Answers for "scrapy web scraping python"

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
3

python web crawler

import scrapy

class BlogSpider(scrapy.Spider):
    name = 'blogspider'
    start_urls = ['https://blog.scrapinghub.com']

    def parse(self, response):
        for title in response.css('.post-header>h2'):
            yield {'title': title.css('a ::text').get()}

        for next_page in response.css('a.next-posts-link'):
            yield response.follow(next_page, self.parse)
Posted by: Guest on March-31-2020
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