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How to "log in" to zalando.de using Python's Requests module?

Regardless of Zalando usually blocking any requests traffic (I already know how to get around this), how can I detect the Post method that Zalando uses to login me in using their form zalando.de/myaccount/? With DevTools I don't seem to find the specific post method.
As far as I could see: with having the data that is needed, I could then perform something like this: How to "log in" to a website using Python's Requests module?
Can anyone show me how such a request would look like? Thanks.

Firstly, please respect their wishes to not send direct requests to their site.
For any other site that allows it:
Make sure you set up your dev tools to not clear the log on redirect
Check the response header (or your browser-storage) for which cookies have been set, since those are used to identify your session (e.g. Service-Client-Id, ...)
Make a request with those cookies from your script to pretend to have a logged in user
NEVER SHARE OR POST THOSE COOKIES ANYWHERE and read up on session-hijacking

How to change link url in a HTTP response?

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.

Python trace URL get requests - using python script

I'm writing a script, to help me do some repetitive testing of a bunch of URLs.
I've written a python method in the script that it opens up the URL and sends a get request. I'm using Requests: HTTP for Humans -http://docs.python-requests.org/en/latest/- api to handle the http calls.
There's the request.history that returns a list of status codes of the directs. I need to be able to access the particular redirects for those list of 301s. There doesn't seem to be a way to do this - to access and trace what my URLS are redirecting to. I want to be able to access the redirected URLS (status code 301)
Can anyone offer any advice?
Thanks

Okay, I'm so silly. Here's the answer I was looking for
r = requests.get("http://someurl")
r.history[1].url will return the URL

Extracting information from AJAX based sites using Python

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...

Using Python to download a document that's not explicitly referenced in a URL

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)