如何确定从'yield scrapy.Request'返回的生成器是否有任何数据?

2024-05-19 16:10:52 发布

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Scrapy Tutorial中,爬行器从class="next"中提取下一页链接并对其进行爬网-

import scrapy

class QuotesSpider(scrapy.Spider):
    name = "quotes"
    start_urls = [
        'http://quotes.toscrape.com/page/1/',
    ]

    def parse(self, response):
        for quote in response.css('div.quote'):
            yield {
                'text': quote.css('span.text::text').get(),
                'author': quote.css('span small::text').get(),
                'tags': quote.css('div.tags a.tag::text').getall(),
            }

        next_page = response.css('li.next a::attr(href)').get()
        if next_page is not None:
            yield response.follow(next_page, callback=self.parse)

就我而言,我在从Web服务器下载的文件中找不到下一页链接,但我知道格式是response.url/page/[page number]/连接的。不产生引号的请求页面仍然返回一个response,例如-No quotes found!。由于下一页的数量通常少于20页,我可以通过将spider的最后3行替换为-

for page_num in range(2, 20):
    yield response.follow(f"/page/{page_num}/", callback=self.parse)

但是,这会迫使爬行器请求不产生引号的页面(例如http://quotes.toscrape.com/page/11到20)。在请求不产生引号的第一页后,如何调整spider以终止page_num循环?(例如http://quotes.toscrape.com/page/11

伪码

    page_num = 2
    while (quotes are yielded from the response):
        yield response.follow(f"/page/{page_num}/", callback=self.parse)
        page_num += 1

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1条回答
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1楼 · 发布于 2024-05-19 16:10:52

您可以使用response.css('..')的结果作为下一页的条件。
在这种情况下,您的代码如下所示:

import scrapy

class QuotesSpider(scrapy.Spider):
    name = "quotes"
    start_urls = [
        'http://quotes.toscrape.com/page/1/',
    ]

    def parse(self, response):
        page_num = get_pagenumber_from_url(response.url)
        
        quotes_sel = response.css('div.quote'):
        # quotes_sel - will be SelectorList if page have item data
        # quotes_sel - will be None if page doesn't have item data
        for quote in quotes_sel:
            yield {
                'text': quote.css('span.text::text').get(),
                'author': quote.css('span small::text').get(),
                'tags': quote.css('div.tags a.tag::text').getall(),
            }

        if quotes_sel:
            next_page_url = f"/page/{str(page_num+1)}"
            yield response.follow(next_page_url , callback=self.parse)

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