Python 使用scrapy爬行时动态启动URL列表

test.py

class SomewebsiteProductSpider(scrapy.Spider):
    name = "somewebsite"
    allowed_domains = ["somewebsite.com"]


start_urls = [

]

def parse(self, response):
    items = somewebsiteItem()

    title = response.xpath('//h1[@id="title"]/span/text()').extract()
    sale_price = response.xpath('//span[contains(@id,"ourprice") or contains(@id,"saleprice")]/text()').extract()
    category = response.xpath('//a[@class="a-link-normal a-color-tertiary"]/text()').extract()
    availability = response.xpath('//div[@id="availability"]//text()').extract()
    items['product_name'] = ''.join(title).strip()
    items['product_sale_price'] = ''.join(sale_price).strip()
    items['product_category'] = ','.join(map(lambda x: x.strip(), category)).strip()
    items['product_availability'] = ''.join(availability).strip()
    fo = open("C:\\Users\\user1\PycharmProjects\\test.txt", "w")
    fo.write("%s \n%s \n%s" % (items['product_name'], items['product_sale_price'], self.start_urls))
    fo.close()
    print(items)
    yield items

在启动爬网过程之前，如何将动态开始URL列表从test.py传递给SomewebsiteProductSpiders对象？任何帮助都将不胜感激。 谢谢。

process.crawl接受传递给spider构造函数的可选参数，因此您可以从spider的uu init_uuu填充开始URL，也可以使用自定义开始请求过程。比如说

test.py

class SomewebsiteProductSpider(scrapy.Spider):
    name = "somewebsite"
    allowed_domains = ["somewebsite.com"]


start_urls = [

]

def parse(self, response):
    items = somewebsiteItem()

    title = response.xpath('//h1[@id="title"]/span/text()').extract()
    sale_price = response.xpath('//span[contains(@id,"ourprice") or contains(@id,"saleprice")]/text()').extract()
    category = response.xpath('//a[@class="a-link-normal a-color-tertiary"]/text()').extract()
    availability = response.xpath('//div[@id="availability"]//text()').extract()
    items['product_name'] = ''.join(title).strip()
    items['product_sale_price'] = ''.join(sale_price).strip()
    items['product_category'] = ','.join(map(lambda x: x.strip(), category)).strip()
    items['product_availability'] = ''.join(availability).strip()
    fo = open("C:\\Users\\user1\PycharmProjects\\test.txt", "w")
    fo.write("%s \n%s \n%s" % (items['product_name'], items['product_sale_price'], self.start_urls))
    fo.close()
    print(items)
    yield items

有点像蜘蛛

---

只需将start_URL作为参数传递，就可以避免@mizghun答案中的额外kwargs解析

class SomewebsiteProductSpider(scrapy.Spider):
    ...
    def __init__(self, *args, **kwargs):
        self.start_urls = kwargs.pop('url_list', [])
        super(SomewebsiteProductSpider, *args, **kwargs)

---

也许这是一个稍微不同的问题，但当我将ITME列表作为字符串传递时，我会将每个url作为一个单独的字母。下面是一个scrapy网站的语句表单如果要从命令行设置start_url属性，则必须使用类似ast.literal_eval或json.loads的内容将其解析为列表，然后将其设置为属性。否则，您将导致对start_url字符串的迭代，这是一个非常常见的python陷阱，导致每个字符都被视为一个单独的url。如何设置属性？我想您的意思是将项目列表作为CLI参数传递。然后只需将其作为逗号分隔的列表传递，并像self.start\u url=comma\u delimited\u string.split'，'

class SomewebsiteProductSpider(scrapy.Spider):
    ...
    def __init__(self, *args, **kwargs):
        self.start_urls = kwargs.pop('url_list', [])
        super(SomewebsiteProductSpider, *args, **kwargs)

import scrapy
from scrapy.crawler import CrawlerProcess

class QuotesSpider(scrapy.Spider):
  name = 'quotes'

  def parse(self, response):
    print(response.url)

 process = CrawlerProcess()
 process.crawl(QuotesSpider, start_urls=["http://example.com", "http://example.org"])
 process.start()