I am trying to scrape a betting site. However, when I check for the retrieved data in scrapy shell, I receive nothing.
The xpath to what I need is: //*[#id="yui_3_5_0_1_1562259076537_31330"] and when I write in the shell this is what I get:
In [18]: response.xpath ( '//*[#id="yui_3_5_0_1_1562259076537_31330"]')
Out[18]: []
The output is [] but I expected to be something from which I could extract the href.
When I use the "inspect" tool from Chrome, while the site is still loading, this id is outlined in purple. Does this mean that the site is using JavaScipt? And if this is true, is this the reason why scrapy does not find the item and returns []?
i try scraping the site just using Scrapy and this is my result.
This the items.py file
import scrapy
class LifeMatchsItem(scrapy.Item):
Event = scrapy.Field() # Name of event
Match = scrapy.Field() # Teams1 vs Team2
Date = scrapy.Field() # Date of Match
This is my Spider code
import scrapy
from LifeMatchesProject.items import LifeMatchsItem
class LifeMatchesSpider(scrapy.Spider):
name = 'life_matches'
start_urls = ['http://www.betfair.com/sport/home#sscpl=ro/']
custom_settings = {'FEED_EXPORT_ENCODING': 'utf-8'}
def parse(self, response):
for event in response.xpath('//div[contains(#class,"events-title")]'):
for element in event.xpath('./following-sibling::ul[1]/li'):
item = LifeMatchsItem()
item['Event'] = event.xpath('./a/#title').get()
item['Match'] = element.xpath('.//div[contains(#class,"event-name-info")]/a/#data-event').get()
item['Date'] = element.xpath('normalize-space(.//div[contains(#class,"event-name-info")]/a//span[#class="date"]/text())').get()
yield item
And this is the result
Related
At the moment have got a bit of the Frankenstein code (consisting of Beautifulsoup and Scrapy parts) that seem to be doing a job in terms of the reading the info from page 1 urls. Shall try to redo everything in Scrapy as soon as pagination issue resolved.
So what codes is meant to do:
Read all subcategories (BeautifulSoup part)
The rest are Scrapy code parts
Using the above urls read sub-subcategories.
Extract the last page number and loop over the above urls.
Extract the necessary product info from the above urls.
All except part 3 do seem to work.
Have tried to use the below code to extract the last page number but not sure how to integrate it into the main code
def parse_paging(self, response):
try:
for next_page in ('?pn=1' + response.xpath('//ul[#class="pagination pull-left"]/noscript/a/text()').extract()[-1]):
print(next_page)
# yield scrapy.Request(url=response.urljoin(next_page))
except:
pass
The below is the main code.
import requests
from bs4 import BeautifulSoup
import pandas as pd
import scrapy
from scrapy.crawler import CrawlerProcess
category_list = []
sub_category_url = []
root_url = 'https://uk.rs-online.com/web'
page = requests.get(root_url)
soup = BeautifulSoup(page.content, 'html.parser')
cat_up = [a.find_all('a') for a in soup.find_all('div',class_='horizontalMenu sectionUp')]
category_up = [item for sublist in cat_up for item in sublist]
cat_down = [a.find_all('a') for a in soup.find_all('div',class_='horizontalMenu sectionDown')]
category_down = [item for sublist in cat_down for item in sublist]
for c_up in category_up:
sub_category_url.append('https://uk.rs-online.com' + c_up['href'])
for c_down in category_down:
sub_category_url.append('https://uk.rs-online.com' + c_down['href'])
# print(k)
class subcategories(scrapy.Spider):
name = 'subcategories'
def start_requests(self):
urls = sub_category_url
for url in urls:
yield scrapy.Request(url=url, callback=self.parse)
def parse(self, response):
products = response.css('div.card.js-title a::href').extract() #xpath("//div[contains(#class, 'js-tile')]/a/#href").
for p in products:
url = urljoin(response.url, p)
yield scrapy.Request(url, callback=self.parse_product)
def parse_product(self, response):
for quote in response.css('tr.resultRow'):
yield {
'product': quote.css('div.row.margin-bottom a::text').getall(),
'stock_no': quote.css('div.stock-no-label a::text').getall(),
'brand': quote.css('div.row a::text').getall(),
'price': quote.css('div.col-xs-12.price.text-left span::text').getall(),
'uom': quote.css('div.col-xs-12.pack.text-left span::text').getall(),
}
process = CrawlerProcess()
process.crawl(subcategories)
process.start()
Would be exceptionally grateful if you could provides any hints on how to deal with the above issue.
Let me know if you have any questions.
I would suggest you to extract next page number by using this
and then construct next page url using this number.
next_page_number = response.css('.nextPage::attr(ng-click)').re_first('\d+')
I'm just starting to use scrapy and this is one of my first few projects. I am trying to scrape some company metadata from https://www.baincapitalprivateequity.com/portfolio/ . I have figured out my selectors but I'm unable to structure the output. I'm currently getting everything in one cell but I want the output to be one row for each company. If someone could help with where I'm going wrong, it'll be really great.
import scrapy
from ..items import BainpeItem
class BainPeSpider(scrapy.Spider):
name = 'Bain-PE'
allowed_domains = ['baincapitalprivateequity.com']
start_urls = ['https://www.baincapitalprivateequity.com/portfolio/']
def parse(self, response):
items = BainpeItem()
all_cos = response.css('div.grid')
for i in all_cos:
company = i.css('ul li::text').extract()
about = i.css('div.companyDetail p').extract()
items['company'] = company
items['about'] = about
yield items
You can just yield each item in the for loop:
for i in all_cos:
item = BainpeItem()
company = i.css('ul li::text').extract()
about = i.css('div.companyDetail p').extract()
item['company'] = company
item['about'] = about
yield item
This way each item will arrive in the pipeline separately.
I've been working on this for a week and am determined to get this working!
My ultimate goal is to write a webscraper where you can insert the county name and the scraper will produce a csv file of information from mugshots - Name, Location, Eye Color, Weight, Hair Color and Height (it's a genetics project I am working on).
The site organization is primary site page --> state page --> county page -- 120 mugshots with name and url --> url with data I am ultimately after and next links to another set of 120.
I thought the best way to do this would be to write a scraper that will grab the URLs and Names from the table of 120 mugshots and then use pagination to grab all the URLs and names from the rest of the county (in some cases there are 10's of thousands). I can get the first 120, but my pagination doesn't work.. so Im ending up with a csv of 120 names and urls.
I closely followed this article which was very helpful
from bs4 import BeautifulSoup
import requests
import lxml
import pandas as pd
county_name = input('Please, enter a county name: /Arizona/Maricopa-County-AZ \n')
print(f'Searching {county_name}. Wait, please...')
base_url = 'https://www.mugshots.com'
search_url = f'https://mugshots.com/US-Counties/{county_name}/'
data = {'Name': [],'URL': []}
def export_table_and_print(data):
table = pd.DataFrame(data, columns=['Name', 'URL'])
table.index = table.index + 1
table.to_csv('mugshots.csv', index=False)
print('Scraping done. Here are the results:')
print(table)
def get_mugshot_attributes(mugshot):
name = mugshot.find('div', attrs={'class', 'label'})
url = mugshot.find('a', attrs={'class', 'image-preview'})
name=name.text
url=mugshot.get('href')
url = base_url + url
data['Name'].append(name)
data['URL'].append(url)
def parse_page(next_url):
page = requests.get(next_url)
if page.status_code == requests.codes.ok:
bs = BeautifulSoup(page.text, 'lxml')
list_all_mugshot = bs.find_all('a', attrs={'class', 'image-preview'})
for mugshot in list_all_mugshot:
get_mugshot_attributes(mugshot)
next_page_text = mugshot.find('a class' , attrs={'next page'})
if next_page_text == 'Next':
next_page_text=mugshot.get_text()
next_page_url=mugshot.get('href')
next_page_url=base_url+next_page_url
print(next_page_url)
parse_page(next_page_url)
else:
export_table_and_print(data)
parse_page(search_url)
Any ideas on how to get the pagination to work and also how to eventually get the data from the list of URLs I scrape?
I appreciate your help! I've been working in python for a few months now, but the BS4 and Scrapy stuff is so confusing for some reason.
Thank you so much community!
Anna
It seems you want to know the logic as to how you can get the content using populated urls derived from each of the page traversing next pages. This is how you can parse all the links from each page including next page and then use those links to get the content from their inner pages.
import requests
from bs4 import BeautifulSoup
from urllib.parse import urljoin
url = "https://mugshots.com/"
base = "https://mugshots.com"
def get_next_pages(link):
print("**"*20,"current page:",link)
res = requests.get(link)
soup = BeautifulSoup(res.text,"lxml")
for item in soup.select("[itemprop='name'] > a[href^='/Current-Events/']"):
yield from get_main_content(urljoin(base,item.get("href")))
next_page = soup.select_one(".pagination > a:contains('Next')")
if next_page:
next_page = urljoin(url,next_page.get("href"))
yield from get_next_pages(next_page)
def get_main_content(link):
res = requests.get(link)
soup = BeautifulSoup(res.text,"lxml")
item = soup.select_one("h1#item-title > span[itemprop='name']").text
yield item
if __name__ == '__main__':
for elem in get_next_pages(url):
print(elem)
Here's an example: [http://www.cincinnati.com/story/money/2016/11/26/see-which-companies-16-deloitte-100/94441104/][1]
Ideally like to see a neatly crawled and extracted output data array with the following fields:
Company Name
2016 Rank
2015 Rank
Years in Business
Business Description
Website
2015 Revenues
2014 Revenues
HQ City
Year Founded
Employees
Is family owned?
from each of the specific company data pages.I'm purely beginner to scrapy i want know how to extract links automatically. Here in this code i'm feeding it manual. Can anyone help me here.
import scrapy
from spy.items import SpyItem
from scrapy.selector import HtmlXPathSelector
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.linkextractors import LinkExtractor
class ProjectSpider(CrawlSpider):
name = "project"
allowed_domains = ["cincinnati.com/story/money/2016/11/26/see-which-companies-16-deloitte-100/94441104/"]
start_urls = [100Links in here]
def parse(self, response):
item = SpyItem()
item['title'] = response.xpath('//*[#id="overlay"]/div[2]/article/div[3]/p[1]/strong/text()').extract()
item['Business'] =response.xpath('//*[#id="overlay"]/div[2]/article/div[3]/p[4]/text()').extract()
item['website'] =response.xpath('//p[5]/a/text()').extract()
item['Ranking']=response.xpath('//*[#id="overlay"]/div[2]/article/div[3]/p[2]/text()[1]').extract()
item['HQ']=response.css('p:nth-child(12)::text').extract()
item['Revenue2015']=response.xpath('//*[#id="overlay"]/div[2]/article/div[3]/p[7]/text()').extract()
item['Revenue2014']=response.css('p:nth-child(10)::text').extract()
item['YearFounded']=response.xpath('//p[11]/text()').extract().encode('utf-8')
item['Employees']=response.xpath('//article/div[3]/p[12]/text()').extract()
item['FamilyOwned']=response.xpath('//*[#id="overlay"]/div[2]/article/div[3]/p[13]/text()').extract()
yield item
There are at least two issues with your code.
allowed_domain has to be a domain. Not more.
You use a CrawlSpider that is meant to be used with Rules. You don't have any rules.
In the following there is some tested code as starting point:
import scrapy
class ProjectItem(scrapy.Item):
title = scrapy.Field()
owned = scrapy.Field()
class ProjectSpider(scrapy.Spider):
name = "cin100"
allowed_domains = ['cincinnati.com']
start_urls = ['http://www.cincinnati.com/story/money/2016/11/26/see-which-companies-16-deloitte-100/94441104/']
def parse(self, response):
# get selector for all 100 companies
sel_companies = response.xpath('//p[contains(.,"Here are the companies")]/following-sibling::p/a')
# create request for every single company detail page from href
for sel_companie in sel_companies:
href = sel_companie.xpath('./#href').extract_first()
url = response.urljoin(href)
request = scrapy.Request(url, callback=self.parse_company_detail)
yield request
def parse_company_detail(self, response):
# On detail page create item
item = ProjectItem()
# get detail information with specific XPath statements
# e.g. title is the first paragraph
item['title'] = response.xpath('//div[#role="main"]/p[1]//text()').extract_first()
# e.g. family owned has a label we can select
item['owned'] = response.xpath('//div[#role="main"]/p[contains(.,"Family owned")]/text()').extract_first()
# find clever XPaths for other fields ...
# ...
# Finally: yield the item
yield item
I am just starting to use Scrapy, and I am learning to use it as I go along. Please can someone explain why there is an error in my code, and what this error is? Is this error related to an invalid URL I have provided, and/or is it connected with invalid xpaths?
Here is my code:
from scrapy.spider import Spider
from scrapy.selector import Selector
class CatswikiSpider(Spider):
name = "catswiki"
allowed_domains = ["http://en.wikipedia.org/wiki/Cat"]
start_urls = [
"http://en.wikipedia.org/wiki/Cat"
]
def parse(self, response):
sel = Selector(response)
sites = sel.xpath('//body/div')
for site in sites:
title = ('//h1/span/text()').extract()
subtitle = ('//h2/span/text()').extract()
boldtext = ('//p/b').extract()
links = ('//a/#href').extract()
imagelinks = ('//img/#src').re(r'.*cat.*').extract()
print title, subtitle, boldtext, links, imagelinks
#filename = response.url.split("/")[-2]
#open(filename, 'wb').write(response.body)
And here are some attachments, showing the errors in the command prompt:
You need a function call before all your extract lines. I'm not familiar with scrapy, but it's probably something like:
title = site.xpath('//h1/span/text()').extract()