I am creating a simple application where I have to follow links from a page and so on...thus building a very basic prototype of a web crawler.
When I was testing it, i came across robot.txt which has hits limit for any external crawlers trying to crawl their site. For example, if a website's robot.txt has a hit limit of not more than 1 hit per second (as that of wikipedia.org) from a given IP, and if I crawl few pages of Wikipedia at the rate of 1 page per second, then how do i estimate how many hits will it incur while i crawl?
Question: if i am downloading one entire page through the urllib of python, how many hits will it account to?
Here is my Example Code:
import urllib.request
opener = urllib.request.FancyURLopener({})
open_url = opener.open(a)
page = open_url.read()
print page
If you download an entire page from a site with urllib, it will account as one (1) hit.
Save the page into a variable, and work with this variable from now on.
Additionally, I'd advise you to use requests instead of urllib. Much easier/better/stronger.
Link to the documentation of Requests.
One thing you can do is put a time gap between two request , this will solve your problem and it also prevent you from get blocked.
Related
Basically, I'm trying to download some images from a site which has been "down for maintenance" for over a year. Attempting to visit any page on the website redirects to the forums. However, visiting image URLs directly will still take you to the specified image.
There's a particular post that I know contains more images than I've been able to brute force by guessing possible file names. I thought, rather than typing every combination of characters possible ad infinitum, I could program a web scraper to do it for me.
However, I'm brand new to Python, and I've been unable to overcome the Javascript redirects. While I've been able to use requests & beautifulsoup to scrape the page it redirects to for 'href', without circumventing the JS I cannot pull from the news article which has links.
import requests
from bs4 import BeautifulSoup
url = 'https://www.he-man.org/news_article.php?id=6136'
reqs = requests.get(url)
soup = BeautifulSoup(reqs.text, 'html.parser')
urls = []
for link in soup.find_all('a'):
print(link.get('href'))
I've added allow_redirects=False to every field to no avail. I've tried searching for 'jpg' instead of 'href'. I'm currently attempting to install Selenium, though it's fighting me, and I expect I'll just be getting 3xx errors anyway.
The images from the news article all begin with the same 62 characters (https://www.he-man.org/assets/images/home_news/justinedantzer_), so I've also thought maybe there's a way to just infinite-monkeys-on-a-keyboard scrape the rest of it? Or the type of file (.jpg)? I'm open to suggestions here, I really have no idea what direction to come at this thing from now that my first six plans have failed. I've seen a few mentions of scraping spiders, but at this point I've sunk a lot of time into dead ends. Are they worth looking into? What would you recommend?
I wanted to read this article online and something popped and I thought that I want to read it offline after I have successfully extracted it... so here I am after 4 weeks of trials and all the problem is down to is I the crawler can't seem to read the content of the webpages even after all of the ruckus...
the initial problem was that all of the info was not present on one page so is used the button to navigate the content of the website itself...
I've tried BeautifulSoup but it can't seem to parse the page very well. I'm using selenium and chromedriver at the moment.
The reason for crawler not being able to read the page seems to be the robot.txt file (the waiting time for crawlers for a single page is 3600 and the article has about 10 pages, which is bearable but what would happen if it were to say 100+)and I don't know how to bypass it or go around it.
Any help??
If robots.txt puts limitations then that's the end of it. You should be web-scraping ethically and this means if the owner of the site wants you to wait 3600 seconds between requests then so be it.
Even if robots.txt doesn't stipulate wait times you should still be mindful. Small business / website owners might not know of this and by you hammering a website constantly it could be costly to them.
I am scraping job posting data from a website using BeautifulSoup. I have working code that does what I need, but it only scrapes the first page of job postings. I am having trouble figuring out how to iteratively update the url to scrape each page. I am new to Python and have looked at a few different solutions to similar questions, but have not figured out how to apply them to my particular url. I think I need to iteratively update the url or somehow click the next button and then loop my existing code through each page. I appreciate any solutions.
url: https://jobs.utcaerospacesystems.com/search-jobs
First, BeautifulSoup doesn't have anything to do with GETing web pages - you get the webpage yourself, then feed it to bs4 for processing.
The problem with the page you linked is that it's javascript - it only renders correctly in a browser (or any other javascript VM).
#Fabricator is on the right track - you'll need to watch the developer console and see what the ajax requests the js is sending to the server. In this case, also take a look at the query string params, which include a param called CurrentPage - that's probably the one you want to focus on.
I am working to scrape the reports from this site, I hit the home page, enter report date and hit submit, it's Ajax enabled and I am not getting how to get that report table . Any help will be really appreciated.
https://www.theice.com/marketdata/reports/176
I tried sending get and post using requests module, but failed as Session Time out or Report not Available.
EDIT:
Steps Taken so far:
URL = "theice.com/marketdata/reports/datawarehouse/..."
with requests.Session() as sess:
f = sess.get(URL,params = {'selectionForm':''}) # Got 'selectionForm' by analyzing GET requests to URL
data = {'criteria.ReportDate':--, ** few more params i got from hitting submit}
f = sess.post(URL,data=data)
f.text # Session timeout / No Reports Found –
Since you've already identified that the data you're looking to scrape is hidden behind some AJAX calls, you're already on your way to solving this problem.
At the moment, you're using python-requests for HTTP, but that is pretty much all it does. It does not handle executing JavaScript or any other items that involve scanning the content and executing code in another language runtime. For that, you'll need to use something like Mechanize or Selenium to load those websites, interact with the JavaScript, and then scrape the data you're looking for.
I'm new to Scrapy, and not too impressive with Python. I've got a scraper set up to scrape data from a website, but although I'm using proxies, if the same proxy is used too many times then my request is shown a page telling me I'm visiting too many pages too quickly (HTTP status code 200).
As my scraper see's the page's status code as okay, it doesn't find the needed data and moves on to the next page.
I can determine when these pages are show via HtmlXPathSelector, but how do i signal Scrapy to retry that page?
Scrapy comes with a built-in retry middleware. You could subclass it and override the process_response method to include a check to see if the page that is telling you that you're visiting too many pages too quickly is showing up