scrape hidden pages if search yields more results than displayed

问题

Some of the search queries entered under https://www.comparis.ch/carfinder/default would yield more than 1'000 results (shown dynamically on the search page). The results however only show a max of 100 pages with 10 results each so I'm trying to scrape the remaining data given a query that yields more than 1'000 results. The code to scrape the IDs of the first 100 pages is (takes approx. 2 minutes to run through all 100 pages):

from bs4 import BeautifulSoup
import requests

# as the max number of pages is limited to 100
number_of_pages = 100

# initiate empty dict
car_dict = {}

# parse every search results page and extract every car ID
for page in range(0, number_of_pages + 1, 1):
    newest_secondhand_cars = 'https://www.comparis.ch/carfinder/marktplatz/occasion'
    newest_secondhand_cars = requests.get(newest_secondhand_cars + str('?page=') + str(page))
    newest_secondhand_cars = newest_secondhand_cars.content
    soup = BeautifulSoup(newest_secondhand_cars, "lxml")

    for car in list(soup.find('div', {'id': 'cf-result-list'}).find_all('h2')):
        car_id = int(car.decode().split('href="')[1].split('">')[0].split('/')[-1])
        car_dict[car_id] = {}

So I obviously tried just passing a str(page) greater than 100 which does not yield additional results. How could I access the remaining results, if at all?

回答1:

It seems that your website loads data when the client is browsing. There are probably a number of ways to fix this. One option could be to utilize Scrapy Splash.

Assuming you use scrapy, you can do the following:

Start a Splash server using docker - make a note of the
In settings.py add SPLASH_URL = <splash-server-ip-address>
In settings.py add to middlewares

this code:

DOWNLOADER_MIDDLEWARES = {
    'scrapy_splash.SplashCookiesMiddleware': 723,
    'scrapy_splash.SplashMiddleware': 725,
    'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810,
}

Import from scrapy_splash import SplashRequest in your spider.py
Set start_url in your spider.py to iterate over the pages

E.g. like this

base_url = 'https://www.comparis.ch/carfinder/marktplatz/occasion'
start_urls = [
     base_url + str('?page=') + str(page) % page for page in range(0,100)      
    ]

Redirect the url to the splash server by modifing def start_requests(self):

E.g. like this

def start_requests(self):
    for url in self.start_urls:
        yield SplashRequest(url, self.parse,
            endpoint='render.html',
            args={'wait': 0.5},
        )

Parse the response like you do now.

Let me know how that works out for you.

来源：https://stackoverflow.com/questions/59575378/scrape-hidden-pages-if-search-yields-more-results-than-displayed

标签

python

web-scraping

beautifulsoup