I am new to python and am having this same issue that was brought up as an issue on the requests GitHub. I am trying to authenticate to a website that redirects you to a security question after the initial login. Both the initial login and subsequent page use the same "action URL" and on the second post request I am receiving a 404, here is my code, any help would be greatly appreciated, after asking the question on GitHub they said it was a question to ask here as it wasn't on their end. (although they had an issue on their GitHub about this):
from bs4 import BeautifulSoup as bs
import requests
import time
sources = ["https://www.dandh.com/v4/view?pageReq=dhMainNS"]
req = requests.Session()
def login():
authentication_url = "https://www.dandh.com/v4/dh"
header = {"user-agent": "Mozilla/5.0 (Windows NT 10.0; rv:60.0) Gecko/20100101 Firefox/60.0"}
payload = {"Login": "12345",
"PW": "12345",
"Request": "Login"}
payload2 = {"securityAnswer": "12345",
"Request": "postForm"}
req.post(authentication_url, data=payload, headers=header)
time.sleep(3)
req.post(authentication_url, data=payload2, headers=header)
time.sleep(3)
def print_object(sources):
for url in sources:
soup_object = bs(req.get(url).text, "html.parser")
print(soup_object.get_text())
def main():
login()
print_object(sources)
main()
Part 1
After looking around the website, that a bad part of the problem relies on the payload2 you just have to add another item to it which is: "formName":"loginChallengeValidation" so overall payload2should look something like this:
payload2 = {"formName":"loginChallengeValidation","securityAnswer": your_security_answer,
"Request": "postForm"}
This will prevent you from getting the status code 404. Hope this helps.
Part 2
Even though this the issue in your questions I doubt that this is what you really want (as the code in Part 1 will redirect you to another verification form). in order to get access to the website itself you would have to add the following lines:
header2 = {"user-agent": "Mozilla/5.0 (Windows NT 10.0; rv:60.0) Gecko/20100101 Firefox/60.0", "Referer":"https://www.dandh.com/v4/view?pageReq=LoginChallengeValidation"}
and
req.post("https://www.dandh.com/v4/view?pageReq=LoginChallengeValidation", headers=header)
so your final code should look like this:
from bs4 import BeautifulSoup as bs
import requests
import time
sources = ["https://www.dandh.com/v4/view?pageReq=dhMainNS"]
req = requests.Session()
def login():
authentication_url = "https://www.dandh.com/v4/dh"
header = {"user-agent": "Mozilla/5.0 (Windows NT 10.0; rv:60.0) Gecko/20100101 Firefox/60.0"}
header2 = {"user-agent": "Mozilla/5.0 (Windows NT 10.0; rv:60.0) Gecko/20100101 Firefox/60.0", "Referer":"https://www.dandh.com/v4/view?pageReq=LoginChallengeValidation"}
payload = {"Login": your_username,
"PW": your_pasword,
"Request": "Login"}
payload2 = {"formName":"loginChallengeValidation","securityAnswer": your_security_answer,
"Request": "postForm", "btContinue": ""}
req.post(authentication_url, data=payload, headers=header)
req.post("https://www.dandh.com/v4/view?pageReq=LoginChallengeValidation", headers=header)
time.sleep(3)
req.post(authentication_url, data=payload2, headers=header2)
time.sleep(3)
def print_object(sources):
for url in sources:
soup_object = bs(req.get(url).text, "html.parser")
print(soup_object.get_text())
def main():
login()
print_object(sources)
main()
(PS: you should replace your_username, your_password and your_security_answer with your credentials)
Also, I would like to note that I don't think that time.sleep(3) is useful in the code.
Really hope this helps.
Related
I am trying to scrape an article from The Wall Street Journal and it requires log-in to view the whole content. So, I have written a code like the below using Python Requests:
import requests
from bs4 import BeautifulSoup
import re
import base64
import json
username= <username>
password= <password>
base_url= "https://accounts.wsj.com"
session = requests.Session()
r = session.get("{}/login".format(base_url))
soup = BeautifulSoup(r.text, "html.parser")
credentials_search = re.search("Base64\.decode\('(.*)'", r.text, re.IGNORECASE)
base64_decoded = base64.b64decode(credentials_search.group(1))
credentials = json.loads(base64_decoded)
connection = <connection_name>
r = session.post(
'https://sso.accounts.dowjones.com/usernamepassword/login',
data = {
"username": username,
"password": password,
"connection": connection,
"client_id": credentials["clientID"],
"state": credentials["internalOptions"]["state"],
"nonce": credentials["internalOptions"]["nonce"],
"scope": credentials["internalOptions"]["scope"],
"tenant": "sso",
"response_type": "code",
"protocol": "oauth2",
"redirect_uri": "https://accounts.wsj.com/auth/sso/login"
})
soup = BeautifulSoup(r.text, "html.parser")
login_result = dict([
(t.get("name"), t.get("value"))
for t in soup.find_all('input')
if t.get("name") is not None
])
r = session.post(
'https://sso.accounts.dowjones.com/login/callback',
data = login_result,
headers={"User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.67 Safari/537.36"},
)
# article get request
r = session.get(
"https://www.wsj.com/articles/singapore-prime-minister-lee-rejects-claims-he-misused-state-powers-in-family-feud-1499094761",
headers={"User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.67 Safari/537.36"}
)
print(r.text)
am able to login through the request but still I am not getting full article to scrape. Can anyone help me with this? Thanks in advance :-)
An easy and reliable solution is using the selenium webdriver.
With selenium you create an automated browser window which opens the website and from there you can make it choose the elements to log in. Then the content loads in as usual, like when you look for it manually on your browser.
Then you can soup that page with BeautifulSoup.
from bs4 import BeautifulSoup
from selenium import webdriver
driver = webdriver.Firefox(executable_path="C:\Program Files (x86)\geckodriver.exe")
# Download the driver for your desired browser and place it in any path
# for Chrome it's: driver = webdriver.Chrome("C:\Program Files (x86)\chromedriver.exe")
# open your website link
driver.get("https://www.your-url.com")
# then soup the page with BS
html = driver.page_source
page_soup = BeautifulSoup(html)
From there you can use the "page_soup" as you usually would.
Any questions? :)
I want to extract all links of this website: https://pflegefinder.bkk-dachverband.de/pflegeheime/searchresult.php?required=1&statistics=1&searchdata%5BmaxDistance%5D=0&searchdata%5BcareType%5D=inpatientCare#/tab/general
The information I want are stored in the tbody: page code
Every time I try to extract the data I get no result.
from bs4 import BeautifulSoup
import requests
from requests_html import HTMLSession
url = "https://pflegefinder.bkk-dachverband.de/pflegeheime/searchresult.php?required=1&statistics=1&searchdata%5BmaxDistance%5D=0&searchdata%5BcareType%5D=inpatientCare#complex-searchresult"
session = HTMLSession()
r = session.get(url)
r.html.render()
soup = BeautifulSoup(r.html.html,'html.parser')
print(r.html.search("Details"))
Thank you for your help!
The site uses a backend api to deliver the info, if you look at your browser's Developer Tools - Network - fetch/XHR and refresh the page you'll see the data load via json in a request with a similar url to the one you posted.
You can scrape that data like this, it returns json which is easy enough to parse:
import requests
headers = {
'Referer':'https://pflegefinder.bkk-dachverband.de/pflegeheime/searchresult.php?required=1&statistics=1&searchdata%5BmaxDistance%5D=0&searchdata%5BcareType%5D=inpatientCare',
'User-Agent':'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/97.0.4692.71 Safari/537.36'
}
for page in range(2):
url = f'https://pflegefinder.bkk-dachverband.de/api/nursing-homes?required=1&statistics=1&maxDistance=0&careType=inpatientCare&limit=20&offset={page*20}'
resp = requests.get(url,headers=headers).json()
print(resp)
The api checks that you have a "Referer" header otherwise you get a 400 response.
i would like to scrape the header-element from these both links -
For me this 2 sites look absolute identical - pics see below
Why is only the scraping for the second link working and not for the first?
import time
import requests
from bs4 import BeautifulSoup
# not working
link = "https://apps.apple.com/us/app/bingo-story-live-bingo-games/id1179108009?uo=4"
page = requests.get (link)
time.sleep (1)
soup = BeautifulSoup (page.content, "html.parser")
erg = soup.find("header")
print(f"First Link: {erg}")
# working
link = "https://apps.apple.com/us/app/jackpot-boom-casino-slots/id1554995201?uo=4"
page = requests.get (link)
time.sleep (1)
soup = BeautifulSoup (page.content, "html.parser")
erg = soup.find("header")
print(f"Second Link: {len(erg)}")
Working:
Not Working:
The page is sometimes loaded by JavaScript, so request won't support it.
You can use a while loop to check if header appears in the soup and then break
import requests
from bs4 import BeautifulSoup
headers = {
"user-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/91.0.4472.77 Safari/537.36"
}
link = "https://apps.apple.com/us/app/bingo-story-live-bingo-games/id1179108009?uo=4"
while True:
soup = BeautifulSoup(requests.get(link).content, "html.parser")
header = soup.find("header")
if header:
break
print(header)
Try this to get whatever fields you wish to grab from those links. curently it fetches the title. You can modify res.json()['data'][0]['attributes']['name'] to grab any field of your interest. Mkae sure to put the urls within this list urls_to_scrape.
import json
import requests
from bs4 import BeautifulSoup
from urllib.parse import unquote
urls_to_scrape = {
'https://apps.apple.com/us/app/bingo-story-live-bingo-games/id1179108009?uo=4',
'https://apps.apple.com/us/app/jackpot-boom-casino-slots/id1554995201?uo=4'
}
base_url = 'https://apps.apple.com/us/app/bingo-story-live-bingo-games/id1179108009?uo=4'
link = 'https://amp-api.apps.apple.com/v1/catalog/US/apps/{}'
params = {
'platform': 'web',
'additionalPlatforms': 'appletv,ipad,iphone,mac',
'extend': 'customPromotionalText,customScreenshotsByType,description,developerInfo,distributionKind,editorialVideo,fileSizeByDevice,messagesScreenshots,privacy,privacyPolicyText,privacyPolicyUrl,requirementsByDeviceFamily,supportURLForLanguage,versionHistory,websiteUrl',
'include': 'genres,developer,reviews,merchandised-in-apps,customers-also-bought-apps,developer-other-apps,app-bundles,top-in-apps,related-editorial-items',
'l': 'en-us',
'limit[merchandised-in-apps]': '20',
'omit[resource]': 'autos',
'sparseLimit[apps:related-editorial-items]': '5'
}
with requests.Session() as s:
s.headers['User-Agent'] = 'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/91.0.4472.101 Safari/537.36'
res = s.get(base_url)
soup = BeautifulSoup(res.text,"lxml")
token_raw = soup.select_one("[name='web-experience-app/config/environment']").get("content")
token = json.loads(unquote(token_raw))['MEDIA_API']['token']
s.headers['Accept'] = 'application/json'
s.headers['Referer'] = 'https://apps.apple.com/'
s.headers['Authorization'] = f'Bearer {token}'
for url in urls_to_scrape:
id_ = url.split("/")[-1].strip("id").split("?")[0]
res = s.get(link.format(id_),params=params)
title = res.json()['data'][0]['attributes']['name']
print(title)
I want to get the first link from the html parser, but I'm getting anything(tried to print).
Also when i inspect the page on browser, the links are under class='r'
But when i print the soup.prettify(), and closely analyse then i find there is no class='r', instead class="BNeawe UPmit AP7Wnd".
Please help, thanks in advance!
import requests
import sys
import bs4
import webbrowser
def open_web(query):
res = requests.get('https://google.com/search?q=' + query)
res.raise_for_status()
soup = bs4.BeautifulSoup(res.text, "html.parser")
link_elements = soup.select('.r a')
link_to_open = min(1, len(link_elements))
for i in range(link_to_open):
webbrowser.open('https://google.com' + link_elements[i].get('href'))
open_web('youtube')
The problem is that google serves different HTML when you don't specify User-Agent in headers. To add User-Agent to your request, put it in the headers= attribute:
import requests
import bs4
def open_web(query):
headers = {'User-Agent':'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:68.0) Gecko/20100101 Firefox/68.0'}
res = requests.get('https://google.com/search?q=' + query, headers=headers)
res.raise_for_status()
soup = bs4.BeautifulSoup(res.text, "html.parser")
link_elements = soup.select('.r a')
print(link_elements)
open_web('youtube')
Prints:
[<a href="https://www.youtube.com/?gl=EE&hl=et" onmousedown="return rwt(this,'','','','1','AOvVaw2lWnw7oOhIzXdoFGYhvwv_','','2ahUKEwjove3h7onkAhXmkYsKHbWPAUYQFjAAegQIBhAC','','',event)"><h3 class="LC20lb">
... and so on.
You received a completely different HTML with different elements and selectors thus the output is empty. The reason why Google blocks your request is because default requests user-agent is python-requests and Google understands it and blocks it. Check what's your user-agent.
User-agent let identifies the browser, its version number, and its host operating system that representing a person (browser) in a Web context that lets servers and network peers identify if it's a bot or not.
Sometimes you can receive a different HTML, with different selectors.
You can pass URL params as a dict() which is more readable and requests do everything for you automatically (same goes for adding user-agent into headers):
params = {
"q": "My query goes here"
}
requests.get("YOUR_URL", params=params)
If you want to get the very first link then use select_one() instead.
Code and full example in the online IDE:
from bs4 import BeautifulSoup
import requests
headers = {
'User-agent':
"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582"
}
params = {
"q": "My query goes here"
}
html = requests.get('https://www.google.com/search', headers=headers, params=params)
soup = BeautifulSoup(html.text, 'lxml')
link = soup.select_one('.yuRUbf a')['href']
print(link)
# https://dev.mysql.com/doc/refman/8.0/en/entering-queries.html
Alternatively, you can do the same thing by using Google Organic Results API from SerpApi. It's a paid API with a free plan.
The difference in your case is that you only need to extract the data you want from JSON string rather than figuring out how to extract, maintain or bypass blocks from Google.
Code to integrate:
import os
from serpapi import GoogleSearch
params = {
"engine": "google",
"q": "My query goes here",
"hl": "en",
"api_key": os.getenv("API_KEY"),
}
search = GoogleSearch(params)
results = search.get_dict()
# [0] means first index of search results
link = results['organic_results'][0]['link']
# https://dev.mysql.com/doc/refman/8.0/en/entering-queries.html
Disclaimer, I work for SerpApi.
In Python 3, the following code obtains the HTML source for a webpage.
import urllib.request
url = "https://docs.python.org/3.4/howto/urllib2.html"
response = urllib.request.urlopen(url)
response.read()
How can I add the following custom header to the request when using urllib.request?
headers = { 'User-Agent' : 'Mozilla/5.0 (Windows NT 6.1; Win64; x64)' }
The request headers can be customized by first creating a request object then supplying it to urlopen.
import urllib.request
url = "https://docs.python.org/3.4/howto/urllib2.html"
hdr = { 'User-Agent' : 'Mozilla/5.0 (Windows NT 6.1; Win64; x64)' }
req = urllib.request.Request(url, headers=hdr)
response = urllib.request.urlopen(req)
response.read()
Source: Python 3.4 Documentation
import urllib.request
opener = urllib.request.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
urllib.request.install_opener(opener)
response = urllib.request.urlopen("url")
response.read()
Should you wish to learn about the details you can refer to the python documentation: https://docs.python.org/3/library/urllib.request.html
#Using urllib.request, with urlopen, allows to open the specified URL.
#Headers can be included inside the urlopen along with the url.
from urllib.request import urlopen
url = "https://docs.python.org/3.4/howto/urllib2.html"
header = { 'User-Agent' : 'Mozilla/5.0 (Windows NT 6.1; Win64; x64)' }
response = urlopen(url, headers=header)
response.read()