I'm starting around with Scrapy, and managed to extract some of the data I need. However, not everything is properly obtained. I'm applying the knowledge from the official tutorial found here, but it's not working. I've Googled around a bit, and also read this SO question but I'm fairly certain this isn't the problem here.
Anyhow, I'm trying to parse the product information from this webshop. I'm trying to obtain the product name, price, rrp, release date, category, universe, author and publisher. Here is the relevant CSS for one product: https://pastebin.com/9tqnjs7A. Here's my code. Everything with a #! at the end isn't working as expected.
import scrapy
import pprint
class ForbiddenPlanetSpider(scrapy.Spider):
name = "fp"
start_urls = [
'https://forbiddenplanet.com/catalog/?q=mortal%20realms&sort=release-date&page=1',
]
def parse(self, response):
for item in response.css("section.zshd-00"):
print(response.css)
name = item.css("h3.h4::text").get() #!
price = item.css("span.clr-price::text").get() + item.css("span.t-small::text").get()
rrp = item.css("del.mqr::text").get()
release = item.css("dd.mzl").get() #!
category = item.css("li.inline-list__item::text").get() #!
universe = item.css("dt.txt").get() #!
authors = item.css("a.SubTitleItems").get() #!
publisher = item.css("dd.mzl").get() #!
pprint.pprint(dict(name=name,
price=price,
rrp=rrp,
release=release,
category=category,
universe=universe,
authors=authors,
publisher = publisher
)
)
I think I need to add some sub-searching (at the moment release and publisher have the same criteria, for example), but I don't know how to word it to search for it (I've tried, but ended up with generic tutorials that don't cover it). Anything pointing me in the right direction is appreciated!
Oh, and I didn't include ' ' spaces because whenever I used one Scrapy immediately failed to find.
Scrapy doesn't render JS, try to disable javascript in your browser and refresh the page, the HTML structure is different for site version without JS.
you should rewrite your selectors with a new HTML structure. Try to use XPATH instead of CSS it's much flexible.
UPD
The easiest way to scrape this website makes a request to https://forbiddenplanet.com/api/products/listing/?q=mortal%20realms&sort=release-date
The response is a JSON object with all necessary data. You may transform the "results" field (or the whole JSON object) to a python dictionary and get all fields with dictionary methods.
A code draft that works and shows the idea.
import scrapy
import json
def get_tags(tags: list):
parsed_tags = []
if tags:
for tag in tags:
parsed_tags.append(tag.get('name'))
return parsed_tags
return None
class ForbiddenplanetSpider(scrapy.Spider):
name = 'forbiddenplanet'
allowed_domains = ['forbiddenplanet.com']
start_urls = ['https://forbiddenplanet.com/api/products/listing/?q=mortal%20realms&sort=release-date']
def parse(self, response):
response_dict = json.loads(response.body)
items = response_dict.get('results')
for item in items:
yield {
'name': item.get('title'),
'price': item.get('site_price'),
'rrp': item.get('rrp'),
'release': item.get('release_date'),
'category': get_tags(item.get('derived_tags').get('type')),
'universe': get_tags(item.get('derived_tags').get('universe')),
'authors': get_tags(item.get('derived_tags').get('author')),
'publisher': get_tags(item.get('derived_tags').get('publisher')),
}
next_page = response_dict.get('next')
if next_page:
yield scrapy.Request(
url=next_page,
callback=self.parse
)
Related
I downloaded scrapy-crawl-once and I am trying to run it in my program. I want to scrape each book's url from the first page of http://books.toscrape.com/ and then scrape the title of the book from that url. I know I can scrape each book title from the first page, but as practice for scrapy-crawl-once, I wanted to do it this way. I already added the middlewares and need to know where to add request.meta. From doing some research, there isn't much codes out there for some example guidance so was hoping someone can help here. I learned the basics of python two weeks ago so struggling right now. I tried this, but the results hasn't changed. Can someone help me out please. I added [:2] so that if I change it to [:3], I can show myself that it works.
def parse(self, response):
all_the_books = response.xpath("//article[#class='product_pod']")
for div in all_the_books[:2]:
book_link = 'http://books.toscrape.com/' + div.xpath(".//h3/a/#href").get()
request = scrapy.Request(book_link, self.parse_book)
request.meta['book_link'] = book_link
yield request
def parse_book(self, response):
name = response.xpath("//div[#class='col-sm-6 product_main']/h1/text()").get()
yield {
'name': name,
}
Its docs says
To avoid crawling a particular page multiple times set
request.meta['crawl_once'] = True
so you need to do
def parse(self, response):
all_the_books = response.xpath("//article[#class='product_pod']")
for div in all_the_books[:2]:
book_link = 'http://books.toscrape.com/' + div.xpath(".//h3/a/#href").get()
request = scrapy.Request(book_link, self.parse_book)
request.meta['crawl_once'] = True
yield request
And it will not crawl that link again
I have built a very simple scraper using Scrapy. For the output table, I would like to show the Google News search term as well as the Google resultstats value.
The information I would like to capture is showing in the source of the Google page as
<input class="gsfi" value="Elon Musk">
and
<div id="resultStats">About 52,300 results</div>
I have already tried to include both through ('input.value::text') and ('id.resultstats::text'), which did not work, however. Does anyone have an idea how to solve this situation?
import scrapy
class QuotesSpider(scrapy.Spider):
name = 'quotes'
allowed_domains = ['google.com']
start_urls = ['https://www.google.com/search?q=elon+musk&source=lnt&tbs=cdr%3A1%2Ccd_min%3A1%2F1%2F2015%2Ccd_max%3A12%2F31%2F2015&tbm=nws']
def parse(self, response):
for quote in response.css('div.quote'):
item = {
'search_title': quote.css('input.value::text').extract(),
'results': quote.css('id.resultstats::text').extract(),
}
yield item
The pages renders differently when you access it with Scrapy.
The search field becomes:
response.css('input#sbhost::attr(value)').get()
The results count is:
response.css('#resultStats::text').get()
Also, there is no quote class on that page.
You can test this in the scrapy shell:
scrapy shell -s ROBOTSTXT_OBEY=False "https://www.google.com/search?q=elon+musk&source=lnt&tbs=cdr%3A1%2Ccd_min%3A1%2F1%2F2015%2Ccd_max%3A12%2F31%2F2015&tbm=nws"
And then run those 2 commands.
[EDIT]
If your goal is to get one item for each URL, then you can do this:
def parse(self, response):
item = {
'search_title': response.css('input#sbhost::attr(value)').get(),
'results': response.css('#resultStats::text').get(),
}
yield item
If your goal is to extract every result on the page, then you need something different.
I have the crawler implemented as below.
It is working and it would go through sites regulated under the link extractor.
Basically what I am trying to do is to extract information from different places in the page:
- href and text() under the class 'news' ( if exists)
- image url under the class 'think block' ( if exists)
I have three problems for my scrapy:
1) duplicating linkextractor
It seems that it will duplicate processed page. ( I check against the export file and found that the same ~.img appeared many times while it is hardly possible)
And the fact is , for every page in the website, there are hyperlinks at the bottom that facilitate users to direct to the topic they are interested in, while my objective is to extract information from the topic's page ( here listed several passages's title under the same topic ) and the images found within a passage's page( you can arrive to the passage's page by clicking on the passage's title found at topic page).
I suspect link extractor would loop the same page over again in this case.
( maybe solve with depth_limit?)
2) Improving parse_item
I think it is quite not efficient for parse_item. How could I improve it? I need to extract information from different places in the web ( for sure it only extracts if it exists).Beside, it looks like that the parse_item could only progress HkejImage but not HkejItem (again I checked with the output file). How should I tackle this?
3) I need the spiders to be able to read Chinese.
I am crawling a site in HK and it would be essential to be capable to read Chinese.
The site:
http://www1.hkej.com/dailynews/headline/article/1105148/IMF%E5%82%B3%E4%BF%83%E4%B8%AD%E5%9C%8B%E9%80%80%E5%87%BA%E6%95%91%E5%B8%82
As long as it belongs to 'dailynews', that's the thing I want.
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.selector import Selector
from scrapy.http import Request, FormRequest
from scrapy.contrib.linkextractors import LinkExtractor
import items
class EconjournalSpider(CrawlSpider):
name = "econJournal"
allowed_domains = ["hkej.com"]
login_page = 'http://www.hkej.com/template/registration/jsp/login.jsp'
start_urls = 'http://www.hkej.com/dailynews'
rules=(Rule(LinkExtractor(allow=('dailynews', ),unique=True), callback='parse_item', follow =True),
)
def start_requests(self):
yield Request(
url=self.login_page,
callback=self.login,
dont_filter=True
)
# name column
def login(self, response):
return FormRequest.from_response(response,
formdata={'name': 'users', 'password': 'my password'},
callback=self.check_login_response)
def check_login_response(self, response):
"""Check the response returned by a login request to see if we are
successfully logged in.
"""
if "username" in response.body:
self.log("\n\n\nSuccessfully logged in. Let's start crawling!\n\n\n")
return Request(url=self.start_urls)
else:
self.log("\n\n\nYou are not logged in.\n\n\n")
# Something went wrong, we couldn't log in, so nothing happens
def parse_item(self, response):
hxs = Selector(response)
news=hxs.xpath("//div[#class='news']")
images=hxs.xpath('//p')
for image in images:
allimages=items.HKejImage()
allimages['image'] = image.xpath('a/img[not(#data-original)]/#src').extract()
yield allimages
for new in news:
allnews = items.HKejItem()
allnews['news_title']=new.xpath('h2/#text()').extract()
allnews['news_url'] = new.xpath('h2/#href').extract()
yield allnews
Thank you very much and I would appreciate any help!
First, to set settings, make it on the settings.py file or you can specify the custom_settings parameter on the spider, like:
custom_settings = {
'DEPTH_LIMIT': 3,
}
Then, you have to make sure the spider is reaching the parse_item method (which I think it doesn't, haven't tested yet). And also you can't specify the callback and follow parameters on a rule, because they don't work together.
First remove the follow on your rule, or add another rule, to check which links to follow, and which links to return as items.
Second on your parse_item method, you are getting incorrect xpath, to get all the images, maybe you could use something like:
images=hxs.xpath('//img')
and then to get the image url:
allimages['image'] = image.xpath('./#src').extract()
for the news, it looks like this could work:
allnews['news_title']=new.xpath('.//a/text()').extract()
allnews['news_url'] = new.xpath('.//a/#href').extract()
Now, as and understand your problem, this isn't a Linkextractor duplicating error, but only poor rules specifications, also make sure you have valid xpath, because your question didn't indicate you needed xpath correction.
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.
I am noob with python, been on and off teaching myself since this summer. I am going through the scrapy tutorial, and occasionally reading more about html/xml to help me understand scrapy. My project to myself is to imitate the scrapy tutorial in order to scrape http://www.gamefaqs.com/boards/916373-pc. I want to get a list of the thread title along with the thread url, should be simple!
My problem lies in not understanding xpath, and also html i guess. When viewing the source code for the gamefaqs site, I am not sure what to look for in order to pull the link and title. I want to say just look at the anchor tag and grab the text, but i am confused on how.
from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector
from tutorial.items import DmozItem
class DmozSpider(BaseSpider):
name = "dmoz"
allowed_domains = ["http://www.gamefaqs.com"]
start_urls = ["http://www.gamefaqs.com/boards/916373-pc"]
def parse(self, response):
hxs = HtmlXPathSelector(response)
sites = hxs.select('//a')
items = []
for site in sites:
item = DmozItem()
item['link'] = site.select('a/#href').extract()
item['desc'] = site.select('text()').extract()
items.append(item)
return items
I want to change this to work on gamefaqs, so what would i put in this path?
I imagine the program returning results something like this
thread name
thread url
I know the code is not really right but can someone help me rewrite this to obtain the results, it would help me understand the scraping process better.
The layout and organization of a web page can change and deep tag based paths can be difficult to deal with. I prefer to pattern match the text of the links. Even if the link format changes, matching the new pattern is simple.
For gamefaqs the article links look like:
http://www.gamefaqs.com/boards/916373-pc/37644384
That's the protocol, domain name, literal 'boards' path. '916373-pc' identifies the forum area and '37644384' is the article ID.
We can match links for a specific forum area using using a regular expression:
reLink = re.compile(r'.*\/boards\/916373-pc\/\d+$')
if reLink.match(link)
Or any forum area using using:
reLink = re.compile(r'.*\/boards\/\d+-[^/]+\/\d+$')
if reLink.match(link)
Adding link matching to your code we get:
import re
reLink = re.compile(r'.*\/boards\/\d+-[^/]+\/\d+$')
def parse(self, response):
hxs = HtmlXPathSelector(response)
sites = hxs.select('//a')
items = []
for site in sites:
link = site.select('a/#href').extract()
if reLink.match(link)
item = DmozItem()
item['link'] = link
item['desc'] = site.select('text()').extract()
items.append(item)
return items
Many sites have separate summary and detail pages or description and file links where the paths match a template with an article ID. If needed, you can parse the forum area and article ID like this:
reLink = re.compile(r'.*\/boards\/(?P<area>\d+-[^/]+)\/(?P<id>\d+)$')
m = reLink.match(link)
if m:
areaStr = m.groupdict()['area']
idStr = m.groupdict()['id']
isStr will be a string which is fine for filling in a URL template, but if you need to calculate the previous ID, etc., then convert it to a number:
idInt = int(idStr)
I hope this helps.