Answers for "build a scraper in 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
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
3

web scraping python

import scrapy
from ..items import SampletestItem #items class

class QuoteTestSpider(scrapy.Spider):
    name = 'quote_test'
    start_urls = ['https://quotes.toscrape.com/']

    def parse(self, response):
        items = SampletestItem() #items class
        quotes = response.css("div.quote")
        for quote in quotes:
            items['title'] = quote.css("span.text::text").get()
            items['author'] = quote.css(".author::text").get()
            items['tags'] = quote.css(".tags .tag::text").getall()
            
            yield items
            next_page = response.css(".next a::attr(href)").get()
            if next_page is not None:
                next_url = response.urljoin(next_page)
                yield scrapy.Request(next_url, callback=self.parse)
Posted by: Guest on December-05-2020
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
-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

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