Question

我有一个抓取程序，我需要在抓取时单击下一步按钮，实际上我大约在一周前在这里问了一个有关如何执行此操作的问题，并获得了一些不错的答复，但是我得到的答案代码仅部分起作用。它将抓取第1页和第2页，但没有跳到第3页，而是跳到了最后一页，第10页，我不知道为什么。

import csv
from scrapy.spiders import Spider
from scrapy_splash import SplashRequest
from ..items import GameItem
def process_csv(csv_file):
    data = []
    reader = csv.reader(csv_file)
    next(reader)
    for fields in reader:
        if fields[0] != "":
            url = fields[0]
        else:
            continue # skip the whole row if the url column is empty
        if fields[1] != "":
            ip = "http://" + fields[1] + ":8050" # adding http and port because this is the needed scheme
        if fields[2] != "":
            useragent = fields[2]
        data.append({"url": url, "ip": ip, "ua": useragent})
    return data
class MySpider(Spider):
    name = 'splash_spider'  # Name of Spider

    # notice that we don't need to define start_urls
    # just make sure to get all the urls you want to scrape inside start_requests function

    # getting all the url + ip address + useragent pairs then request them
    def start_requests(self):

        # get the file path of the csv file that contains the pairs from the settings.py
        with open(self.settings["PROXY_CSV_FILE"], mode="r") as csv_file:
           # requests is a list of dictionaries like this -> {url: str, ua: str, ip: str}
            requests = process_csv(csv_file)

        for req in requests:
            # no need to create custom middlewares  # just pass useragent using the headers param, and pass proxy using the meta param

            yield SplashRequest(url=req["url"], callback=self.parse, args={"wait": 3},
                    headers={"User-Agent": req["ua"]},
                    splash_url = req["ip"],
                    )

    # Scraping
    def parse(self, response):
        item = GameItem()
        for game in response.css("tr"):
            # Card Name
            yield {
                    'card_name':  game.css("a.card_popup::text").get(),
                    }

           next_page = response.css('table+ div a:nth-child(8)::attr("href")').get()
            if next_page is not None:
                yield response.follow(next_page, self.parse)

Answer 1

next_page = response.css('table+ div a:nth-child(8)::attr("href")').get()

您肯定不想要nth-child(8)，而是想要最后一个div及其最后一个a，它包含一个href属性，即：

response.css("#content > div:last-of-type > a[href]:last-of-type')

如果您想更加勤奋，请检查匹配的<a>的文本，以确保其中包含短语Next

Answer 2

这是正确的代码，需要使用xpath而不是CSS。现在工作正常。

next_page = response.xpath('//a[contains(., "- Next>>")]/@href').get()
        if next_page is not None:
            yield response.follow(next_page, self.parse)

尝试在抓取时单击下一步按钮

2 个答案: