Develop Reference

Reading in Content From URLS in a File

I'm trying to get other subset URLs from a main URL. However,as I print to see if I get the content, I noticed that I am only getting the HTML, not the URLs within it.
import urllib
file = 'http://example.com'
with urllib.request.urlopen(file) as url:
collection = url.read().decode('UTF-8')

I think this is what you are looking for.
You can use beautiful soup library of python and this code should work with python3
import urllib
from urllib.request import urlopen
from bs4 import BeautifulSoup
def get_all_urls(url):
open = urlopen(url)
url_html = BeautifulSoup(open, 'html.parser')
for link in url_html.find_all('a'):
links = str(link.get('href'))
if links.startswith('http'):
print(links)
else:
print(url + str(links))
get_all_urls('url.com')

Having trouble in getting page source with beautifulsoup

I am trying to get the HTML source of a web page using beautifulsoup.
import bs4 as bs
import requests
import urllib.request
sourceUrl='https://www.pakwheels.com/forums/t/planing-a-trip-from-karachi-to-lahore-by-road-in-feb-2017/414115/2.html'
source=urllib.request.urlopen(sourceUrl).read()
soup=bs.BeautifulSoup(source,'html.parser')
print(soup)
I want the HTML source of the page. This is what I am getting now:
'ps.store("siteSettings", {"title":"PakWheels Forums","contact_email":"sami.ullah#pakeventures.com","contact_url":"https://www.pakwheels.com/main/contact_us","logo_url":"https://www.pakwheels.com/assets/logo.png","logo_small_url":"/images/d-logo-sketch-small.png","mobile_logo_url":"data:image/svg+xml;base64,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'

Have a look at this code:
from urllib import request
from bs4 import BeautifulSoup
url_1 = "http://www.google.com"
page = request.urlopen(url_1)
soup = BeautifulSoup(page)
print(soup.prettify())
Import everything you need correctly. Read this.

BeautifulSoup url scraping

Trying out BeautifulSoup for the first time.
I have this link http://www.mediafire.com/download/alv8dq6k35n4m2k/For+You.zip
I want to catch the direct download url from the download button which is
http://download2110.mediafire.com/niz8p9iu6r9g/alv8dq6k35n4m2k/For+You.zip
What I have tried so far.
r = requests.get(url)
soup = BeautifulSoup(r.content)
links = soup.findAll('a')
I think the last function findAll('a')would find all the links from that page, but I could not find the direct download url in my linkslist.
Am I doing something wrong here? If so, how can I grab that link with beautifulsoup. I inspect the element in Chrome Developer Console and I see that the link is there.

You can try this to extract the url from the javascript:
from bs4 import BeautifulSoup
import requests
r = requests.get("http://www.mediafire.com/download/alv8dq6k35n4m2k/For+You.zip")
soup = BeautifulSoup(r.content)
link = soup.find("div",{"class":"download_link"})
import re
url = re.findall("http.*.zip?",link.text)[0]

How to find and extract link from webpage?

I have website , e.g http://site.com
I would like fetch main page and extract only links that match the regular expression, e.g .*somepage.*
The format of links in html code can be:
url
url
url
I need the output format:
http://site.com/my-somepage
http://site.com/my-somepage.html
http://site.com/my-somepage.htm
Output url must contain domain name always.
What is the fast python solution for this?

You could use lxml.html:
from lxml import html
url = "http://site.com"
doc = html.parse(url).getroot() # download & parse webpage
doc.make_links_absolute(url)
for element, attribute, link, _ in doc.iterlinks():
if (attribute == 'href' and element.tag == 'a' and
'somepage' in link): # or e.g., re.search('somepage', link)
print(link)
Or the same using beautifulsoup4:
import re
try:
from urllib2 import urlopen
from urlparse import urljoin
except ImportError: # Python 3
from urllib.parse import urljoin
from urllib.request import urlopen
from bs4 import BeautifulSoup, SoupStrainer # pip install beautifulsoup4
url = "http://site.com"
only_links = SoupStrainer('a', href=re.compile('somepage'))
soup = BeautifulSoup(urlopen(url), parse_only=only_links)
urls = [urljoin(url, a['href']) for a in soup(only_links)]
print("\n".join(urls))

Use an HTML Parsing module, like BeautifulSoup.
Some code(only some):
from bs4 import BeautifulSoup
import re
html = '''url
url
url'''
soup = BeautifulSoup(html)
links = soup.find_all('a',{'href':re.compile('.*somepage.*')})
for link in links:
print link['href']
Output:
http://site.com/my-somepage
/my-somepage.html
my-somepage.htm
You should be able to get the format you want from this much data...

Scrapy is the simplest way to do what you want. There is actually link extracting mechanism built-in.
Let me know if you need help with writing the spider to crawl links.
Please, also see:
How do I use the Python Scrapy module to list all the URLs from my website?
Scrapy tutorial

BeautifulSoup does not work for some web sites

I have this sript:
import urrlib2
from bs4 import BeautifulSoup
url = "http://www.shoptop.ru/"
page = urllib2.urlopen(url).read()
soup = BeautifulSoup(page)
divs = soup.findAll('a')
print divs
For this web site, it prints empty list? What can be problem? I am running on Ubuntu 12.04

Actually there are quite couple of bugs in BeautifulSoup which might raise some unknown errors. I had a similar issue when working on apache using lxml parser
So, just try to use other couple of parsers mentioned in the documentation
soup = BeautifulSoup(page, "html.parser")
This should work!

It looks like you have a few mistakes in your code urrlib2 should be urllib2, I've fixed the code for you and this works using BeautifulSoup 3
import urllib2
from BeautifulSoup import BeautifulSoup
url = "http://www.shoptop.ru/"
page = urllib2.urlopen(url).read()
soup = BeautifulSoup(page)
divs = soup.findAll('a')
print divs