Spider code is wrong. I created a demon project, but it does not work, kindly check my vs code shot cut & I have no idea for all my spider code and problems.
import scrapy
class EmailSpider(scrapy.Spider):
name='Email'
start_url = [
'http://jsjy.114chn.com/'
]
def parse(self,response):
for Email in response.xpath("//span[#id='lblEmail']"):
yiel{
'email_text': Email.xpath(".//span[#id='lblEmail_text']/p").extract_first()
}
next_page= response.xpath("//li[#class='next']/a/#href").extract_first()
if next_page is not None:
next_page_link= response.urljoin(next_page)
yield scrapy.Request(url=next_page_link, callback=self.parse)
You have problems with indentation and yield function. Also made some code-style corrections:
import scrapy
class EmailSpider(scrapy.Spider):
name = 'Email'
start_url = ['http://jsjy.114chn.com/']
def parse(self, response):
for email in response.xpath("//span[#id='lblEmail']"):
yield {
'email_text': email.xpath(".//span[#id='lblEmail_text']/p").get()
}
next_page = response.xpath("//li[#class='next']/a/#href").get()
if next_page:
yield scrapy.Request(response.urljoin(next_page))
But since you don't have any #lblEmail elements on page, this spider will not output anything.
Related
Can anyone help me? I'm practicing and I can't understand what I did wrong on pagination! It only returns the first page to me and sometimes an error comes up. When it works, it just returns the first page.
"The source list for the Content Security Policy directive 'frame-src' contains an invalid source '*trackcmp.net' It will be ignored", source: https://naturaldaterra.com.br/hortifruti.html?page=2"
import scrapy
from scrapy_selenium import SeleniumRequest
class ComputerdealsSpider(scrapy.Spider):
name = 'produtos'
def start_requests(self):
yield SeleniumRequest(
url='https://naturaldaterra.com.br/hortifruti.html?page=1',
wait_time=3,
callback=self.parse
)
def parse(self, response):
for produto in response.xpath("//div[#class='gallery-items-1IC']/div"):
yield {
'nome_produto': produto.xpath(".//div[#class='item-nameContainer-1kz']/span/text()").get(),
'valor_produto': produto.xpath(".//span[#class='itemPrice-price-1R-']/text()").getall(),
}
next_page = response.xpath("//button[#class='tile-root-1uO'][1]/text()").get()
if next_page:
absolute_url = f"https://naturaldaterra.com.br/hortifruti.html?page={next_page}"
yield SeleniumRequest(
url=absolute_url,
wait_time=3,
callback=self.parse
)
The problem is that your xpath selector returns None instead of the next page number. Consider changing it from
next_page = response.xpath("//button[#class='tile-root-1uO'][1]/text()").get()
to
next_page = response.xpath("//button[#class='tile-root_active-TUl tile-root-1uO']/following-sibling::button[1]/text()").get()
For your future projects consider using scrapy-playwright to scrape js rendered websites. It is faster and simple to use. See a sample implementation of your scraper using scrapy-playwright
import scrapy
from scrapy.crawler import CrawlerProcess
class ComputerdealsSpider(scrapy.Spider):
name = 'produtos'
def start_requests(self):
yield scrapy.Request(
url='https://naturaldaterra.com.br/hortifruti.html?page=1',
meta={"playwright": True}
)
def parse(self, response):
for produto in response.xpath("//div[#class='gallery-items-1IC']/div"):
yield {
'nome_produto': produto.xpath(".//div[#class='item-nameContainer-1kz']/span/text()").get(),
'valor_produto': produto.xpath(".//span[#class='itemPrice-price-1R-']/text()").getall(),
}
# scrape next page
next_page = response.xpath(
"//button[#class='tile-root_active-TUl tile-root-1uO']/following-sibling::button[1]/text()").get()
yield scrapy.Request(
url='https://naturaldaterra.com.br/hortifruti.html?page=' + next_page,
meta={"playwright": True}
)
if __name__ == "__main__":
process = CrawlerProcess(settings={
"TWISTED_REACTOR": "twisted.internet.asyncioreactor.AsyncioSelectorReactor",
"DOWNLOAD_HANDLERS": {
"https": "scrapy_playwright.handler.ScrapyPlaywrightDownloadHandler",
}, })
process.crawl(ComputerdealsSpider)
process.start()
Here's my code to crawl RSS BBC but it returned nothing.
I checked xpath interactively using "Inspect" in Chrome and it seemed OK.
import scrapy
class BbcSpider(scrapy.Spider):
name = "bbc"
allowed_domains = ["feeds.bbci.co.uk/news/world/rss.xml"]
start_urls = ["https://feeds.bbci.co.uk/news/world/rss.xml"]
def parse(self, response):
all_rss = response.xpath('//div[#id="item"]/ul/li')
for rss in all_rss:
rss_url = rss.xpath('//a/#href').extract_first()
rss_title = rss.xpath('//a/text()').extract_first()
rss_short_content = rss.xpath('//div/text()').extract_first()
yield {
"URL": rss_url,
"Title": rss_title,
"Short Content": rss_short_content
}
Any help would be greatly appreciated!
Response is a .txt file so you can parse it in following way:
import scrapy
class BbcSpider(scrapy.Spider):
name = "bbc"
allowed_domains = ["feeds.bbci.co.uk/news/world/rss.xml"]
start_urls = ["https://feeds.bbci.co.uk/news/world/rss.xml"]
def parse(self, response):
rss_url = response.xpath('//link/text()').extract()[2:]
rss_title = response.xpath('//title/text()').extract()[2:]
rss_short_content = response.xpath('//description/text()').extract()
for i in range(len(rss_url)):
yield {
"URL": rss_url[i],
"Title": rss_title[i],
"Short Content": rss_short_content[i],
}
The first two URLs and titles had nothing to do with news so I dropped them.
The main reason for this crawler not yielding any data, because all_rss list is empty. Secondly, In Scrapy you have access to only the first GET request so if you open-source code using ctrl/cmd + U, you will be unable to find item id. So your
response.xpath('//div[#id="item"]/ul/li') selector returns empty list and for loop didn't execute.
Try this
for rss in response.css('item'):
rss_url = rss.css('link::text').extract_first()
rss_title = rss.css('title::text').extract_first()
rss_short_content = response.css('description::text').extract_first()
I'm trying to scrape data from amazon India website. I am not able collect response and parse the elements using the yield() method when:
1) I have to move from product page to review page
2) I have to move from one review page to another review page
Product page
Review page
Code flow:
1) customerReviewData() calls the getCustomerRatingsAndComments(response)
2) The getCustomerRatingsAndComments(response)
finds the URL of the review page and call the yield request method with getCrrFromReviewPage(request) as callback method, with url of this review page
3) getCrrFromReviewPage() gets new response of the firstreview page and scrape all the elements from the first review page (page loaded) and add it to customerReviewDataList[]
4) get URL of the next page if it exists and recursively call getCrrFromReviewPage() method, and crawl elements from next page, until all the review page is crawled
5) All the reviews gets added to the customerReviewDataList[]
I have tried playing around with yield() changing the parameters and also looked up the scrapy documentation for yield() and Request/Response yield
# -*- coding: utf-8 -*-
import scrapy
import logging
customerReviewDataList = []
customerReviewData = {}
#Get product name in <H1>
def getProductTitleH1(response):
titleH1 = response.xpath('normalize-space(//*[#id="productTitle"]/text())').extract()
return titleH1
def getCustomerRatingsAndComments(response):
#Fetches the relative url
reviewRelativePageUrl = response.css('#reviews-medley-footer a::attr(href)').extract()[0]
if reviewRelativePageUrl:
#get absolute URL
reviewPageAbsoluteUrl = response.urljoin(reviewRelativePageUrl)
yield Request(url = reviewPageAbsoluteUrl, callback = getCrrFromReviewPage())
self.log("yield request complete")
return len(customerReviewDataList)
def getCrrFromReviewPage():
userReviewsAndRatings = response.xpath('//div[#id="cm_cr-review_list"]/div[#data-hook="review"]')
for userReviewAndRating in userReviewsAndRatings:
customerReviewData[reviewTitle] = response.css('#cm_cr-review_list .review-title span ::text').extract()
customerReviewData[reviewDescription] = response.css('#cm_cr-review_list .review-text span::text').extract()
customerReviewDataList.append(customerReviewData)
reviewNextPageRelativeUrl = response.css('#cm_cr-pagination_bar .a-pagination .a-last a::attr(href)')[0].extract()
if reviewNextPageRelativeUrl:
reviewNextPageAbsoluteUrl = response.urljoin(reviewNextPageRelativeUrl)
yield Request(url = reviewNextPageAbsoluteUrl, callback = getCrrFromReviewPage())
class UsAmazonSpider(scrapy.Spider):
name = 'Test_Crawler'
allowed_domains = ['amazon.in']
start_urls = ['https://www.amazon.in/Philips-Trimmer-Cordless-Corded-QT4011/dp/B00JJIDBIC/ref=sr_1_3?keywords=philips&qid=1554266853&s=gateway&sr=8-3']
def parse(self, response):
titleH1 = getProductTitleH1(response),
customerReviewData = getCustomerRatingsAndComments(response)
yield{
'Title_H1' : titleH1,
'customer_Review_Data' : customerReviewData
}
I'm getting the following response:
{'Title_H1': (['Philips Beard Trimmer Cordless and Corded for Men QT4011/15'],), 'customer_Review_Data': <generator object getCustomerRatingsAndComments at 0x048AC630>}
The "Customer_review_Data" should be a list of dict of title and review
I am not able to figure out as to what mistake I am doing here.
When I use the log() or print() to see what data is captured in customerReviewDataList[], unable to see the data in the console either.
I am able to scrape all the reviews in customerReviewDataList[], if they are present in the product page,
In this scenario where I have to use the yield function I am getting the output stated above like this [https://ibb.co/kq8w6cf]
This is the kind of output I am looking for:
{'customerReviewTitle': ['Difficult to find a charger adapter'],'customerReviewComment': ['I already have a phillips trimmer which was only cordless. ], 'customerReviewTitle': ['Good Product'],'customerReviewComment': ['Solves my need perfectly HK']}]}
Any help is appreciated. Thanks in advance.
You should complete the Scrapy tutorial. The Following links section should be specially helpful to you.
This is a simplified version of your code:
def data_request_iterator():
yield Request('https://example.org')
class MySpider(Spider):
name = 'myspider'
start_urls = ['https://example.com']
def parse(self, response):
yield {
'title': response.css('title::text').get(),
'data': data_request_iterator(),
}
Instead, it should look like this:
class MySpider(Spider):
name = 'myspider'
start_urls = ['https://example.com']
def parse(self, response):
item = {
'title': response.css('title::text').get(),
}
yield Request('https://example.org', meta={'item': item}, callback=self.parse_data)
def parse_data(self, response):
item = response.meta['item']
# TODO: Extend item with data from this second response as needed.
yield item
I have ran across an issue in which my Lua script refuses to execute. The returned response from the ScrapyRequest call seems to be an HTML body, while i'm expecting a document title. I am assuming that the Lua script is never being called as it seems to have no apparent effect on the response. I have dug a lot through the documentation and can't quite seem to figure out what is missing here. Does anyone have any suggestions?
from urlparse import urljoin
import scrapy
from scrapy_splash import SplashRequest
GOOGLE_BASE_URL = 'https://www.google.com/'
GOOGLE_QUERY_PARAMETERS = '#q={query}'
GOOGLE_SEARCH_URL = urljoin(GOOGLE_BASE_URL, GOOGLE_QUERY_PARAMETERS)
GOOGLE_SEARCH_QUERY = 'example search query'
LUA_SCRIPT = """
function main(splash)
assert(splash:go(splash.args.url))
return splash:evaljs("document.title")
end
"""
SCRAPY_CRAWLER_NAME = 'google_crawler'
SCRAPY_SPLASH_ENDPOINT = 'render.html'
SCRAPY_ARGS = {
'lua_source': LUA_SCRIPT
}
def get_search_url(query):
return GOOGLE_SEARCH_URL.format(query=query)
class GoogleCrawler(scrapy.Spider):
name=SCRAPY_CRAWLER_NAME
search_url = get_search_url(GOOGLE_SEARCH_QUERY)
def start_requests(self):
response = SplashRequest(self.search_url,
self.parse, endpoint=SPLASH_ENDPOINT, args=SCRAPY_ARGS)
yield response
def parse(self, response):
doc_title = response.body_as_unicode()
print doc_title
'endpoint' argument of SplashRequest must be 'execute' in order to execute a Lua script; it is 'render.html' in the example.
LUA_SCRIPT = """
function main(splash)
assert(splash:go(splash.args.url))
return title = splash:evaljs("document.title")
end
"""
def start_requests(self):
SplashRequest(self.search_url,self.parse, endpoint='execute',args=SCRAPY_ARGS)
You can recover the value with response.data['title']
I need help to convert relative URL to absolute URL in Scrapy spider.
I need to convert links on my start pages to absolute URL to get the images of the scrawled items, which are on the start pages. I unsuccessfully tried different ways to achieve this and I'm stuck. Any suggestion?
class ExampleSpider(scrapy.Spider):
name = "example"
allowed_domains = ["example.com"]
start_urls = [
"http://www.example.com/billboard",
"http://www.example.com/billboard?page=1"
]
def parse(self, response):
image_urls = response.xpath('//div[#class="content"]/section[2]/div[2]/div/div/div/a/article/img/#src').extract()
relative_url = response.xpath(u'''//div[contains(concat(" ", normalize-space(#class), " "), " content ")]/a/#href''').extract()
for image_url, url in zip(image_urls, absolute_urls):
item = ExampleItem()
item['image_urls'] = image_urls
request = Request(url, callback=self.parse_dir_contents)
request.meta['item'] = item
yield request
There are mainly three ways to achieve that:
Using urljoin function from urllib:
from urllib.parse import urljoin
# Same as: from w3lib.url import urljoin
url = urljoin(base_url, relative_url)
Using the response's urljoin wrapper method, as mentioned by Steve.
url = response.urljoin(relative_url)
If you also want to yield a request from that link, you can use the handful response's follow method:
# It will create a new request using the above "urljoin" method
yield response.follow(relative_url, callback=self.parse)