Now I have a http response from website A, I need to change all the link urls in this http response to the url of website B, so that when users get this http response in browser, click on links, they will be directed to website B not A.
I'm using python and django. Is there a package or tool can do this trick?
Thanks in advance.
Depending upon the nature of the response you get from website A, what you want to do with it, and on how important it is that the replacement be efficient, there are a few possible ways of doing things. I'm not 100% clear on your situation and what you want to achieve.
If the links in the response from website A start with website A's hostname, then just get the response as a string and do response = response.replace('http://website-a.com', 'http://website-b.com') before you present the response to the user.
If the response is HTML, and the links are relative, the easiest solution to code would probably be to use lxml.rewrite_links (see http://lxml.de/lxmlhtml.html#working-with-links). I suspect this is what you're looking for.
If you've got some other situation, well, then I dunno what's appropriate. Maybe a regex. Maybe a custom algorithm of your own design. It depends upon what kind of content you're getting back from website A, how you can recognise links in it, and how you want to change them.
If you use Apache as Webserver you could use a module to replace Text in the response like http://mod-replace.sourceforge.net/. This seems to be more reasonable than invoking perl or python for every request. But you have to be aware that all the text might be replaced - not only the links which have an efect. Therefore this would be a very dirty solution.
Related
I want to build a api that accepts a string and returns html code.
Here is my scraping code that i want as a web-service.
Code
from selenium import webdriver
import bs4
import requests
import time
url = "https://www.pnrconverter.com/"
browser = webdriver.Firefox()
browser.get(url)
string = "3 PS 232 M 03FEB 7 JFKKBP HK2 1230A 420P 03FEB E
PS/JPIX8U"
button =
browser.find_element_by_xpath("//textarea[#class='dataInputChild']")
button.send_keys(string) #accept string
button.submit()
time.sleep(5)
soup = bs4.BeautifulSoup(browser.page_source,'html.parser')
html = soup.find('div',class_="main-content") #returns html
print(html)
Can anyone tell me the best possible solution to wrap up my code as a api/web-service.
There's no best possible solution in general, because a solution has to fit the problem and the available resources.
Right now it seems like you're trying to wrap someone else's website. If that's the problem you're actually trying to solve, and you want to give credit, you should probably just forward people to their site. Have your site return a 302 Redirect with their URL in the Location field in your header.
If what you're trying to do is get the response from this one sample check you have hardcoded, and and make that result available, I would suggest you put it in a static file behind nginx.
If what you're trying to do is use their backend to turn itineraries you have into responses you can return, you can do that by using their backend API, once that becomes available. Read the documentation, use the requests library to hit the API endpoint that you want, and get the JSON result back, and format it to your desires.
If you're trying to duplicate their site by making yourself a man-in-the-middle, that may be illegal and you should reconsider what you're doing.
For hosting purposes, you need to figure out how often your API will be hit. You can probably start on Heroku or something similar fairly easily, and scale up if you need to. You'll probably want WebObj or Flask or something similar sitting at the website where you intend to host this application. You can use those to process what I presume will be a simple request into the string you wish to hit their API with.
I am the owner of PNR Converter, so I can shed some light on your attempt to scrape content from our site. Unfortunately scraping from PNR Converter is not recommended. We are developing an API which looks like it would suit your needs, and should be ready in the not too distant future. If you contact us through the site we would be happy to work with you should you wish to use PNR Converter legitimately. PNR Converter gets at least one complete update per year and as such we change all the code on a regular basis. We also monitor all requests to our site, and we will block any requests which are deemed as improper usage. Our filter has already picked up your IP address (ends in 250.144) as potential misuse.
Like I said, should you wish to work with us at PNR Converter legitimately and not scrape our content then we would be happy to do so! please keep checking https://www.pnrconverter.com/api-introduction for information relating to our API.
We are releasing a backend upgrade this weekend, which will have a different HTML structure, and dynamically named elements which will cause a serious issue for web scrapers!
I have a rather strange question regarding urls which point to another url. So, for example, I have a url:
http://mywebpage/this/is/a/forward
which ultimately points to another url:
http://mynewpage/this/is/new
My question is, when I use for example urllib2 in python to fetch the first page, it ultimately fetches the second page. I would like to know if its possible to know what the original link is pointing to. Is there something like a "header" which tells me the second link when I request the first link?
Sorry if this is a really silly question!
When you issue a GET request for the first URL, the web server will return a 300-series reply code, with a Location header whose value is the second URL. You can find out what the second URL was from Python with the geturl method of the object returned by urlopen. If there is more than one redirection involved, it appears that urllib will tell you the last hop and there's no way to get the others.
This will not handle redirections via JavaScript or meta http-equiv="refresh", but you probably aren't in that situation or you wouldn't have asked the question the way you did.
It's most commonly done via a redirection response code (3xx) as defined in RFC2616 although a "pseudo redirect effect" cann be achieved with some javascript in the original page.
This SO question is about how to prevent urllib2 from following redirects, it looks like something you might be able to use.
You can do this using requests:
>>> url = 'http://ofa.bo/foagK7'
>>> r = requests.head(url)
>>> r.headers['location']
'https://my.barackobama.com/page/s/what-does-2000-mean-to-you'
I am trying to retrieve query results on sites based on ajax like www.snapbird.org using Python. Since it doesn't show in the page source, I am not sure how to proceed.
I am a Python newbie and hence it would be great if I could get a pointer in the right direction.
I am also open to some other approach to the task if that is easier
This is going to be complex but as a start, ppen firebug and find the URL that gets called when the AJAX request is handled. You can call that directly in your Python program and parse the output.
You could use Selenium's Python client driver to parse the page source. I usually use this in conjunction with PyQuery to make web scraping easier.
Here's the basic tutorial for Selenium's Python driver. Be sure to follow the instructions for Selenium version 2 instead of version 1 (unless you're using version 1 for some reason).
You could also configure chrome/firefox to an HTTP proxy and then log/extract the necessary content with the proxy. I've tinkered with python proxies to save/log the requests/content based on content-type or URI globs.
For other projects I've written site-specific javascript bookmarklets which poll for new data and then POST it to my server (by dynamically creating both a form and iframe, and setting myform.target=myiframe;
Other javascript scripts/bookmarklets simulate a user interacting with sites, so instead of polling every few seconds the javascript automates clicking buttons and form submissions, etc. These scripts are always very site-specific of course but they've been hugely useful for me, especially when iterating over all the paginated results for a given search.
Here is a stripped down version of walking over a list of "paginated" results and preparing to send the data off to my server (which then further parses it with BeautifulSoup). In particular this was designed for Youtube's Sent/Inbox messages.
var tables = [];
function process_and_repeat(){
if(!(inbox && inbox.message_pane_ && inbox.message_pane_.innerHTML)){
alert("We've got no data!");
}
if(inbox.message_pane_.innerHTML.indexOf('<table') === 0)
{
tables.push(inbox.message_pane_.innerHTML);
inbox.next_page();
setTimeout("process_and_repeat()",3000);
}
else{
alert("Fininshed, [" + tables.length + " processed]");
document.write('<form action=http://curl.sente.cc method=POST><textarea name=sent.html>'+escape(tables.join('\n'))+'</textarea><input type=submit></form>')
}
}
process_and_repeat(); // now we wait and watch as all the paginated pages are viewed :)
This is a stripped down example without any fancy iframes/non-essentials which just add complexity.
Adding to what Liam said, Selenium is a great tool, too, which has aided in my various scraping needs. I'd be more than happy to help you out with this if you'd like.
One easy solution might be using a browser like Mechanize. So you can browse site, follow links, make searches and nearly everything that you can do with a browser with user interface.
But for a very sepcific job, you may not even need a such library, you can use urllib and urllib2 python libraries to make a connection and read response... You can use Firebug to see data structure of a search and response body. Then use urllib to make a request with relevant parameters...
With an example...
I made a search with joyvalencia and check the request url with firebug to see:
http://api.twitter.com/1/statuses/user_timeline.json?screen_name=joyvalencia&count=100&page=2&include_rts=true&callback=twitterlib1321017083330
So calling this url with urllib2.urlopen() will be the same with making the query on Snapbird. Response body is:
twitterlib1321017083330([{"id_str":"131548107799396357","place":null,"geo":null,"in_reply_to_user_id_str":null,"coordinates":.......
When you use urlopen() and read the response, the upper string is what you get... Then you can use json library of python to read the data and parse it to a pythonic data structure...
I've got a link that I know redirects to another end url, and I'm trying to get the address for that end url using python. But the original link is a little weird, and doesn't work like a normal redirect, and I can't figure out why. When I post the link (the link's below for you try, if you'd like) into a browser, it redirects perfectly. But when I run the following code, it doesn't.
import urllib2
request = urllib2.Request('http://www.facebook.com/ajax/emu/end.php?eid=AQJSWpZ3e4cCTHoNdahpJzPYzmzHOENzbTWBVlW4SgIxX0rL9bo6NXmS3q06cjeh5jO9wbsmr3IyGrpbXPSj0GPLbRJl4VUH-EBnmSy_R4j7iYzpMe1ooZ6IEqSEIlBl0-5SEldIhxI82m75YPa5nOhuBdokiwTw79hoiRB-Zn1auxN-6WLVe3e5WNSt3HLAEjZL-2e4ox_7yAyLcBo1nkamEvShTyZ-GfIf0A9oFXylwRnV8oNaqNmUnqrFYqDbUhzh7d6LSm3jbv1ue2coS3w8N7OxTKVwODHa-Hd3qRbYskB9weio8eKdDFtkvDKuzSSq5hjr711UjlDsgpxLuAmdD95xVwpomxeEsBsMCYJoUEQYa-cM7q3W1aiIYBHlyn2__t74qHWVvzK5zaLKFMKjRFQqphDlUMgMni6AP1VHSn1wli_3lgeVD8TzcJMSlJIF7DC_O44WdjBIMY8OufER3ZB_mm2NqwUe6cvV9oV9SNyYHE4UUURYjW_Z6sUxz3SpHG8c6QxJ-ltSeShvU3mIwAhFE3M0jGTg7AQ7nIoOUfC8PDainFZ1NV8g31aqaqDsF7UxdlOmBT6w-Y8TPmHOXfSlWB-M3MQYUBmcWS3UzlbSsavQG8LXPqYbyKfvkAfncSnZS3_tkoqbTksFirQWlSxJ3mgXrO5PqopH63Esd9ynCbFQM1q_3_wgkYvTeGS9XK6G63_Ag3N9dCHsO_bCJToJT4jeHQCSQ83cb1U5Qpe_7EWbw1ilzgyL-LBVrpH424dwK-4AoaL00W-gWzShSdOynjcoGeB7KE0pHbg-XhuaVribSodriSGybNdADBosnddVvZldY22-_97MqEuA&&c=4&&f=4&&ui=6003071106023-id_4e0b51323f9d01393198225&&en=1&&a=0&&sig=78154')
opener = urllib2.build_opener()
f = opener.open(request)
f.geturl()
I simply get my original url back. I encounter the same problem when I save cookies and use mechanize. Any help would be much appreciated! Thanks!
It looks like this is using Javascript to perform the redirect. You'll either have to figure out exactly how the Javascript is performing the redirects and pull out the appropriate urls, or you'll have to actually run the Javascript. As far as I know, running Javascript from python is not an easy task.
(original answer deleted)
If you look at the contents of f.read() you'll see what's going on here. Instead of returning a 301 or 302 that redirects to the new URL, Facebook actually returns a real HTML document - which contains a piece of Javascript that uses document.location.replace to change the URL in the browser.
There's no easy way of replicating that with Python - the best thing to do is to parse the document with something like BeautifulSoup to find the Javascript, and somehow extract the new URL. It won't be pretty.
I wrote a web crawler in Python 2.6 using the Bing API that searches for certain documents and then downloads them for classification later. I've been using string methods and urllib.urlretrieve() to download results whose URL ends in .pdf, .ps etc., but I run into trouble when the document is 'hidden' behind a URL like:
http://www.oecd.org/officialdocuments/displaydocument/?cote=STD/CSTAT/WPNA(2008)25&docLanguage=En
So, two questions. Is there a way in general to tell if a URL has a pdf/doc etc. file that it's linking to if it's not doing so explicitly (e.g. www.domain.com/file.pdf)? Is there a way to get Python to snag that file?
Edit:
Thanks for replies, several of which suggest downloading the file to see if it's of the correct type. Only problem is... I don't know how to do that (see question #2, above). urlretrieve(<above url>) gives only an html file with an href containing that same url.
There's no way to tell from the URL what it's going to give you. Even if it ends in .pdf it could still give you HTML or anything it likes.
You could do a HEAD request and look at the content-type, which, if the server isn't lying to you, will tell you if it's a PDF.
Alternatively you can download it and then work out whether what you got is a PDF.
In this case, what you refer to as "a document that's not explicitly referenced in a URL" seems to be what is known as a "redirect". Basically, the server tells you that you have to get the document at another URL. Normally, python's urllib will automatically follow these redirects, so that you end up with the right file. (and - as others have already mentioned - you can check the response's mime-type header to see if it's a pdf).
However, the server in question is doing something strange here. You request the url, and it redirects you to another url. You request the other url, and it redirects you again... to the same url! And again... And again... At some point, urllib decides that this is enough already, and will stop following the redirect, to avoid getting caught in an endless loop.
So how come you are able to get the pdf when you use your browser? Because apparently, the server will only serve the pdf if you have cookies enabled. (why? you have to ask the people responsible for the server...) If you don't have the cookie, it will just keep redirecting you forever.
(check the urllib2 and cookielib modules to get support for cookies, this tutorial might help)
At least, that is what I think is causing the problem. I haven't actually tried doing it with cookies yet. It could also be that the server is does not "want" to serve the pdf, because it detects you are not using a "normal" browser (in which case you would probably need to fiddle with the User-Agent header), but it would be a strange way of doing that. So my guess is that it is somewhere using a "session cookie", and in the case you haven't got one yet, keeps on trying to redirect.
As has been said there is no way to tell content type from URL. But if you don't mind getting the headers for every URL you can do this:
obj = urllib.urlopen(URL)
headers = obj.info()
if headers['Content-Type'].find('pdf') != -1:
# we have pdf file, download whole
...
This way you won't have to download each URL just it's headers. It's still not exactly saving network traffic, but you won't get better than that.
Also you should use mime-types instead of my crude find('pdf').
No. It is impossible to tell what kind of resource is referenced by a URL just by looking at it. It is totally up to the server to decide what he gives you when you request a certain URL.
Check the mimetype with the urllib.info() function. This might not be 100% accurate, it really depends on what the site returns as a Content-Type header. If it's well behaved it'll return the proper mime type.
A PDF should return application/pdf, but that may not be the case.
Otherwise you might just have to download it and try it.
You can't see it from the url directly. You could try to only download the header of the HTTP response and look for the Content-Type header. However, you have to trust the server on this - it could respond with a wrong Content-Type header not matching the data provided in the body.
Detect the file type in Python 3.x and webapp with url to the file which couldn't have an extension or a fake extension. You should install python-magic, using
pip3 install python-magic
For Mac OS X, you should also install libmagic using
brew install libmagic
Code snippet
import urllib
import magic
from urllib.request import urlopen
url = "http://...url to the file ..."
request = urllib.request.Request(url)
response = urlopen(request)
mime_type = magic.from_buffer(response.read())
print(mime_type)