python - scrapy obtains the specified content of the web page, then flips to the next page to continue, with a fixed number of cycles. . question

Question

import scrapy
from movie.items import MovieItem
 
class MeijuSpider(scrapy.Spider):
    name = "meiju"
    allowed_domains = ["alexa.cn"]
    start_urls = ['www.alexa.cn/siterank']
 
def parse(self, response):
    movies = response.xpath('//ul[@class="siterank-sitelist"]/li')
    for each_movie in movies:
        item = MovieItem()
        item['name'] =each_movie.xpath('.//p[@class="infos"]').extract()[0]
        yield item

The code is like this. What I want to capture in a loop is:

www.alexa.cn/siterank/2
www.alexa.cn/siterank/3
www.alexa.cn/siterank/4
...

I think the loop should be like this for i in range(2,10):
yield scrapy.Request('www.alexa.cn/siterank/%d'%i), but I don’t know how to fill it in Go in. Help

Answer 1

If you are sure about the scope, it is better to start with start_urls

start_urls = ['http://www.alexa.cn/siterank/{n}'.format(n=x) for x in range(2,10)] 

Answer 2

There are examples on the official website. Regarding tracking the next page, the examples on the official website use recursion. The code on the official website is as follows:

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('p.quote'):
            yield {
                'text': quote.css('span.text::text').extract_first(),
                'author': quote.css('small.author::text').extract_first(),
                'tags': quote.css('p.tags a.tag::text').extract(),
            }
        
        # next_page是用css选择器获取到的下一页， 在下面它递归地调用了parse方法来不断地追踪下一页
        next_page = response.css('li.next a::attr(href)').extract_first()
        if next_page is not None:
            next_page = response.urljoin(next_page)
            yield scrapy.Request(next_page, callback=self.parse)

I used Scrapy to write a Tieba crawler. I also used this recursive method to get the next page. The code is as follows:

import scrapy
from tieba_crawler.items import ImageItem


class TiebaSpider(scrapy.Spider):
    name = 'tbimg'

    def start_requests(self):
        url = 'http://tieba.baidu.com/f?kw=%E6%B8%A1%E8%BE%B9%E9%BA%BB%E5%8F%8B'
        yield scrapy.Request(url=url, callback=self.parse_post)

    def parse_post(self, response):
        post_list = response.css('ul#thread_list li.j_thread_list')
        for item in post_list:
            title = item.css('a.j_th_tit::text').extract_first()
            url = 'http://tieba.baidu.com' \
              + item.css('a.j_th_tit::attr(href)').extract_first()
            yield scrapy.Request(url=url, callback=self.parse_image)
        page_list = response.css('p#frs_list_pager a::attr(href)').extract()
        if not page_list:
            return
        else:
            next_page = page_list[-2]
            if next_page:
                yield response.follow(next_page, callback=self.parse_post)

    def parse_image(self, response):
        img_urls = response.css('p#j_p_postlist img.BDE_Image::attr(src)').extract()
        yield ImageItem(image_urls=img_urls)
        page_list = response.css('ul.l_posts_num li.pb_list_pager a::attr(href)').extract()
        if not page_list:
            return
        else:
            next_page = page_list[-2]
            if next_page:
                yield response.follow(next_page, callback=self.parse_image)