A website url www.example.com/abc does not change when iterating over different pages. After inspecting using DEV TOOLS on CHROME browser XHR tab it is noticed that POST request is posted to url www.example.com/abc-data and based on which response source code of url www.example.com/abc changes.
However 90% of the data is being returned and can be scraped from XHR response, but 10% of the data is present in dynamic source code which is updated depending on XHR response.
I've tried all the possible available solutions on internet but not able to crack the solution for this problem.
Env:
Mac OS X Ventura
Python 3.7.3
Note: Using BeautifulSoup
Short code snippet
url1 = www.example.com/abc
url2 = www.example.com/abc-data
with requests.Session() as s:
r = s.get(url1) # Extract token from this URL
# SOME CODE HERE
r = s.post(url2, data=payload) # Use token from above for this URL and session
soup = BeautifulSoup(r.text, 'html.parser')
After POST request as above, HTML SOURCE CODE is updated and I am not able to get that using BeautifulSoup.
What I am receiving is just JSON response.
Any help would be much appreciated!!!
As I understood, you're trying to get the dynamic content of a web page using BeautifulSoup. That is not possible to do. BeautifulSoup only scrapes static web content.
If you really want to get the Dynamic Content, I recommend using Selenium.
Related
I tried to getting the title of a web page by web scraping using Beautifulsoup4 python module and it's returning a string "Not Acceptable!" as the title, but when I open the webpage via browser the title is different. I tried looping through list of links and extract titles of all the webpages but it's returning the same string "Not Acceptable!" for all the links.
here is the python code
from bs4 import BeautifulSoup
import requests
URL = 'https://insights.blackcoffer.com/how-is-login-logout-time-tracking-for-employees-in-office-done-by-ai/'
result = requests.get(URL)
doc = BeautifulSoup(result.text, 'html.parser')
tag = doc.title
print(tag.get_text())
here is link to the corresponding web page webpage link
I don't know if it is a problem with Beautifulsoup4 or with requests library, is it because the site has enabled bot protection and not returning the HTML when sending the requests?
The server expects the User-Agent header. Interestingly, it is happy with any User-Agent, even a fictitious one:
result = requests.get(URL, headers = {'User-Agent': 'My User Agent 1.0'})
An easy way to debug this kind of issue is to print (or write to a file) the request.text. This is because some servers don't allow scraping. Some websites generate HTML using JavaScript at runtime (e.g. YouTube). These are some of the scenarios where the request.text can be different than the source HTML we see in the browser. The below text has been returned by the server.
<head><title>Not Acceptable!</title></head><body><h1>Not Acceptable!</h1><p>An appropriate representation of the requested resource could not be found on this server. This error was generated by Mod_Security.</p></body></html>
Edit:
As pointed by DYZ, this is a 406 error and User Agent in the request header was missing.
https://www.exai.com/blog/406-not-acceptable
The 406 Not Acceptable status code is a client-side error. It's part
of the HTTP response status codes in the 4xx category, which are
considered client error responses
Here is the URL that I'm trying to scrape: https://www.sec.gov/ix?doc=/Archives/edgar/data/320193/000032019319000076/a10-qq320196292019.htm
I'm trying to scrape the webpage using Python which mean I will require the XHR request for this page as it is loaded via JavaScript.
Upon inspection of the Network under Developer Tools, I can see the XHR request: a10-qq320196292019.htm which produces the request URL: https://www.sec.gov/Archives/edgar/data/320193/000032019319000076/a10-qq320196292019.htm
My question is two-fold,
How can I automatically get this request URL if i am only accessing using the URL given initially and,
How do I know that this is THE XHR request I need? This particular URL works for my needs but I noticed that there were many other XHR reqeusts as well. How does one differentiate?
In this case, I don't think you need to go that route. The link you're using is an ixbrl view of the actual html document. The url for the html doc is embedded in that first link. All you have to do is extract it:
url = 'https://www.sec.gov/ix?doc=/Archives/edgar/data/320193/000032019319000076/a10-qq320196292019.htm'
html_url = url.replace('/ix?doc=','')
html_url
Output:
'https://www.sec.gov/Archives/edgar/data/320193/000032019319000076/a10-qq320196292019.htm
I'm trying to scrape multiple financial websites (Wells Fargo, etc.) to pull my transaction history for data analysis purposes. I can do the scraping part once I get to the page I need; the problem I'm having is getting there. I don't know how to pass my username and password and then navigate from there. I would like to do this without actually opening a browser.
I found Michael Foord's article "HOWTO Fetch Internet Resources Using The urllib Package" and tried to adapt one of the examples to meet my needs but can't get it to work (I've tried adapting to several other search results as well). Here's my code:
import bs4
import urllib.request
import urllib.parse
##Navigate to the website.
url = 'https://www.wellsfargo.com/'
values = {'j_username':'USERNAME', 'j_password':'PASSWORD'}
data = urllib.parse.urlencode(values)
data = data.encode('ascii')
req = urllib.request.Request(url, data)
with urllib.request.urlopen(req) as response:
the_page = response.read()
soup = bs4.BeautifulSoup(the_page,"html.parser")
The 'j_username' and 'j_password' both come from inspecting the text boxes on the login page.
I just don't think I'm pointing to the right place or passing my credentials correctly. The URL I'm using is just the login page so is it actually logging me in? When I print the URL from response it returns https://wellsfargo.com/. If I'm ever able to successfully login, it just takes me to a summary page of my accounts. I would then need to follow another link to my checking, savings, etc.
I really appreciate any help you can offer.
I'm using mechanize library to log in website. I checked, it works well. But problem is i can't use response.read() with BeautifulSoup and 'lxml'.
#BeautifulSoup
response = browser.open(url)
source = response.read()
soup = BeautifulSoup(source) #source.txt doesn't work either
for link in soup.findAll('a', {'class':'someClass'}):
some_list.add(link)
This doesn't work, actually doesn't find any tag. It works well when i use requests.get(url).
#lxml->html
response = browser.open(url)
source = response.read()
tree = html.fromstring(source) #souce.txt doesn't work either
print tree.text
like_pages = buyers = tree.xpath('//a[#class="UFINoWrap"]') #/text() doesn't work either
print like_pages
Doesn't print anything. I know it has problem with return type of response, since it works well with requests.open(). What could i do? Could you, please, provide sample code where response.read() used in html parsing?
By the way, what is difference between response and requests objects?
Thank you!
I found solution. It is because mechanize.browser is emulated browser, and it gets only raw html. The page i wanted to scrape adds class to tag with help of JavaScript, so those classes were not on raw html. Best option is to use webdriver. I used Selenium for Python. Here is code:
from selenium import webdriver
profile = webdriver.FirefoxProfile()
profile.set_preference('network.http.phishy-userpass-length', 255)
driver = webdriver.Firefox(firefox_profile=profile)
driver.get(url)
list = driver.find_elements_by_xpath('//a[#class="someClass"]')
Note: You need to have Firefox installed. Or you can choose another profile according to browser you want to use.
A request is what a web client sends to a server, with details about what URL the client wants, what http verb to use (get / post, etc), and if you are submitting a form the request typically contains the data you put in the form.
A response is what a web server sends back in reply to a request from a client. The response has a status code which indicates if the request was successful (code 200 usually if there were no problems, or an error code like 404 or 500). The response usually contains data, like the html in a page, or the binary data in a jpeg. The response also has headers that give more information about what data is in the response (e.g. the "Content-Type" header which says what format the data is in).
Quote from #davidbuxton's answer on this link.
Good luck!
I am using cookielib and some times opening a url in browser downloads many other files by browser making many other requests. Can I replicate the same behaviour using cookie lib or any other python library?
For example: To get all the required information from page https://applicant.keybank.com/psp/hrsappl/EMPLOYEE/HRMS/c/HRS_HRAM.HRS_CE.GBL?Page=HRS_CE_HM_PRE&Action=A&SiteId=1
I have to make more than 1 GET requests from my python script. I got the request urls of all the requests browser makes by analysing the network requests when I opened the page.
I am seeing if there is any way I can just make 1 request and it fetches all the related requests by itself like browser.
I am not very much interested in the js or css but the main html.
I tried with the following code but it couldn't download whole page
cj = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj))
response = opener.open('https://applicant.keybank.com/psp/hrsappl/EMPLOYEE/HRMS/c/HRS_HRAM.HRS_CE.GBL?Page=HRS_CE_HM_PRE&Action=A&SiteId=1')
html = response.read()
but when I fetched 3 other GET urls in sequence it is able to give me the required html in the third GET response. I got these urls by examining network tab of the browser
'https://applicant.keybank.com/psc/hrsappl/EMPLOYEE/EMPL/s/WEBLIB_PT_NAV.ISCRIPT1.FieldFormula.IScript_UniHeader_Frame?c=NNTCgkqGs001AcPaisqGbYpTu%2fbGx4jx&Page=HRS_CE_HM_PRE&Action=A&SiteId=1&PortalActualURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentProvider=HRMS&PortalRegistryName=EMPLOYEE&PortalServletURI=https%3a%2f%2fapplicant.keybank.com%2fpsp%2fhrsappl%2f&PortalURI=https%3a%2f%2fapplicant.keybank.com%2fpsc%2fhrsappl%2f&PortalHostNode=EMPL&PortalIsPagelet=true&NoCrumbs=yes')
'https://applicant.keybank.com/psc/hrsappl/EMPLOYEE/EMPL/s/WEBLIB_PTPPB.ISCRIPT2.FieldFormula.IScript_TemplatePageletBuilder?PTPPB_PAGELET_ID=KC_LNAV_APPLICANT&target=KCNV_KC_LNAV_APPLICANT_TMPL&Page=HRS_CE_HM_PRE&Action=A&SiteId=1&PortalActualURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentProvider=HRMS&PortalRegistryName=EMPLOYEE&PortalServletURI=https%3a%2f%2fapplicant.keybank.com%2fpsp%2fhrsappl%2f&PortalURI=https%3a%2f%2fapplicant.keybank.com%2fpsc%2fhrsappl%2f&PortalHostNode=EMPL&PortalIsPagelet=true&NoCrumbs=yes&PortalTargetFrame=TargetContent'
'https://hronline.keybank.com/psc/hrshrm/EMPLOYEE/HRMS/c/HRS_HRAM.HRS_CE.GBL?Page=HRS_CE_HM_PRE&Action=A&SiteId=1&PortalActualURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentProvider=HRMS&PortalCRefLabel=Careers&PortalRegistryName=EMPLOYEE&PortalServletURI=https%3a%2f%2fapplicant.keybank.com%2fpsp%2fhrsappl%2f&PortalURI=https%3a%2f%2fapplicant.keybank.com%2fpsc%2fhrsappl%2f&PortalHostNode=EMPL&NoCrumbs=yes&PortalKeyStruct=yes'
and following is the complete code for the other fetches I am making
response = opener.open('https://applicant.keybank.com/psc/hrsappl/EMPLOYEE/EMPL/s/WEBLIB_PT_NAV.ISCRIPT1.FieldFormula.IScript_UniHeader_Frame?c=NNTCgkqGs001AcPaisqGbYpTu%2fbGx4jx&Page=HRS_CE_HM_PRE&Action=A&SiteId=1&PortalActualURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentProvider=HRMS&PortalRegistryName=EMPLOYEE&PortalServletURI=https%3a%2f%2fapplicant.keybank.com%2fpsp%2fhrsappl%2f&PortalURI=https%3a%2f%2fapplicant.keybank.com%2fpsc%2fhrsappl%2f&PortalHostNode=EMPL&PortalIsPagelet=true&NoCrumbs=yes')
response.read()
response = opener.open('https://applicant.keybank.com/psc/hrsappl/EMPLOYEE/EMPL/s/WEBLIB_PTPPB.ISCRIPT2.FieldFormula.IScript_TemplatePageletBuilder?PTPPB_PAGELET_ID=KC_LNAV_APPLICANT&target=KCNV_KC_LNAV_APPLICANT_TMPL&Page=HRS_CE_HM_PRE&Action=A&SiteId=1&PortalActualURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentProvider=HRMS&PortalRegistryName=EMPLOYEE&PortalServletURI=https%3a%2f%2fapplicant.keybank.com%2fpsp%2fhrsappl%2f&PortalURI=https%3a%2f%2fapplicant.keybank.com%2fpsc%2fhrsappl%2f&PortalHostNode=EMPL&PortalIsPagelet=true&NoCrumbs=yes&PortalTargetFrame=TargetContent')
response.read()
response = opener.open('https://hronline.keybank.com/psc/hrshrm/EMPLOYEE/HRMS/c/HRS_HRAM.HRS_CE.GBL?Page=HRS_CE_HM_PRE&Action=A&SiteId=1&PortalActualURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentURL=https%3a%2f%2fhronline.keybank.com%2fpsc%2fhrshrm%2fEMPLOYEE%2fHRMS%2fc%2fHRS_HRAM.HRS_CE.GBL%3fPage%3dHRS_CE_HM_PRE%26Action%3dA%26SiteId%3d1&PortalContentProvider=HRMS&PortalCRefLabel=Careers&PortalRegistryName=EMPLOYEE&PortalServletURI=https%3a%2f%2fapplicant.keybank.com%2fpsp%2fhrsappl%2f&PortalURI=https%3a%2f%2fapplicant.keybank.com%2fpsc%2fhrsappl%2f&PortalHostNode=EMPL&NoCrumbs=yes&PortalKeyStruct=yes')
required_html = response.read()
requests can handle cookies, as you can see here.
It's a great library, far more powerful that urllib2, and yet simpler-looking.
>>> import requests
>>> r = requests.get('https://applicant.keybank.com/psp/hrsappl/EMPLOYEE/HRMS/c/HRS_HRAM.HRS_CE.GBL?Page=HRS_CE_HM_PRE&Action=A&SiteId=1')
>>> r.cookies
Edit: This answer dos not really address the problem, I read too fast. Sorry about that.
As suggested by #J.F.Sebastian, I'm adding a link to a python webkit client, Ghost.py, that could emulate a browser, as you requested.