scrape a website python code example
Example 1: python web scraping
import requests
from bs4 import BeautifulSoup
URL = 'https://www.monster.com/jobs/search/?q=Software-Developer&where=Australia'
page = requests.get(URL)
soup = BeautifulSoup(page.content, 'html.parser')
Example 2: web scraper python
def get_hits_on_name(name):
"""
Accepts a `name` of a mathematician and returns the number
of hits that mathematician's Wikipedia page received in the
last 60 days, as an `int`
"""
url_root = 'URL_REMOVED_SEE_NOTICE_AT_START_OF_ARTICLE'
response = simple_get(url_root.format(name))
if response is not None:
html = BeautifulSoup(response, 'html.parser')
hit_link = [a for a in html.select('a')
if a['href'].find('latest-60') > -1]
if len(hit_link) > 0:
link_text = hit_link[0].text.replace(',', '')
try:
return int(link_text)
except:
log_error("couldn't parse {} as an `int`".format(link_text))
log_error('No pageviews found for {}'.format(name))
return None
Example 3: web scraper python
def simple_get(url):
"""
Attempts to get the content at `url` by making an HTTP GET request.
If the content-type of response is some kind of HTML/XML, return the
text content, otherwise return None.
"""
try:
with closing(get(url, stream=True)) as resp:
if is_good_response(resp):
return resp.content
else:
return None
except RequestException as e:
log_error('Error during requests to {0} : {1}'.format(url, str(e)))
return None
def is_good_response(resp):
"""
Returns True if the response seems to be HTML, False otherwise.
"""
content_type = resp.headers['Content-Type'].lower()
return (resp.status_code == 200
and content_type is not None
and content_type.find('html') > -1)
def log_error(e):
"""
It is always a good idea to log errors.
This function just prints them, but you can
make it do anything.
"""
print(e)