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List links of xls files using Beautifulsoup

I'm trying to retrieve a list of downloadable xls files on a website.
I'm a bit reluctant to provide full links to the website in question.
Hopefully I'm able to provide all necessary details all the same.
If this is useless, please let me know.
Download .xls files from a webpage using Python and BeautifulSoup is a very similar question, but the details below will show that the solution most likely will have to be different since the links on that particular site are tagged with a href anchor:
And the ones I'm trying to get are not tagged the same way.
On the webpage, the files that are available for downloading are listed like this:
A simple mousehover gives these further details:
I'm following the setup here with a few changes to produce the snippet below that provides a list of some links, but not to any of the xls files:
from bs4 import BeautifulSoup
import urllib
import re
def getLinks(url):
with urllib.request.urlopen(url) as response:
html = response.read()
soup = BeautifulSoup(html, "lxml")
links = []
for link in soup.findAll('a', attrs={'href': re.compile("^http://")}):
links.append(link.get('href'))
return links
links1 = getLinks("https://SOMEWEBSITE")
A further inspection using ctrl+shift+I in Google Chrome reveals that those particular links do not have a href anchor tag, but rather a ng-href anchor tag:
So I tried changing that in the snippet above, but with no success.
And I've tried different combinations with e.compile("^https://"), attrs={'ng-href' and links.append(link.get('ng-href')), but still with no success.
So I'm hoping someone has a better suggestion!
EDIT - Further details
It seems it's a bit problematic to read these links directly.
When I use ctrl+shift+I and the Select an element in the page to inspect it Ctrl+Shift+C, this is what I can see when I hover over one of the links listed above:
And what I'm looking to extract here is the information associated with the ng-href tag. But If I right-click the page and select Show Source, the same tag only appears once along with som metadata(?):
And I guess this is why my rather basic approach is failing in the first place.
I'm hoping this makes sense to some of you.

Update:
using selenium
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
driver = webdriver.Chrome()
driver.get('http://.....')
# wait max 15 second until the links appear
xls_links = WebDriverWait(driver, 15).until(lambda d: d.find_elements_by_xpath('//a[contains(#ng-href, ".xls")]'))
# Or
# xls_links = WebDriverWait(driver, 15).until(lambda d: d.find_elements_by_xpath('//a[contains(#href, ".xls")]'))
links = []
for link in xls_links:
url = "https://SOMEWEBSITE" + link.get_attribute('ng-href')
print(url)
links.append(url)
Assume ng-href is not dynamically generated, from your last image I see that the URL is not starts with https:// but the slash / you can try with regex URL contains .xls
for link in soup.findAll('a', attrs={'ng-href': re.compile(r"\.xls")}):
xls_link = "https://SOMEWEBSITE" + link['ng-href']
print(xls_link)
links.append(xls_link)

My guess is that the data you are trying to crawl is created dynamically: ng-href is one of AngularJs's constructs. You could try using Google Chrome's Network inspection as you already did (ctrl+shift+I) and see if you can find the url that is queried (open the network tab and reload the page). The query should typically return a JSON with the links to the xls-files.
There is a thread about a similar problem here. Perhaps that helps you: Unable to crawl some href in a webpage using python and beautifulsoup

Why is the html in view-source different from what I see in the terminal when I call prettify()?

I have decided to view a website's source code, and chose a class, which is "expanded" (I found it using view-source, prettify() shows different code). I wanted to print out all of its contents, with this code:
import requests
from bs4 import BeautifulSoup
page = requests.get("https://www.quora.com/How-can-I-write-a-bot-using-Python")
soup = BeautifulSoup(page.content, 'html.parser')
print soup.find_all(class_='expanded')
but it simply prints out:
[]
Please help me detect what's wrong.
I already saw this thread and tried following what the answer said but it did not help me since this error appears in the terminal:
bs4.FeatureNotFound: Couldn't find a tree builder with the features you requested: lxml. Do you need to install a parser library?

I had a look at the site in question and the only class similar was actually named ui_qtext_expanded
When you use findAll / find_all you have to iterate over it to return each item as it is a list of items using .text.. That is if you want the text and not the actual HTML..
import requests
from bs4 import BeautifulSoup
page = requests.get("https://www.quora.com/How-can-I-write-a-bot-using-Python")
soup = BeautifulSoup(page.content, 'html.parser')
res = soup.find_all(class_='ui_qtext_expanded')
for i in res:
print i.text
The beginning of the output from your link is
A combination of mechanize, Requests and BeautifulSoup works pretty good for the basic stuff.Learn about mechanize here.Mechanize is sufficient for basic form filling, form submission and that sort of stuff, but for real browser emulation (like dealing with Javascript rendered HTML) you should look into selenium.

Python scraping deep nested divs whose classes change

I'm somewhat new to python, and working on this 1st part of a project where i need to get the link(s) on a FanDuel page, and i've been spinning my tires trying get the 'href'.
Here's what the Inspect Element shows:
What i'm trying to get to is highlighted above.
I see that the seems to be the parent, but as you go down the tree, the classes listed with lettering (ie - "_a _ch _al _nr _dq _ns _nt _nu") changes from day to day.
What I noticed is that the 'href' that i need has a constant "data-test-id" that does not change, so i was trying to use that as my way to find what i need, but it does not seem to be working.
I'm not sure how far, or if, I need to drill down farther to get what I need, or if my code is totally off. Thanks for your help in advance!
import requests
from bs4 import BeautifulSoup
url = "https://www.fanduel.com/contests/mlb/96"
#authentication might not be necessary, it was a test, still getting the same results
site = requests.get(url, cookies={'X-Auth-Token':'MY TOKEN IS HERE'})
soup = BeautifulSoup(site.content, 'lxml')
game = soup.find_all('a', {'data-test-id':"ContestCardEnterLink"})
#If i use this, i get an error
game = soup.find_all('a', {'data-test-id':"ContestCardEnterLink"})[('href')]
print(game)

The HTML is constructed by javascript, to check this, instead of using inspect element, use view source-page and see if the HTML is already constructed there ( this is the html that you get when you do requests.get() ) ,i've already checked this and this is true. To resolve this, you should have to use Selenium to render the javascript on the page, and then you can get the source page code by selenium after he constructed the elements from DOM.

How to scrape Price from a site that has a changing structure?

I want to scrape the pricing data from an eCommerce site called flipkart, I tried using Beautifulsoup with casperjs(nodejs utility) and similar libraries but none of them is good enough.
Here's the URL and the structure.
https://www.flipkart.com/redmi-note-4-gold-32-gb/p/itmer37fmekafqct?
the problem is the layout...What are some ways to get around this?
P.S : Is there anyway I could apply machine learning for getting the pricing data without knowing complex math? Like where do i even start?

You should probably construct your XPath in a way so it does not rely on the class, but rather on the content (node()) of the element you want to match. Alternatively you could match the data-reactid if that doesn't change?
For matching the div by data-reactid:
//div[#data-reactid=220]
Or for matching the div based on its location:
//span[child::img[#src="//img1a.flixcart.com/www/linchpin/fk-cp-zion/img/fa_8b4b59.png"]]/preceding-sibling::div
Assuming the img_path doesn't change you're on the safe side.

Since you can't use xpath due to dynamic changing you probably could try to use a regex for finding a price in the script tag on the page.
Something like this:
import requests
import re
url = "https://www.flipkart.com/redmi-note-4-gold-32-gb/p/itmer37fmekafqct"
r = requests.get(url)
pattern = re.compile('prexoAvailable\":[\w]+,\"price\":(\d+)')
result = pattern.search(r.text)
print(result.group(1))

from bs4 import BeatifulSoup
page = request.get(url, headers)
soup = BeautifulSoup(page.content, 'html.parser')
for a in soup.findAll('a', href=True, attrs={'class': '_31qSD5'}):
price = a.find('div', attrs={'class': '_1vC4OE _2rQ-NK'})
print(price.text)

E-commerce have does not allow anymore to scrape data like before, every entity of the product like product price, specification, reviews are now enclosed in a separate “Dynamic” class name.
And scraping certain data from the webpage you need to use specific class name which is dynamic. So using request.get() or soup() won't work.

web scraping in python

I'd like to scrape all the ~62000 names from this petition, using python. I'm trying to use the beautifulsoup4 library.
However, it's just not working.
Here's my code so far:
import urllib2, re
from bs4 import BeautifulSoup
soup = BeautifulSoup(urllib2.urlopen('http://www.thepetitionsite.com/104/781/496/ban-pesticides-used-to-kill-tigers/index.html').read())
divs = soup.findAll('div', attrs={'class' : 'name_location'})
print divs
[]
What am I doing wrong? Also, I want to somehow access the next page to add the next set of names to the list, but I have no idea how to do that right now. Any help is appreciated, thanks.

You could try something like this:
import urllib2
from bs4 import BeautifulSoup
html = urllib2.urlopen('http://www.thepetitionsite.com/xml/petitions/104/781/496/signatures/latest.xml?1374861495')
# uncomment to try with a smaller subset of the signatures
#html = urllib2.urlopen('http://www.thepetitionsite.com/xml/petitions/104/781/496/signatures/00/00/00/05.xml')
results = []
while True:
# Read the web page in XML mode
soup = BeautifulSoup(html.read(), "xml")
try:
for s in soup.find_all("signature"):
# Scrape the names from the XML
firstname = s.find('firstname').contents[0]
lastname = s.find('lastname').contents[0]
results.append(str(firstname) + " " + str(lastname))
except:
pass
# Find the next page to scrape
prev = soup.find("prev_signature")
# Check if another page of result exists - if not break from loop
if prev == None:
break
# Get the previous URL
url = prev.contents[0]
# Open the next page of results
html = urllib2.urlopen(url)
print("Extracting data from {}".format(url))
# Print the results
print("\n")
print("====================")
print("= Printing Results =")
print("====================\n")
print(results)
Be warned though there is a lot of data there to go through and I have no idea if this is against the terms of service of the website so you would need to check it out.

In most cases it is extremely inconsiderate to simply scrape a site. You put a fairly large load on a site in a short amount of time slowing down legitimate users requests. Not to mention stealing all of their data.
Consider an alternate approach such as asking (politely) for a dump of the data (as mentioned above).
Or if you do absolutely need to scrape:
Space your requests using a timer
Scrape smartly
I took a quick glance at that page and it appears to me they use AJAX to request the signatures. Why not simply copy their AJAX request, it'll most likely be using some sort of REST call. By doing this you lessen the load on their server by only requesting the data you need. It will also be easier for you to actually process the data because it will be in a nice format.
Reedit, I looked at their robots.txt file. It dissallows /xml/ Please respect this.

what do you mean by not working? empty list or error?
if you are receiving an empty list, it is because the class "name_location" does not exist in the document. also checkout bs4's documentation on findAll