Don't get data from soup - python

I created bs4 web-scraping app with python. My program return empty list for review. For soup program runs normally.
from bs4 import BeautifulSoup
import requests
import pandas as pd
data = []
usernames = []
titles = []
comments = []
result = requests.get('https://www.kupujemprodajem.com/review.php?action=list')
soup = BeautifulSoup(result.text, 'html.parser')
review = soup.findAll('div', class_="single-review")
print(review)
for i in review:
header = i.find('div', class_="single-review__header")
footer = i.find('div', class_="comment-holder")
username = header.find('a', class_="single-review__username").text
title = header.find('div', class_="single-review__related-to").text
comment = footer.find('div', class_="single-review__comment").text
usernames.append(username)
titles.append(title)
comments.append(comment)
data.append(usernames)
data.append(titles)
data.append(comments)
print(data)
It isn't problem with class.

It looks like the reason this doesn't work is because the website needs a login in order to access that page. If in a private tab in a browser you where to visit https://www.kupujemprodajem.com/review.php?action=list, it would just take you to a login page.
There's 2 paths I can think of that you could take here:
Reverse engineer how the login process works and use the requests library to make a request to login and get (most likely) the session cookie from that in order to be able to request pages that require sign in.
(much simpler) use selenium instead. Selenium is a library that allows you to control a full browser instance, so you would be able to easily input credentials using this method. Beautiful soup on the other hand simply just parses html, so doing things like authenticating often take much more work in Beautiful Soup then they do in Selenium. I'd definitely suggest looking into it if you haven't already.

Related

When I scrape data from a website it only returns a newline

I've tried the code with different websites and elements, but nothing was working.
import requests
from lxml import html
page = requests.get('https://www.instagram.com/username.html')
tree = html.fromstring(page.content)
follow = tree.xpath('//span[#class="g47SY"]/text()')
print(follow)
input()
Above is the code I tried to use to aquire the number of instagram followers someone had.
One issue with web scraping Instagram is that a lot of content, including tag attribute values, is rendered dynamically. So the class you are using to fetch followers may change.
If you are able to use the Beautiful Soup library in Python, you might have an easier time parsing the page and getting the data. You can install it using pip install bs4. You can then search for the og:description descriptor, which follows the Open Graph protocol, and parse it to get follower counts.
Here's an example script that should get the follower count for a particular user:
import requests
from bs4 import BeautifulSoup
username = 'google'
html = requests.get('https://www.instagram.com/' + username)
bs = BeautifulSoup(html.text, 'lxml')
item = bs.select_one("meta[property='og:description']")
name = item.find_previous_sibling().get("content").split("•")[0]
follower_count = item.get("content").split(",")[0]
print(follower_count)

How do I scrape a full instagram page in python?

Long story short, I'm trying to create an Instagram python scraper, that loads the entire page and grabs all the links to the images. I have it working, only problem is, it only loads the original 12 photos that Instagram shows. Is there anyway I can tell requests to load the entire page?
Working code;
import json
import requests
from bs4 import BeautifulSoup
import sys
r = requests.get('https://www.instagram.com/accountName/')
soup = BeautifulSoup(r.text, 'lxml')
script = soup.find('script', text=lambda t: t.startswith('window._sharedData'))
page_json = script.text.split(' = ', 1)[1].rstrip(';')
data = json.loads(page_json)
non_bmp_map = dict.fromkeys(range(0x10000, sys.maxunicode + 1), 0xfffd)
for post in data['entry_data']['ProfilePage'][0]['graphql']['user']['edge_owner_to_timeline_media']['edges']:
image_src = post['node']['display_url']
print(image_src)
As Scratch already mentioned, Instagram uses "infinite scrolling" which won't allow you to load the entire page. But you can check the total amount of messages at the top of the page (within the first span with the _fd86t class). Then you can check if the page already contains all of the messages. Otherwise, you'll have to use a GET request to get a new JSON response. The benefit to this is that this request contains the first field, which seems to allow you to modify how many messages you get. You can modify this from its standard 12 to get all of the remaining messages (hopefully).
The necessary request looks similar to the following (where I've anonymised the actual entries, and with some help from the comments):
https://www.instagram.com/graphql/query/?query_hash=472f257a40c653c64c666ce877d59d2b&variables={"id":"XXX","first":12,"after":"XXX"}
parse_ig.py
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from bs4 import BeautifulSoup
from InstagramAPI import InstagramAPI
import time
c = webdriver.Chrome()
# load IG page here, whether a hashtag or a public user's page using c.get(url)
for i in range(10):
c.send_keys(Keys.END)
time.sleep(1)
soup = BeautifulSoup(c.page_source, 'html.parser')
ids = [a['href'].split('/') for a in soup.find_all('a') if 'tagged' in a['href']]
Once you have the ids, you can use Instagram's old API to get data for those. I'm not sure if it still works, but there was an API that I used -- which was limited by how much FB has slowly deprecated parts of the old API. Here's the link, in case you don't want to access Instagram API on your own :)
You can also add improvements to this simple code. Like instead of a "for" loop, you could do a "while" loop instead (i.e. while page is still loading, keep pressing END button.)
#zero's answer is incomplete (at least as of 1/15/19). c.send_keys is not a valid method. Instead, this is what I did:
c = webdriver.Chrome()
c.get(some_url)
element = c.find_element_by_tag_name('body') # or whatever tag you're looking to scrape from
for i in range(10):
element.send_keys(Keys.END)
time.sleep(1)
soup = BeautifulSoup(c.page_source, 'html.parser')
Here is a link to good tutorial for scraping Instagram profile info and posts that also handles pagination and works in 2022: Scraping Instagram
In summary, you have to use Instagram GraphQL API endpoint that requires user identifier and cursor from the previous page response: https://instagram.com/graphql/query/?query_id=17888483320059182&id={user_id}&first=24&after={end_cursor}

Unable to navigate Amazon pagination with Python and BS4

I've been trying to create a simple web scraper program to scrape the book titles of a 100 bestseller list on Amazon. I've used this code before on another site with no problems. But for some reason, it scraps the first page fine but then posts the same results for the following iterations.
I'm not sure if it's something to do with how Amazon creates its urls or not. When I manually enter the "#2" (and beyond) at the end of the url in the browser it navigates fine.
(Once the scrape is working I plan on dumping the data in csv files. But for now, print to the terminal will do.)
import requests
from bs4 import BeautifulSoup
for i in range(5):
url = "https://smile.amazon.com/Best-Sellers-Kindle-Store-Dystopian-Science-Fiction/zgbs/digital-text/6361470011/ref=zg_bs_nav_kstore_4_158591011#{}".format(i)
r = requests.get(url)
soup = BeautifulSoup(r.content, "lxml")
for book in soup.find_all('div', class_='zg_itemWrapper'):
title = book.find('div', class_='p13n-sc-truncate')
name = book.find('a', class_='a-link-child')
price = book.find('span', class_='p13n-sc-price')
print(title)
print(name)
print(price)
print("END")
This is a common problem that you have to face, some sites load the data asynchronous(with ajax) those are XMLHttpRequest that you can see in the tab networking of your DOM inspector. Usually the websites load the data from a different endpoint with POST method to solve that you can use urllib or requests library.
In this case the request is through a GET method and you can scrape it from this url with no need of extend your code https://www.amazon.com/Best-Sellers-Kindle-Store-Dystopian-Science-Fiction/zgbs/digital-text/6361470011/ref=zg_bs_pg_3?_encoding=UTF8&pg=3&ajax=1 where you only change the pg parameter

Web Scraping Python (BeautifulSoup,Requests)

I am learning web scraping using python but I can't get the desired result. Below is my code and the output
code
import bs4,requests
url = "https://twitter.com/24x7chess"
r = requests.get(url)
soup = bs4.BeautifulSoup(r.text,"html.parser")
soup.find_all("span",{"class":"account-group-inner"})
[]
Here is what I was trying to scrape
https://i.stack.imgur.com/tHo5S.png
I keep on getting an empty array. Please Help.
Sites like Twitter load the content dynamically, which sometimes depends upon the browser you are using etc. And due to dynamic loading there could be some elements in the webpage which are lazily loaded, which means that the DOM is inflated dynamically, depending upon the user actions, The tag you are inspecting in your browser Inspect element, is inspected the fully dynamically inflated HTML, But the response you are getting using requests, is inflated HTML, or a simple DOM waiting to load the elements dynamically on the user actions which in your case while fetching from requests module is None.
I would suggest you to use selenium webdriver for scraping dynamic javascript web pages.
Try this. It will give you the items you probably look for. Selenium with BeautifulSoup is easy to handle. I've written it that way. Here it is.
from bs4 import BeautifulSoup
from selenium import webdriver
driver = webdriver.Chrome()
driver.get("https://twitter.com/24x7chess")
soup = BeautifulSoup(driver.page_source,"lxml")
driver.quit()
for title in soup.select("#page-container"):
name = title.select(".ProfileHeaderCard-nameLink")[0].text.strip()
location = title.select(".ProfileHeaderCard-locationText")[0].text.strip()
tweets = title.select(".ProfileNav-value")[0].text.strip()
following = title.select(".ProfileNav-value")[1].text.strip()
followers = title.select(".ProfileNav-value")[2].text.strip()
likes = title.select(".ProfileNav-value")[3].text.strip()
print(name,location,tweets,following,followers,likes)
Output:
akul chhillar New Delhi, India 214 44 17 5
You could have done the whole thing with requests rather than selenium
import requests
from bs4 import BeautifulSoup as bs
import re
r = requests.get('https://twitter.com/24x7chess')
soup = bs(r.content, 'lxml')
bio = re.sub(r'\n+',' ', soup.select_one('[name=description]')['content'])
stats_headers = ['Tweets', 'Following', 'Followers', 'Likes']
stats = [item['data-count'] for item in soup.select('[data-count]')]
data = dict(zip(stats_headers, stats))
print(bio, data)

Scrape multiple pages with BeautifulSoup and Python

My code successfully scrapes the tr align=center tags from [ http://my.gwu.edu/mod/pws/courses.cfm?campId=1&termId=201501&subjId=ACCY ] and writes the td elements to a text file.
However, there are multiple pages available at the site above in which I would like to be able to scrape.
For example, with the url above, when I click the link to "page 2" the overall url does NOT change. I looked at the page source and saw a javascript code to advance to the next page.
How can my code be changed to scrape data from all the available listed pages?
My code that works for page 1 only:
import bs4
import requests
response = requests.get('http://my.gwu.edu/mod/pws/courses.cfm?campId=1&termId=201501&subjId=ACCY')
soup = bs4.BeautifulSoup(response.text)
soup.prettify()
acct = open("/Users/it/Desktop/accounting.txt", "w")
for tr in soup.find_all('tr', align='center'):
stack = []
for td in tr.findAll('td'):
stack.append(td.text.replace('\n', '').replace('\t', '').strip())
acct.write(", ".join(stack) + '\n')
The trick here is to check the requests that are coming in and out of the page-change action when you click on the link to view the other pages. The way to check this is to use Chrome's inspection tool (via pressing F12) or installing the Firebug extension in Firefox. I will be using Chrome's inspection tool in this answer. See below for my settings.
Now, what we want to see is either a GET request to another page or a POST request that changes the page. While the tool is open, click on a page number. For a really brief moment, there will only be one request that will appear, and it's a POST method. All the other elements will quickly follow and fill the page. See below for what we're looking for.
Click on the above POST method. It should bring up a sub-window of sorts that has tabs. Click on the Headers tab. This page lists the request headers, pretty much the identification stuff that the other side (the site, for example) needs from you to be able to connect (someone else can explain this muuuch better than I do).
Whenever the URL has variables like page numbers, location markers, or categories, more often that not, the site uses query-strings. Long story made short, it's similar to an SQL query (actually, it is an SQL query, sometimes) that allows the site to pull the information you need. If this is the case, you can check the request headers for query string parameters. Scroll down a bit and you should find it.
As you can see, the query string parameters match the variables in our URL. A little bit below, you can see Form Data with pageNum: 2 beneath it. This is the key.
POST requests are more commonly known as form requests because these are the kind of requests made when you submit forms, log in to websites, etc. Basically, pretty much anything where you have to submit information. What most people don't see is that POST requests have a URL that they follow. A good example of this is when you log-in to a website and, very briefly, see your address bar morph into some sort of gibberish URL before settling on /index.html or somesuch.
What the above paragraph basically means is that you can (but not always) append the form data to your URL and it will carry out the POST request for you on execution. To know the exact string you have to append, click on view source.
Test if it works by adding it to the URL.
Et voila, it works. Now, the real challenge: getting the last page automatically and scraping all of the pages. Your code is pretty much there. The only things remaining to be done are getting the number of pages, constructing a list of URLs to scrape, and iterating over them.
Modified code is below:
from bs4 import BeautifulSoup as bsoup
import requests as rq
import re
base_url = 'http://my.gwu.edu/mod/pws/courses.cfm?campId=1&termId=201501&subjId=ACCY'
r = rq.get(base_url)
soup = bsoup(r.text)
# Use regex to isolate only the links of the page numbers, the one you click on.
page_count_links = soup.find_all("a",href=re.compile(r".*javascript:goToPage.*"))
try: # Make sure there are more than one page, otherwise, set to 1.
num_pages = int(page_count_links[-1].get_text())
except IndexError:
num_pages = 1
# Add 1 because Python range.
url_list = ["{}&pageNum={}".format(base_url, str(page)) for page in range(1, num_pages + 1)]
# Open the text file. Use with to save self from grief.
with open("results.txt","wb") as acct:
for url_ in url_list:
print "Processing {}...".format(url_)
r_new = rq.get(url_)
soup_new = bsoup(r_new.text)
for tr in soup_new.find_all('tr', align='center'):
stack = []
for td in tr.findAll('td'):
stack.append(td.text.replace('\n', '').replace('\t', '').strip())
acct.write(", ".join(stack) + '\n')
We use regular expressions to get the proper links. Then using list comprehension, we built a list of URL strings. Finally, we iterate over them.
Results:
Processing http://my.gwu.edu/mod/pws/courses.cfm?campId=1&termId=201501&subjId=ACCY&pageNum=1...
Processing http://my.gwu.edu/mod/pws/courses.cfm?campId=1&termId=201501&subjId=ACCY&pageNum=2...
Processing http://my.gwu.edu/mod/pws/courses.cfm?campId=1&termId=201501&subjId=ACCY&pageNum=3...
[Finished in 6.8s]
Hope that helps.
EDIT:
Out of sheer boredom, I think I just created a scraper for the entire class directory. Also, I update both the above and below codes to not error out when there is only a single page available.
from bs4 import BeautifulSoup as bsoup
import requests as rq
import re
spring_2015 = "http://my.gwu.edu/mod/pws/subjects.cfm?campId=1&termId=201501"
r = rq.get(spring_2015)
soup = bsoup(r.text)
classes_url_list = [c["href"] for c in soup.find_all("a", href=re.compile(r".*courses.cfm\?campId=1&termId=201501&subjId=.*"))]
print classes_url_list
with open("results.txt","wb") as acct:
for class_url in classes_url_list:
base_url = "http://my.gwu.edu/mod/pws/{}".format(class_url)
r = rq.get(base_url)
soup = bsoup(r.text)
# Use regex to isolate only the links of the page numbers, the one you click on.
page_count_links = soup.find_all("a",href=re.compile(r".*javascript:goToPage.*"))
try:
num_pages = int(page_count_links[-1].get_text())
except IndexError:
num_pages = 1
# Add 1 because Python range.
url_list = ["{}&pageNum={}".format(base_url, str(page)) for page in range(1, num_pages + 1)]
# Open the text file. Use with to save self from grief.
for url_ in url_list:
print "Processing {}...".format(url_)
r_new = rq.get(url_)
soup_new = bsoup(r_new.text)
for tr in soup_new.find_all('tr', align='center'):
stack = []
for td in tr.findAll('td'):
stack.append(td.text.replace('\n', '').replace('\t', '').strip())
acct.write(", ".join(stack) + '\n')

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