Answers for "python website scraping"

30

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')
Posted by: Guest on June-16-2020
13

web scraping python

#pip install beautifulsoup4

import os
import requests
from bs4 import BeautifulSoup

url = "https://www.google.com/"
reponse = requests.get(url)

if reponse.ok:
	soup = BeautifulSoup(reponse.text, "lxml")
	title = str(soup.find("title"))

	title = title.replace("<title>", "")
	title = title.replace("</title>", "")
	print("The title is : " + str(title))

os.system("pause")

#python (code name).py
Posted by: Guest on January-09-2021
0

web scraper python

>>> from bs4 import BeautifulSoup
>>> raw_html = open('contrived.html').read()
>>> html = BeautifulSoup(raw_html, 'html.parser')
>>> for p in html.select('p'):
...     if p['id'] == 'walrus':
...         print(p.text)

'I am the walrus'
Posted by: Guest on August-09-2020
-1

web scraper python

>>> raw_html = simple_get('http://www.fabpedigree.com/james/mathmen.htm')
>>> html = BeautifulSoup(raw_html, 'html.parser')
>>> for i, li in enumerate(html.select('li')):
        print(i, li.text)

0  Isaac Newton
 Archimedes
 Carl F. Gauss
 Leonhard Euler
 Bernhard Riemann

1  Archimedes
 Carl F. Gauss
 Leonhard Euler
 Bernhard Riemann

2  Carl F. Gauss
 Leonhard Euler 
 Bernhard Riemann

 3  Leonhard Euler
 Bernhard Riemann

4  Bernhard Riemann

# 5 ... and many more...
Posted by: Guest on August-09-2020
0

web scraper python

def get_names():
    """
    Downloads the page where the list of mathematicians is found
    and returns a list of strings, one per mathematician
    """
    url = 'http://www.fabpedigree.com/james/mathmen.htm'
    response = simple_get(url)

    if response is not None:
        html = BeautifulSoup(response, 'html.parser')
        names = set()
        for li in html.select('li'):
            for name in li.text.split('n'):
                if len(name) > 0:
                    names.add(name.strip())
        return list(names)

    # Raise an exception if we failed to get any data from the url
    raise Exception('Error retrieving contents at {}'.format(url))
Posted by: Guest on August-09-2020
-2

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)
Posted by: Guest on August-09-2020

Python Answers by Framework

Browse Popular Code Answers by Language