首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >如何从需要使用scrapy-selenium单击的选项卡中爬行页

如何从需要使用scrapy-selenium单击的选项卡中爬行页
EN

Stack Overflow用户
提问于 2021-10-25 21:11:41
回答 2查看 329关注 0票数 0

因此,我想从这个网站,特别是从公司的详细信息部分的数据:

站点爬行

我得到了一个人的帮助,让它与python剧作家一起工作,但是我需要用python的scrapy-selenium来完成这个任务。

我想把答案中的代码重写成一种刮除硒的方法。

原始问题

我试着这样做,就像在这个问题上建议的那样。

刮硒

但运气不佳

我的守则:

资源/搜索结果_searchpage.yml:

代码语言:javascript
复制
products:
    css: 'div[data-content="productItem"]'
    multiple: true
    type: Text
    children:
        link:
            css: a.elements-title-normal 
            type: Link

crawler.py:

代码语言:javascript
复制
import scrapy
import csv
from scrapy_selenium import SeleniumRequest
import os
from selectorlib import Extractor
from scrapy import Selector

class Spider(scrapy.Spider):
    name = 'alibaba_crawler'
    allowed_domains = ['alibaba.com']
    start_urls = ['http://alibaba.com/']
    link_extractor = Extractor.from_yaml_file(os.path.join(os.path.dirname(__file__), "../resources/search_results_searchpage.yml"))

    def start_requests(self):
        search_text="Headphones"
        url="https://www.alibaba.com/trade/search?fsb=y&IndexArea=product_en&CatId=&SearchText={0}&viewtype=G".format(search_text)

        yield SeleniumRequest(url=url, callback = self.parse, meta = {"search_text": search_text})


    def parse(self, response):
        data = self.link_extractor.extract(response.text, base_url=response.url)
        for product in data['products']:
            parsed_url=product["link"]

            yield SeleniumRequest(url=parsed_url, callback=self.crawl_mainpage)
    
    def crawl_mainpage(self, response):
        driver = response.request.meta['driver']
        button = driver.find_element_by_xpath( "//span[@title='Company Profile']")
        button.click()
        driver.quit()

        yield {
            'name': response.xpath("//h1[@class='module-pdp-title']/text()").extract(),
            'Year of Establishment': response.xpath("//td[contains(text(), 'Year Established')]/following-sibling::td/div/div/div/text()").extract()
         }

使用以下代码运行代码:

代码语言:javascript
复制
scrapy crawl alibaba_crawler -o out.csv -t csv

公司名称将正确返回。成立之年仍然是空的,应该还一年。

EN

回答 2

Stack Overflow用户

发布于 2021-10-26 08:38:30

我没有正确使用选择器。这件事现在是正确的

代码语言:javascript
复制
def crawl_mainpage(self, response):
    driver = response.request.meta['driver']
    driver.find_element_by_xpath( "//span[@title='Company Profile']").click()
    sel = Selector(text=driver.page_source)
    driver.quit()

    yield {
    sel.xpath("//td[contains(text(), 'Year Established')]/following-sibling::td/div/div/div/text()").extract()
    }
票数 0
EN

Stack Overflow用户

发布于 2021-10-26 17:51:32

请参见下面使用刮硒库实现。Selenium对于for抓取来说非常慢。最好使用替代方法,如scrapy-splashscrapy-playwright。只刮两页,就花了22秒钟,而剧作家只花了不到5秒。

代码语言:javascript
复制
import scrapy
from scrapy.crawler import CrawlerProcess
import os
from selectorlib import Extractor
from scrapy_selenium import SeleniumRequest
from shutil import which


class Spider(scrapy.Spider):
    name = 'alibaba_crawler'
    allowed_domains = ['alibaba.com']
    start_urls = ['http://alibaba.com/']
    link_extractor = Extractor.from_yaml_file(os.path.join(
        os.path.dirname(__file__), "../resources/search_results_searchpage.yml"))

    def start_requests(self):
        search_text = "Headphones"
        url = "https://www.alibaba.com/trade/search?fsb=y&IndexArea=product_en&CatId=&SearchText={0}&viewtype=G".format(
            search_text)
        yield scrapy.Request(url, callback=self.parse, meta={"search_text": search_text})

    def parse(self, response):
        data = self.link_extractor.extract(
            response.text, base_url=response.url)
        for product in data['products']:
            parsed_url = product["link"]

            yield SeleniumRequest(url=parsed_url, callback=self.crawl_mainpage, script='document.querySelector("span[title=\'Company Profile\']").click();')

    def crawl_mainpage(self, response):
        yield {
            'name': response.xpath("//h1[@class='module-pdp-title']/text()").extract_first(),
            'Year of Establishment': response.xpath("//td[contains(text(), 'Year Established')]/following-sibling::td/div/div/div/text()").extract_first()
        }

if __name__ == "__main__":
    process = CrawlerProcess(settings={'DOWNLOADER_MIDDLEWARES': {
        'scrapy_selenium.SeleniumMiddleware': 800
    },
        'SELENIUM_DRIVER_NAME': 'chrome',
        'SELENIUM_DRIVER_EXECUTABLE_PATH': which('chromedriver'),
        'SELENIUM_DRIVER_ARGUMENTS': ['--headless']
    })
    process.crawl(Spider)
    process.start()

下面是一个样例爬行作业。注意,为了返回字符串而不是列表,我将您的extract()方法更改为extract_first()

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/69714821

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档