How can I get the source of chat without using selenium? - python

So my issue is that, I want to get user's id info from the chat.
The chat area what I'm looking for, looks like this...
<div id="chat_area" class="chat_area" style="will-change: scroll-position;">
<dl class="" user_id="asdf1234"><dt class="user_m"><em class="pc"></em> :</dt><dd id="1">blah blah</dd></dl>
asdf1234
...
What I want do is to,
Get the part starting with <a href='javascript:'' user_id='asdf1234' ...
so that I can parse this and do some other stuffs.
But this webpage is the one I'm currently using, and it can not be proxy(webdriver by selenium).
How can I extract that data from the chat?

It looks like you've got two separate problems here. I'd use both the requests and BeautifulSoup libraries to accomplish this.
Use your browser's developer tools, the network tab, to refresh the page and look for the request which responds with the HTML you want. Use the requests library to emulate this request exactly.
import requests
headers = {"name": "value"}
# Get case example.
response = requests.get("some_url", headers=headers)
# Post case example.
data = {"key": "value"}
response = requests.post("some_url", headers=headers, data=data)
Web-scraping is always finicky, if this doesn't work you're most likely going to need to use a requests session. Or a one-time hacky solution is just to set your cookies from the browser.
Once you have made the request you can use BeautifulSoup to scrape your user id very easily.
from bs4 import BeautifulSoup
# Create BS parser.
soup = BeautifulSoup(response.text, 'lxml')
# Find all elements with the attribute "user_id".
find_results = soup.findAll("a", {"user_id" : True})
# Iterate results. Could also just index if you want the single user_id.
for result in find_results:
user_id = result["user_id"]

Related

How do I fix fix getting "None" as a response when web scraping?

So I am trying to create a small code that gets the views from a youtube video and prints them. However using this code when printing the text var I just get the response "None". Is there a way to get a response of the actual view count using these libraries?
import requests
from bs4 import BeautifulSoup
url = requests.get("https://www.youtube.com/watch?v=dQw4w9WgXcQ")
soup = BeautifulSoup(url.text, 'html.parser')
text = soup.find('span', {'class': "view-count style-scopeytd-video-view-count-renderer"})
print(text)
To see why, you should use wget or curl to fetch a copy of that page and look at it, or use "view source" from your browser. That's what requests sees. None of those classes appear in the HTML you get back. That's why you get None -- because there ARE none.
YouTube builds all of its pages dynamically, through Javascript. requests doesn't interpret Javascript. If you need to do this, you'll need to use something like Selenium to run a real browser with a Javascript interpreter built in.

Python Requests Library - Scraping separate JSON and HTML responses from POST request

I'm new to web scraping, programming, and StackOverflow, so I'll try to phrase things as clearly as I can.
I'm using the Python requests library to try to scrape some info from a local movie theatre chain. When I look at the Chrome developer tools response/preview tabs in the network section, I can see what appears to be very clean and useful JSON:
However, when I try to use requests to obtain this same info, instead I get the entire page content (pages upon pages of html). Upon further inspection of the cascade in the Chrome developer tools, I can see there are two events called GetNowPlayingByCity: One contains the JSON info while the other seems to be the HTML.
JSON Response
HTML Response
How can I separate the two and only obtain the JSON response using the Python requests library?
I have already tried modifying the headers within requests.post (the Chrome developer tools indicate this is a post method) to include "accept: application/json, text/plain, */*" but didn't see a difference in the response I was getting with requests.post. As it stands I can't parse any JSON from the response I get with requests.post and get the following error:
"json.decoder.JSONDecodeError: Expecting value: line 4 column 1 (char 3)"
I can always try to parse the full HTML, but it's so long and complex I would much rather work with friendly JSON info. Any help would be much appreciated!
This is probably because the javascript the page sends to your browser is making a request to an API to get the json info about the movies.
You could either try sending the request directly to their API (see edit 2), parse the html with a library like Beautiful Soup or you can use a dedicated scraping library in python. I've had great experiences with scrapy. It is much faster than requests
Edit:
If the page uses dynamically loaded content, which I think is the case, you'd have to use selenium with the PhantomJS browser instead of requests. here is an example:
from bs4 import BeautifulSoup
from selenium import webdriver
url = "your url"
browser = webdriver.PhantomJS()
browser.get(url)
html = browser.page_source
soup = BeautifulSoup(html, 'lxml')
# Then parse the html code here
Or you could load the dynamic content with scrapy
I recommend the latter if you want to get into scraping. It would take a bit more time to learn but it is a better solution.
Edit 2:
To make a request directly to their api you can just reproduce the request you see. Using google chrome, you can see the request if you click on it and go to 'Headers':
After that, you simply reproduce the request using the requests library:
import requests
import json
url = 'http://paste.the.url/?here='
response = requests.get(url)
content = response.content
# in my case content was byte string
# (it looks like b'data' instead of 'data' when you print it)
# if this is you case, convert it to string, like so
content_string = content.decode()
content_json = json.loads(content_string)
# do whatever you like with the data
You can modify the url as you see fit, for example if it is something like http://api.movies.com/?page=1&movietype=3 you could modify movietype=3 to movietype=2 to see a different type of movie, etc

Beautiful Soup returns empty list

I am new to webscraping. So I have been given a task to extract data from : Here
I am choosing dataset of "comments". Below is my code for scraping.
import requests
from bs4 import BeautifulSoup
url = 'https://www.kaggle.com/hacker-news/hacker-news'
headers = {'User-Agent' : 'Mozilla/5.0'}
response = requests.get(url, headers = headers)
response.status_code
response.content
soup = BeautifulSoup(response.content, 'html.parser')
soup.find_all('tbody', class_ = 'TableBody-kSbjpE jGqIxa')
When I try to execute the last command it returns : [].
So, I am stuck here. I know we can get the data from kernel, but just for practice purpose where am I going wrong? Am I choosing wrong class? I want to scrape the data and probably save it to a CSV file or to a No-SQL Database, preferred Cassandra.
you are getting this [] because data you want to scrape is coming from API which loads after you web page load so page you are accessing does not contain that class
you can open you browser console and check in network as given in screenshot there you find data you want to scrape so you have to make request to that URL to get data
you can retrive data in this URL in preview tab you can see all data.
also if you have good knowledge of python you can also use this to scrape data
https://doc.scrapy.org/en/latest/intro/overview.html
Even though you were able to see the 'tbody', class_ = 'TableBody-kSbjpE jGqIxa' in the element inspector, the request that you make does not contain this class. See for yourself print(soup.prettify()). This is most likely because you're not requesting the correct url.
This may be not something you're aware of, but as a fyi:
You don't actually need to scrape using BeautifulSoup, you can get a list of all the available datasets from the API. Once you have it installed and configured, you can get the dataset: kaggle datasets download -d . Here's more info if you wish to proceed with the API instead: https://github.com/Kaggle/kaggle-api

Problems accessing page with python requests

I'm trying to extract the sector of a stock for a ML classification project. If I go to the following page:
https://www.six-swiss-exchange.com/shares/security_info_en.html?id=CH0012221716CHF4
I get (on the screen) some information about this stock (it changes, with the id code - I just pick the first one of the list). However, none of the information is available on a regular request. (The html page contains mostly javascript functions)
What I need is on the "Shares Details" tab (ICB Supersector at the bottom of the page). Once again nothing is available with a regular requests. I looked into what happens when I click this tab and the desired request is inside the url:
http://www.six-swiss-exchange.com/shares/info_details_en.html?id=CH0210483332CHF4&portalSegment=EQ&dojo.preventCache=1520360103852 HTTP/1.1
However, if I use this url directly, I get an 403 error from requests but work from a browser. I usually don't have any problems with this sort of things but in this case, do I have to submit cookies or any other information to access that page - no login is required and it can be easily accessed from any browser.
I am thinking 1) make a first request to the url that works, 2) store the cookie they send you (I don't know how to do that really) and 3) make a second request to the desired url. Would this work?
I tried using request.session() but I'm not sure if this is the solution or if I implemented it properly.
If anyone has dealt with that sort of problem, I would love any pointers in solving this. Thanks.
from urllib.parse import urljoin
import requests
from bs4 import BeautifulSoup
BASE_URL = 'https://www.six-swiss-exchange.com'
def get_page_html(isin):
params = {
'id': isin,
'portalSegment': 'EQ'
}
r = requests.get(
'{}/shares/info_details_en.html'.format(BASE_URL),
params=params
)
r.raise_for_status()
return r.text
def get_supersector_info(soup):
supersector = soup.find('td', text='ICB Supersector').next_sibling.a
return {
'link': urljoin(BASE_URL, supersector['href']),
'text': supersector.text
}
if __name__ == '__main__':
page_html = get_page_html('CH0012221716CHF4')
soup = BeautifulSoup(page_html, 'lxml')
supersector_info = get_supersector_info(soup)
Console:
https://www.six-swiss-exchange.com/search/quotes_en.html?security=C2700T
Industrial Goods & Services

Scraping second page in Python Yields Different Data Than Browsing to Second Page

I'm attempting to scrape some data from www.ksl.com/auto/ using Python Requests and Beautiful Soup. I'm able to get the results from the first search page but not subsequent pages. When I request the second page using the same URL Chrome constructs when I click the "Next" button on the page, I get a set of results that no longer matches my search query. I've found other questions on Stack Overflow that discuss Ajax calls that load subsequent pages, and using Chrome's Developer tools to examine the request. But, none of that has helped me with this problem -- which I've had on other sites as well.
Here is an example query that returns only Acuras on the site. When you advance in the browser to the second page, the URL is simply this: https://www.ksl.com/auto/search/index?page=1. When I use Requests to hit those two URLs, the second search results are not Acuras. Is there, perhaps a cookie that my browser is passing back to the server to preserve my filters?
I would appreciate any advice someone can give about how to get subsequent pages of the results I searched for.
Here is the simple code I'm using:
from requests import get
from bs4 import BeautifulSoup
page1 = get('https://www.ksl.com/auto/search/index?keyword=&make%5B%5D=Acura&yearFrom=&yearTo=&mileageFrom=&mileageTo=&priceFrom=&priceTo=&zip=&miles=25&newUsed%5B%5D=All&page=0&sellerType=&postedTime=&titleType=&body=&transmission=&cylinders=&liters=&fuel=&drive=&numberDoors=&exteriorCondition=&interiorCondition=&cx_navSource=hp_search&search.x=63&search.y=8&search=Search+raquo%3B').text
page2 = get('https://www.ksl.com/auto/search/index?page=2').text
soup = BeautifulSoup(page1, 'html.parser')
listings = soup.findAll("div", { "class" : "srp-listing-body-right" })
listings[0] # An Acura - success!
soup2 = BeautifulSoup(page2, 'html.parser')
listings2 = soup2.findAll("div", { "class" : "srp-listing-body-right" })
listings2[0] # Not an Acura. :(
Try this. Create a Session object and then call the links. This will maintain your session with the server when you send a call to the next link.
import requests
from bs4 import BeautifulSoup
s = requests.Session() # Add this line
page1 = s.get('https://www.ksl.com/auto/search/index?keyword=&make%5B%5D=Acura&yearFrom=&yearTo=&mileageFrom=&mileageTo=&priceFrom=&priceTo=&zip=&miles=25&newUsed%5B%5D=All&page=0&sellerType=&postedTime=&titleType=&body=&transmission=&cylinders=&liters=&fuel=&drive=&numberDoors=&exteriorCondition=&interiorCondition=&cx_navSource=hp_search&search.x=63&search.y=8&search=Search+raquo%3B').text
page2 = s.get('https://www.ksl.com/auto/search/index?page=1').text
Yes, the website uses cookies so that https://www.ksl.com/auto/search/index shows or extends your last search. More specifically, the search parameters are stored on the server for you particular session cookie, that is, the value of the PHPSESSID cookie.
However, instead of passing that cookie around, you can simply always do full queries (in the sense of the request parameters), each time using a different value for the page parameter.
https://www.ksl.com/auto/search/index?keyword=&make%5B%5D=Acura&yearFrom=&yearTo=&mileageFrom=&mileageTo=&priceFrom=&priceTo=&zip=&miles=25&newUsed%5B%5D=All&page=0&sellerType=&postedTime=&titleType=&body=&transmission=&cylinders=&liters=&fuel=&drive=&numberDoors=&exteriorCondition=&interiorCondition=&cx_navSource=hp_search&search.x=63&search.y=8&search=Search+raquo%3B
https://www.ksl.com/auto/search/index?keyword=&make%5B%5D=Acura&yearFrom=&yearTo=&mileageFrom=&mileageTo=&priceFrom=&priceTo=&zip=&miles=25&newUsed%5B%5D=All&page=1&sellerType=&postedTime=&titleType=&body=&transmission=&cylinders=&liters=&fuel=&drive=&numberDoors=&exteriorCondition=&interiorCondition=&cx_navSource=hp_search&search.x=63&search.y=8&search=Search+raquo%3B

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