I have the following code for a web crawler in Python 3:
import requests
from bs4 import BeautifulSoup
import re
def get_links(link):
return_links = []
r = requests.get(link)
soup = BeautifulSoup(r.content, "lxml")
if r.status_code != 200:
print("Error. Something is wrong here")
else:
for link in soup.findAll('a', attrs={'href': re.compile("^http")}):
return_links.append(link.get('href')))
def recursive_search(links)
for i in links:
links.append(get_links(i))
recursive_search(links)
recursive_search(get_links("https://www.brandonskerritt.github.io"))
The code basically gets all the links off of my GitHub pages website, and then it gets all the links off of those links, and so on until the end of time or an error occurs.
I want to recreate this code in Scrapy so it can obey robots.txt and be a better web crawler overall. I've researched online and I can only find tutorials / guides / stackoverflow / quora / blog posts about how to scrape a specific domain (allowed_domains=["google.com"], for example). I do not want to do this. I want to create code that will scrape all websites recursively.
This isn't much of a problem but all the blog posts etc only show how to get the links from a specific website (for example, it might be that he links are in list tags). The code I have above works for all anchor tags, regardless of what website it's being run on.
I do not want to use this in the wild, I need it for demonstration purposes so I'm not going to suddenly annoy everyone with excessive web crawling.
Any help will be appreciated!
There is an entire section of scrapy guide dedicated to broad crawls. I suggest you to fine-grain your settings for doing this succesfully.
For recreating the behaviour you need in scrapy, you must
set your start url in your page.
write a parse function that follow all links and recursively call itself, adding to a spider variable the requested urls
An untested example (that can be, of course, refined):
class AllSpider(scrapy.Spider):
name = 'all'
start_urls = ['https://yourgithub.com']
def __init__(self):
self.links=[]
def parse(self, response):
self.links.append(response.url)
for href in response.css('a::attr(href)'):
yield response.follow(href, self.parse)
If you want to allow crawling of all domains, simply don't specify allowed_domains, and use a LinkExtractor which extracts all links.
A simple spider that follows all links:
class FollowAllSpider(CrawlSpider):
name = 'follow_all'
start_urls = ['https://example.com']
rules = [Rule(LinkExtractor(), callback='parse_item', follow=True)]
def parse_item(self, response):
pass
Related
Hello everyone I'm a beginner at scraping and i try to scrape all iPhones in https://www.electroplanet.ma/
this is the scripts i wrote
import scrapy
from ..items import EpItem
class ep(scrapy.Spider):
name = "ep"
start_urls = ["https://www.electroplanet.ma/smartphone-tablette-gps/smartphone/iphone?p=1",
"https://www.electroplanet.ma/smartphone-tablette-gps/smartphone/iphone?p=2"
]
def parse(self, response):
products = response.css("ol li") # to find all items in the page
for product in products :
try:
lien = product.css("a.product-item-link::attr(href)").get() # get the link of each item
image= product.css("a.product-item-photo::attr(href)").get() # get the image
# and to get in each item page and scrap it, i use follow method
# i passed image as argument to parse_item cauz i couldn't scrap the image from item's page
# i think it's hidden
yield response.follow(lien,callback = self.parse_item,cb_kwargs={"image":image})
except: pass
def parse_item(self,response,image):
item = EpItem()
item["Nom"]= response.css(".ref::text").get()
pattern = re.compile(r"\s*(\S+(?:\s+\S+)*)\s*")
item["Catégorie"]= pattern.search(response.xpath("//h1/a/text()").get()).group(1)
item["Marque"]=pattern.search(response.xpath("//*[#data-th='Marque']/text()").get()).group(1)
try :
item["RAM"]= pattern.search(response.xpath("//*[#data-th='MÉMOIRE RAM']/text()").get()).group(1)
except:
pass
item["ROM"]=pattern.search(response.xpath("//*[#data-th='MÉMOIRE DE STOCKAGE']/text()").get()).group(1)
item["Couleur"]=pattern.search(response.xpath("//*[#data-th='COULEUR']/text()").get()).group(1)
item["lien"]=response.request.url
item["image"]=image
item["état"]="neuf"
item["Market"]= "Electro Planet"
yield item
i found problems to scrape all the pages, because it uses javascript to follow pages so i write all pages links in start urls and i believe it's not the best practice so i ask you to give some advices to improve my code
you can use the scrapy-playwright plugin to scrape the interactive websites, and for the start_urls, just add the main website index URL if there is just one website, and check this link in the scrapy docs to make the spider follow the pages links automatically instead of written them manually
I’m trying to use Scrapy to log into a website, then navigate within than website, and eventually download data from it. Currently I’m stuck in the middle of the navigation part. Here are the things I looked into to solve the problem on my own.
Datacamp course on Scrapy
Following Pagination Links with Scrapy
http://scrapingauthority.com/2016/11/22/scrapy-login/
Scrapy - Following Links
Relative URL to absolute URL Scrapy
However, I do not seem to connect the dots.
Below is the code I currently use. I manage to log in (when I call the "open_in_browser" function, I see that I’m logged in). I also manage to "click" on the first button on the website in the "parse2" part (if I call "open_in_browser" after parse 2, I see that the navigation bar at the top of the website has gone one level deeper.
The main problem is now in the "parse3" part as I cannot navigate another level deeper (or maybe I can, but the "open_in_browser" does not open the website any more - only if I put it after parse or parse 2). My understanding is that I put multiple "parse-functions" after another to navigate through the website.
Datacamp says I always need to start with a "start request function" which is what I tried but within the YouTube videos, etc. I saw evidence that most start directly with parse functions. Using "inspect" on the website for parse 3, I see that this time href is a relative link and I used different methods (See source 5) to navigate to it as I thought this might be the source of error.
import scrapy
from scrapy.http import FormRequest
from scrapy.utils.response import open_in_browser
from scrapy.crawler import CrawlerProcess
class LoginNeedScraper(scrapy.Spider):
name = "login"
start_urls = ["<some website>"]
def parse(self, response):
loginTicket = response.xpath('/html/body/section/div/div/div/div[2]/form/div[3]/input[1]/#value').extract_first()
execution = response.xpath('/html/body/section/div/div/div/div[2]/form/div[3]/input[2]/#value').extract_first()
return FormRequest.from_response(response, formdata={
'loginTicket': loginTicket,
'execution': execution,
'username': '<someusername>',
'password': '<somepassword>'},
callback=self.parse2)
def parse2(self, response):
next_page_url = response.xpath('/html/body/nav/div[2]/ul/li/a/#href').extract_first()
yield scrapy.Request(url=next_page_url, callback=self.parse3)
def parse3(self, response):
next_page_url_2 = response.xpath('/html//div[#class = "headerPanel"]/div[3]/a/#href').extract_first()
absolute_url = response.urljoin(next_page_url_2)
yield scrapy.Request(url=absolute_url, callback=self.start_scraping)
def start_scraping(self, response):
open_in_browser(response)
process = CrawlerProcess()
process.crawl(LoginNeedScraper)
process.start()
You need to define rules in order to scrape a website completely. Let's say you want to crawl all links in the header of the website and then open that link in order to see the main page to which that link was referring.
In order to achieve this, firstly identify what you need to scrape and mark CSS or XPath selectors for those links and put them in a rule. Every rule has a default callback to parse or you can also assign it to some other method. I am attaching a dummy example of creating rules, and you can map it accordingly to your case:
rules = (
Rule(LinkExtractor(restrict_css=[crawl_css_selectors])),
Rule(LinkExtractor(restrict_css=[product_css_selectors]), callback='parse_item')
)
I am a newbie and I've written a script in python scrapy to get information recursively.
Firstly, it scrapes links of city including information of tours then it tracks down each cities and reach their pages. Next, it get needed information of tours related to city before move to next pages then so on. Pagination is running on java-script without visible link.
The command I used to get the result along with a csv output is:
scrapy crawl pratice -o practice.csv -t csv
The expected result is csv file:
title, city, price, tour_url
t1, c1, p1, url_1
t2, c2, p2, url_2
...
The problem is that csv file is empty. The running is stopped at "parse_page" and callback="self.parse_item" doesn't work. I don't know how to fix it. Maybe my workflow is invalid or my code has issues. Thanks for your help.
name = 'practice'
start_urls = ['https://www.klook.com/vi/search?query=VI%E1%BB%86T%20NAM%20&type=country',]
def parse(self, response): # Extract cities from country
hxs = HtmlXPathSelector(response)
urls = hxs.select("//div[#class='swiper-wrapper cityData']/a/#href").extract()
for url in urls:
url = urllib.parse.urljoin(response.url, url)
self.log('Found city url: %s' % url)
yield response.follow(url, callback=self.parse_page) # Link to city
def parse_page(self, response): # Move to next page
url_ = response.request.url
yield response.follow(url_, callback=self.parse_item)
# I will use selenium to move next page because of next button is running
# on javascript without fixed url.
def parse_item(self, response): # Extract tours
for block in response.xpath("//div[#class='m_justify_list m_radius_box act_card act_card_lg a_sd_move j_activity_item js-item ']"):
article = {}
article['title'] = block.xpath('.//h3[#class="title"]/text()').extract()
article['city'] = response.xpath(".//div[#class='g_v_c_mid t_mid']/h1/text()").extract()# fixed
article['price'] = re.sub(" +","",block.xpath(".//span[#class='latest_price']/b/text()").extract_first()).strip()
article['tour_url'] = 'www.klook.com'+block.xpath(".//a/#href").extract_first()
yield article
hxs = HtmlXPathSelector(response) #response is already in Selector, use direct `response.xpath`
url = urllib.parse.urljoin(response.url, url)
use as:
url = response.urljoin(url)
yes it will stop as its a duplicate request to prev. url, you need to add dont_filter=True check
Instead of using Selenium, figure out what request the website performs using JavaScript (watch the Network tab of the developer tools of your browser while you navigate) and reproduce a similar request.
The website uses JSON requests undernead to fetch the items, which is much easier to parse than the HTML.
Also, if you are not familiar with Scrapy’s asynchronous nature, you are likely to get unexpected issues while using it in combination with Selenium.
Solutions like Splash or Selenium are only meant to be used as last resource, when everything else fails.
Although I've seen several similar questions here regarding this, none seem to precisely define the process for achieving this task. I borrowed largely from the Scrapy script located here but since it is over a year old I had to make adjustments to the xpath references.
My current code looks as such:
import scrapy
from tripadvisor.items import TripadvisorItem
class TrSpider(scrapy.Spider):
name = 'trspider'
start_urls = [
'https://www.tripadvisor.com/Hotels-g29217-Island_of_Hawaii_Hawaii-Hotels.html'
]
def parse(self, response):
for href in response.xpath('//div[#class="listing_title"]/a/#href'):
url = response.urljoin(href.extract())
yield scrapy.Request(url, callback=self.parse_hotel)
next_page = response.xpath('//div[#class="unified pagination standard_pagination"]/child::*[2][self::a]/#href')
if next_page:
url = response.urljoin(next_page[0].extract())
yield scrapy.Request(url, self.parse)
def parse_hotel(self, response):
for href in response.xpath('//div[starts-with(#class,"quote")]/a/#href'):
url = response.urljoin(href.extract())
yield scrapy.Request(url, callback=self.parse_review)
next_page = response.xpath('//div[#class="unified pagination "]/child::*[2][self::a]/#href')
if next_page:
url = response.urljoin(next_page[0].extract())
yield scrapy.Request(url, self.parse_hotel)
def parse_review(self, response):
item = TripadvisorItem()
item['headline'] = response.xpath('translate(//div[#class="quote"]/text(),"!"," ")').extract()[0][1:-1]
item['review'] = response.xpath('translate(//div[#class="entry"]/p,"\n"," ")').extract()[0]
item['bubbles'] = response.xpath('//span[contains(#class,"ui_bubble_rating")]/#alt').extract()[0]
item['date'] = response.xpath('normalize-space(//span[contains(#class,"ratingDate")]/#content)').extract()[0]
item['hotel'] = response.xpath('normalize-space(//span[#class="altHeadInline"]/a/text())').extract()[0]
return item
When running the spider in its current form, I scrape the first page of reviews for each hotel listed on the start_urls page but the pagination doesn't flip to the next page of reviews. From what I suspect, this is because of this line:
next_page = response.xpath('//div[#class="unified pagination "]/child::*[2][self::a]/#href')
Since these pages load dynamically, there is no existing href for the next page on the current page. Investigating further I've read that these requests are sending a POST request using XHR. By exploring the "Network" tab in Firefox "Inspect" I can see both a Request URL and Form Data that might be needed to flip the page according to other posts on SO regarding the same topic.
However, it seems that the other posts refer to a static URL starting point when trying to pass a FormRequest using Scrapy. With TripAdvisor, the URL will always change based on the name of the hotel we're looking at so I'm not sure how to chose a URL when using FormRequest to submit the form data: reqNum=1&changeSet=REVIEW_LIST (this form data also never seems to change from page to page).
Alternatively, there doesn't appear to be a way to extract the URL shown in the "Network" tab's "Request URL". These pages do have URLs that change from page to page but the way TripAdvisor is set up, I cannot seem to extract them from the source code. The review pages change by incrementing the part of the URL that is -orXX- where "XX" is a number. For example:
https://www.tripadvisor.com/Hotel_Review-g2312116-d113123-Reviews-Fairmont_Orchid_Hawaii-Puako_Kohala_Coast_Island_of_Hawaii_Hawaii.html
https://www.tripadvisor.com/Hotel_Review-g2312116-d113123-Reviews-or5-Fairmont_Orchid_Hawaii-Puako_Kohala_Coast_Island_of_Hawaii_Hawaii.html
https://www.tripadvisor.com/Hotel_Review-g2312116-d113123-Reviews-or10-Fairmont_Orchid_Hawaii-Puako_Kohala_Coast_Island_of_Hawaii_Hawaii.html
https://www.tripadvisor.com/Hotel_Review-g2312116-d113123-Reviews-or15-Fairmont_Orchid_Hawaii-Puako_Kohala_Coast_Island_of_Hawaii_Hawaii.html
So, my question is whether or not it is possible to paginate using the XHR request/form data or do I need to manually build a list of URLs for each hotel that adds the -orXX-?
Well I ended up discovering an xpath that apparently allowed pagination of the reviews, but it's funny because every time I checked the underlying HTML the href link never changed from referring to /Hotel_Review-g2312116-d113123-Reviews-or5-Fairmont_Orchid_Hawaii-Puako_Kohala_Coast_Island_of_Hawaii_Hawaii.html even if I was on page 10 for example. It seems the "-orXX-" part of the link always increments the XX by 5 so I'm not sure why this works.
All I did was change the line:
next_page = response.xpath('//div[#class="unified pagination "]/child::*[2][self::a]/#href')
to:
next_page = response.xpath('//link[#rel="next"]/#href')
and have >41K extracted reviews. Would love to get other's opinions on handling this problem in other situations.
I'm trying to scrap an e-commerce web site, and I'm doing it in 2 steps.
This website has a structure like this:
The homepage has the links to the family-items and subfamily-items pages
Each family & subfamily page has a list of products paginated
Right now I have 2 spiders:
GeneralSpider to get the homepage links and store them
ItemSpider to get elements from each page
I'm completely new to Scrapy, I'm following some tutorials to achieve this. I'm wondering how complex can be the parse functions and how rules works. My spiders right now looks like:
GeneralSpider:
class GeneralSpider(CrawlSpider):
name = 'domain'
allowed_domains = ['domain.org']
start_urls = ['http://www.domain.org/home']
def parse(self, response):
links = LinksItem()
links['content'] = response.xpath("//div[#id='h45F23']").extract()
return links
ItemSpider:
class GeneralSpider(CrawlSpider):
name = 'domain'
allowed_domains = ['domain.org']
f = open("urls.txt")
start_urls = [url.strip() for url in f.readlines()]
# Each URL in the file has pagination if it has more than 30 elements
# I don't know how to paginate over each URL
f.close()
def parse(self, response):
item = ShopItem()
item['name'] = response.xpath("//h1[#id='u_name']").extract()
item['description'] = response.xpath("//h3[#id='desc_item']").extract()
item['prize'] = response.xpath("//div[#id='price_eur']").extract()
return item
Wich is the best way to make the spider follow the pagination of an url ?
If the pagination is JQuery, meaning there is no GET variable in the URL, Would be possible to follow the pagination ?
Can I have different "rules" in the same spider to scrap different parts of the page ? or is better to have the spiders specialized, each spider focused in one thing?
I've also googled looking for any book related with Scrapy, but it seems there isn't any finished book yet, or at least I couldn't find one.
Does anyone know if some Scrapy book that will be released soon ?
Edit:
This 2 URL's fits for this example. In the Eroski Home page you can get the URL's to the products page.
In the products page you have a list of items paginated (Eroski Items):
URL to get Links: Eroski Home
URL to get Items: Eroski Fruits
In the Eroski Fruits page, the pagination of the items seems to be JQuery/AJAX, because more items are shown when you scroll down, is there a way to get all this items with Scrapy ?
Which is the best way to make the spider follow the pagination of an url ?
This is very site-specific and depends on how the pagination is implemented.
If the pagination is JQuery, meaning there is no GET variable in the URL, Would be possible to follow the pagination ?
This is exactly your use case - the pagination is made via additional AJAX calls that you can simulate inside your Scrapy spider.
Can I have different "rules" in the same spider to scrape different parts of the page ? or is better to have the spiders specialized, each spider focused in one thing?
Yes, the "rules" mechanism that a CrawlSpider provides is a very powerful piece of technology - it is highly configurable - you can have multiple rules, some of them would follow specific links that match specific criteria, or located in a specific section of a page. Having a single spider with multiple rules should be preferred comparing to having multiple spiders.
Speaking about your specific use-case, here is the idea:
make a rule to follow categories and subcategories in the navigation menu of the home page - this is there restrict_xpaths would help
in the callback, for every category or subcategory yield a Request that would mimic the AJAX request sent by your browser when you open a category page
in the AJAX response handler (callback) parse the available items and yield an another Request for the same category/subcategory but increasing the page GET parameter (getting next page)
Example working implementation:
import re
import urllib
import scrapy
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors import LinkExtractor
class ProductItem(scrapy.Item):
description = scrapy.Field()
price = scrapy.Field()
class GrupoeroskiSpider(CrawlSpider):
name = 'grupoeroski'
allowed_domains = ['compraonline.grupoeroski.com']
start_urls = ['http://www.compraonline.grupoeroski.com/supermercado/home.jsp']
rules = [
Rule(LinkExtractor(restrict_xpaths='//div[#class="navmenu"]'), callback='parse_categories')
]
def parse_categories(self, response):
pattern = re.compile(r'/(\d+)\-\w+')
groups = pattern.findall(response.url)
params = {'page': 1, 'categoria': groups.pop(0)}
if groups:
params['grupo'] = groups.pop(0)
if groups:
params['familia'] = groups.pop(0)
url = 'http://www.compraonline.grupoeroski.com/supermercado/ajax/listProducts.jsp?' + urllib.urlencode(params)
yield scrapy.Request(url,
meta={'params': params},
callback=self.parse_products,
headers={'X-Requested-With': 'XMLHttpRequest'})
def parse_products(self, response):
for product in response.xpath('//div[#class="product_element"]'):
item = ProductItem()
item['description'] = product.xpath('.//span[#class="description_1"]/text()').extract()[0]
item['price'] = product.xpath('.//div[#class="precio_line"]/p/text()').extract()[0]
yield item
params = response.meta['params']
params['page'] += 1
url = 'http://www.compraonline.grupoeroski.com/supermercado/ajax/listProducts.jsp?' + urllib.urlencode(params)
yield scrapy.Request(url,
meta={'params': params},
callback=self.parse_products,
headers={'X-Requested-With': 'XMLHttpRequest'})
Hope this is a good starting point for you.
Does anyone know if some Scrapy book that will be released soon?
Nothing specific that I can recall.
Though I heard that some publisher has some plans to may be release a book about web-scraping, but I'm not supposed to tell you that.