Answers for "Implementing Web Scraping in Python with Scrapy"

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
1

web scraper python

from requests import get
from requests.exceptions import RequestException
from contextlib import closing
from bs4 import BeautifulSoup
Posted by: Guest on August-09-2020
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
0

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 is a template string that is used to build a URL.
    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:
            # Strip commas
            link_text = hit_link[0].text.replace(',', '')
            try:
                # Convert to integer
                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
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

Code answers related to "Implementing Web Scraping in Python with Scrapy"

Python Answers by Framework

Browse Popular Code Answers by Language