I want a python script that opens a link and print the email address from that page.
E.g
Go to some site like example.com
Search for email in that.
Search in all the pages in that link.
I was tried below code
import requests
from bs4 import BeautifulSoup
r = requests.get('http://www.digitalseo.in/')
data = r.text
soup = BeautifulSoup(data)
for rate in soup.find_all('#'):
print rate.text
I take this website for reference.
Anyone help me to get this?
Because find_all() will only search Tags. From document:
Signature: find_all(name, attrs, recursive, string, limit, **kwargs)
The find_all() method looks through a tag’s descendants and retrieves all descendants that match your filters.
So you need add a keyword argument like this:
import re
import requests
from bs4 import BeautifulSoup
r = requests.get('http://www.digitalseo.in/')
data = r.text
soup = BeautifulSoup(data, "html.parser")
for i in soup.find_all(href=re.compile("mailto")):
print i.string
Demo:
contact#digitalseo.in
contact#digitalseo.in
From document:
Any argument that’s not recognized will be turned into a filter on one of a tag’s attributes. If you pass in a value for an argument called id, Beautiful Soup will filter against each tag's 'id' attribute:
soup.find_all(id='link2')
# [<a class="sister" href="http://example.com/lacie" id="link2">Lacie</a>]
If you pass in a value for href, Beautiful Soup will filter against each tag's 'href' attribute:
soup.find_all(href=re.compile("elsie"))
# [<a class="sister" href="http://example.com/elsie" id="link1">Elsie</a>]
You can see the document for more info: http://www.crummy.com/software/BeautifulSoup/bs4/doc/#find-all
And if you'd like find the email address from a document, regex is a good choice.
For example:
import re
re.findall( '[^#]+#[^#]+\.[^#]+ ', text) # remember change `text` variable
And if you'd like find a link in a page by keyword, just use .get like this:
import re
import requests
from bs4 import BeautifulSoup
def get_link_by_keyword(keyword):
links = set()
for i in soup.find_all(href=re.compile(r"[http|/].*"+str(keyword))):
links.add(i.get('href'))
for i in links:
if i[0] == 'h':
yield i
elif i[0] == '/':
yield link+i
else:
pass
global link
link = raw_input('Please enter a link: ')
if link[-1] == '/':
link = link[:-1]
r = requests.get(link, verify=True)
data = r.text
soup = BeautifulSoup(data, "html.parser")
for i in get_link_by_keyword(raw_input('Enter a keyword: ')):
print i
Related
#!/usr/bin/python3
import requests
from bs4 import BeautifulSoup
import re
url = input("Please enter a URL to scrape: ")
r = requests.get(url)
html = r.text
print(html)
soup = BeautifulSoup(html, "html.parser")
for link in soup.find_all('a', attrs={'href': re.compile("^https://")}):
print(link.get('href'))
down at the bottom, where it prints the link... I know it'll go in there, but I can't think of a way to remove duplicate entries there. Can someone help me with that please?
Use a set to remove duplicates. You call add() to add an item and if the item is already present then it won't be added again.
Try this:
#!/usr/bin/python3
import requests
from bs4 import BeautifulSoup
import re
url = input("Please enter a URL to scrape: ")
r = requests.get(url)
html = r.text
print(html)
soup = BeautifulSoup(html, "html.parser")
urls = set()
for link in soup.find_all('a', attrs={'href': re.compile(r"^https://")}):
urls.add(link.get('href'))
print(urls) # urls contains unique set of URLs
Note some URLs might start with http:// so may want to use the regexp ^https?:// to catch both http and https URLs.
You can also use set comprehension syntax to rewrite the assignment and for statements like this.
urls = {
link.get("href")
for link in soup.find_all("a", attrs={"href": re.compile(r"^https://")})
}
instead of printing it you need to catch is somehow to compare.
Try this:
you get a list with all result by find_all and make it a set.
data = set(link.get('href') for link in soup.find_all('a', attrs={'href': re.compile("^https://")}))
for elem in data:
print(elem)
Seems like i can scrape any tag and class, except h3 on this page. It keeps returning None or an empty list. I'm trying to get this h3 tag:
...on the following webpage:
https://www.empireonline.com/movies/features/best-movies-2/
And this is the code I use:
from bs4 import BeautifulSoup
import requests
from pprint import pprint
from bs4 import BeautifulSoup
URL = "https://www.empireonline.com/movies/features/best-movies-2/"
response = requests.get(URL)
web_html = response.text
soup = BeautifulSoup(web_html, "html.parser")
movies = soup.findAll(name = "h3" , class_ = "jsx-4245974604")
movies_text=[]
for item in movies:
result = item.getText()
movies_text.append(result)
print(movies_text)
Can you please help with the solution for this problem?
As other people mentioned this is dynamic content, which needs to be generated first when opening/running the webpage. Therefore you can't find the class "jsx-4245974604" with BS4.
If you print out your "soup" variable you actually can see that you won't find it. But if simply you want to get the names of the movies you can just use another part of the html in this case.
The movie name is in the alt tag of the picture (and actually also in many other parts of the html).
import requests
from pprint import pprint
from bs4 import BeautifulSoup
URL = "https://www.empireonline.com/movies/features/best-movies-2/"
response = requests.get(URL)
web_html = response.text
soup = BeautifulSoup(web_html, "html.parser")
movies = soup.findAll("img", class_="jsx-952983560")
movies_text=[]
for item in movies:
result = item.get('alt')
movies_text.append(result)
print(movies_text)
If you run into this issue in the future, remember to just print out the initial html you can get with soup and just check by eye if the information you need can be found.
I am writing a python script using BeautifulSoup. I need to scrape a website and count unique links ignoring the links starting with '#'.
Example if the following links exist on a webpage:
https://www.stackoverflow.com/questions
https://www.stackoverflow.com/foo
https://www.cnn.com/
For this example, the only two unique links will be (The link information after the main domain name is removed):
https://stackoverflow.com/ Count 2
https://cnn.com/ Count 1
Note: this is my first time using python and web scraping tools.
I appreciate all the help in advance.
This is what I have tried so far:
from bs4 import BeautifulSoup
import requests
url = 'https://en.wikipedia.org/wiki/Beautiful_Soup_(HTML_parser)'
r = requests.get(url)
soup = BeautifulSoup(r.text, "html.parser")
count = 0
for link in soup.find_all('a'):
print(link.get('href'))
count += 1
There is a function named urlparse from urllib.parse which you can get netloc of urls. And there is a new awesome HTTP library named requests_html which can help you get all links in source file.
from requests_html import HTMLSession
from collections import Counter
from urllib.parse import urlparse
session = HTMLSession()
r = session.get("the link you want to crawl")
unique_netlocs = Counter(urlparse(link).netloc for link in r.html.absolute_links)
for link in unique_netlocs:
print(link, unique_netlocs[link])
You could also do this:
from bs4 import BeautifulSoup
from collections import Counter
import requests
soup = BeautifulSoup(requests.get("https://en.wikipedia.org/wiki/Beautiful_Soup_(HTML_parser)").text, "html.parser")
foundUrls = Counter([link["href"] for link in soup.find_all("a", href=lambda href: href and not href.startswith("#"))])
foundUrls = foundUrls.most_common()
for item in foundUrls:
print ("%s: %d" % (item[0], item[1]))
The soup.find_all line checks if every atag has an href set and if it doesn't start with the # character.
The Counter method counts the occurrences of each list entry and the most_common orders by the value.
The for loop just prints the results.
My way to do this is to find all links using beautiful soup and then determine which link redirects to which location:
def get_count_url(url): # get the umber of links having the same domain and suffix
r = requests.get(url)
soup = BeautifulSoup(r.text, "html.parser")
count = 0
urls={} #dictionary for the domains
# input_domain=url.split('//')[1].split('/')[0]
#library to extract the exact domain( ex.- blog.bbc.com and bbc.com have the same domains )
input_domain=tldextract.extract(url).domain+"."+tldextract.extract(url).suffix
for link in soup.find_all('a'):
word =link.get('href')
# print(word)
if word:
# Same website or domain calls
if "#" in word or word[0]=="/": #div call or same domain call
if not input_domain in urls:
# print(input_domain)
urls[input_domain]=1 #if first encounter with the domain
else:
urls[input_domain]+=1 #multiple encounters
elif "javascript" in word:
# javascript function calls (for domains that use modern JS frameworks to display information)
if not "JavascriptRenderingFunctionCall" in urls:
urls["JavascriptRenderingFunctionCall"]=1
else:
urls["JavascriptRenderingFunctionCall"]+=1
else:
# main_domain=word.split('//')[1].split('/')[0]
main_domain=tldextract.extract(word).domain+"." +tldextract.extract(word).suffix
# print(main_domain)
if main_domain.split('.')[0]=='www':
main_domain = main_domain.replace("www.","") # removing the www
if not main_domain in urls: # maintaining the dictionary
urls[main_domain]=1
else:
urls[main_domain]+=1
count += 1
for key, value in urls.items(): # printing the dictionary in a paragraph format for better readability
print(key,value)
return count
tld extract finds the correct url name and soup.find_all('a') finds a tags. The if statements check for same domain redirect, javascript redirect or other domain redirects.
I am using python, I need regex to get contacts link of web page. So, I made <a (.*?)>(.*?)Contacts(.*?)</a> and result is:
href="/ru/o-nas.html" id="menu263" title="About">About</a></li><li>Photo</li><li class="last"><a href="/ru/kontakt.html" class="last" id="menu583" title="">Contacts
,but I need on last <a ... like
href="/ru/kontakt.html" class="last" id="menu583" title="">Contacts
What regex pattern should I use?
python code:
match = re.findall('<a (.*?)>(.*?)Contacts(.*?)</a>', body)
if match:
for m in match:
print ''.join(m)
Since you are parsing HTML, I would suggest to use BeautifulSoup
# sample html from question
html = '<li>About</li><li>Photo</li><li class="last">Contacts</li>'
from bs4 import BeautifulSoup
doc = BeautifulSoup(html)
aTag = doc.find('a', id='menu583') # id for Contacts link
print(aTag['href'])
# '/ru/kontakt.html'
Try BeautifulSoup
from BeautifulSoup import BeautifulSoup
import urllib2
import re
links = []
urls ['www.u1.com','www.u2.om'....]
for url in urls:
page = urllib2.urlopen(url)
soup = BeautifulSoup(page)
for link in soup.findAll('a'):
if link.string.lower() == 'contact':
links.append(link.get('href'))
I am trying to use Python and Beautifulsoup to get this page from sfglobe website: http://sfglobe.com/2015/04/28/stirring-pictures-from-the-riots-in-baltimore.
This is the code:
import urllib2
from bs4 import BeautifulSoup
url = 'http://sfglobe.com/2015/04/28/stirring-pictures-from-the-riots-in-baltimore'
req = urllib2.urlopen(url)
html = req.read()
soup = BeautifulSoup(html)
desc = soup.find('span', class_='articletext intro')
Could anyone help me to solve this problem?
From the question title, I assuming that the only thing you want is the description of the article, which can be found in the <meta> tag within the HTML <head>.
You were on the right track, but I'm not exactly sure why you did:
desc = soup.find('span', class_='articletext intro')
Regardless, I came up with something using requests (see http://stackoverflow.com/questions/2018026/should-i-use-urllib-or-urllib2-or-requests) rather than urllib2
import requests
from bs4 import BeautifulSoup
url = 'http://sfglobe.com/2015/04/28/stirring-pictures-from-the-riots-in-baltim\
ore'
req = requests.get(url)
html = req.text
soup = BeautifulSoup(html)
tag = soup.find(attrs={'name':'description'}) # find meta tag w/ description
desc = tag['value'] # get value of attribute 'value'
print desc
If that isn't what you are looking for, please clarify so I can try and help you more.
EDIT: after some clarification, I pieced together why you were originally using desc = soup.find('span', class_='articletext intro').
Maybe this is what you are looking for:
import requests
from bs4 import BeautifulSoup, NavigableString
url = 'http://sfglobe.com/2015/04/28/stirring-pictures-from-the-riots-in-baltimore'
req = requests.get(url)
html = req.text
soup = BeautifulSoup(html)
body = soup.find('span', class_='articletext intro')
# remove script tags
[s.extract() for s in body('script')]
text = ""
# iterate through non-script elements in the content body
for stuff in body.select('*'):
# get contents of tags, .contents returns a list
content = stuff.contents
# check if the list has the text content a.k.a. isn't empty AND is a NavigableString, not a tag
if len(content) == 1 and isinstance(content[0], NavigableString):
text += content[0]
print text