("video").get("src")
also with the site you are using i noticed that to get the embed link just change details in the link to embed so it looks like this:
https://archive.org/embed/20070519_detroit2
so if you want to do it to multiple urls without having to parse just do something like this:
url = '''https://archive.org/details/20070519_detroit2'''
spl = url.split('/')
spl[3] = 'embed'
embed = "/".join(spl)
print embed
EDIT
to get the embed link for the other links you provided in your edit you need to look through the html of the page you are parsing, look until you fint the link then get the tag its in then the attribute
for
'''http://www.kumby.com/avatar-the-last-airbender-book-3-chapter-5/'''
just do
soup.find("iframe").get("src")
the iframe becuase the link is in the iframe tag and the .get("src") because the link is the src attribute
You can try the next one because you should learn how to do it if you want to be able to do it in the future :)
Good luck!
You can't parse a URL. BeautifulSoup is used to parse an html page. Retrieve the page first:
import urllib2
data = urllib2.ulropen("https://archive.org/details/20070519_detroit2")
html = data.read()
Then you can use find, and then take the src attribute:
soup = BeautifulSoup(html)
video = soup.find('video')
src = video['src']
this is a one liner to get all the downloadable MP4 file in that page, in case you need it.
import bs4, urllib2
url = 'https://archive.org/details/20070519_detroit2'
soup = bs4.BeautifulSoup(urllib2.urlopen(url))
links = [a['href'] for a in soup.find_all(lambda tag: tag.name == "a" and '.mp4' in tag['href'])]
print links
Here are the output:
['/download/20070519_detroit2/20070519_detroit_jungleearth.mp4',
'/download/20070519_detroit2/20070519_detroit_sweetkissofdeath.mp4',
'/download/20070519_detroit2/20070519_detroit_goodman.mp4',
...
'/download/20070519_detroit2/20070519_detroit_wilson_512kb.mp4']
These are relative links and you and put them together with the domain and you get absolute path.