问如何使用Scrapy和Splash处理分页，如果按钮的href为javascript:void(0)
EN

Stack Overflow用户

提问于 2022-05-12 12:48:26

回答 1查看 55关注 0票数 -1

我正试图从这个网站：https://www.topuniversities.com/university-rankings/world-university-rankings/2021中抓取大学的名称和链接，并且在处理分页时遇到了一个问题，因为指向下一页的按钮的href是javascript:void(0)，所以我无法用scrapy.Request()或response.follow()到达下一页，有什么方法可以这样处理分页吗？

网站屏幕截图

标签和href的屏幕截图

此网站的URL不包含params，如果单击下一个页面按钮，URL将保持不变，因此我无法通过更改URL来处理分页。

下面的代码片段只能在第一页和第二页获取大学的名称和链接：

import scrapy
from scrapy_splash import SplashRequest


class UniSpider(scrapy.Spider):
    name = 'uni'
    allowed_domains = ['www.topuniversities.com']

    script = """
    function main(splash, args)
      splash:set_user_agent("Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.54 Safari/537.36")
      splash.private_mode_enabled = false
      assert(splash:go(args.url))
      assert(splash:wait(3))

      return {
        html = splash:html()
      }
    end
    """

    next_page = """
    function main(splash, args)
        splash:set_user_agent("Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.54 Safari/537.36")
        splash.private_mode_enabled = false
        assert(splash:go(args.url))
        assert(splash:wait(3))

        local btn = assert(splash:jsfunc([[
        function(){
        document.querySelector("#alt-style-pagination a.page-link.next").click()
      }
        ]]))
        assert(splash:wait(2))
        btn()

        splash:set_viewport_full()
        assert(splash:wait(3))

        return {
          html = splash:html()
        }
    end
    """

    def start_requests(self):
        yield SplashRequest(
            url="https://www.topuniversities.com/university-rankings/world-university-rankings/2021",
            callback=self.parse, endpoint="execute",
            args={"lua_source": self.script})

    def parse(self, response):
        for uni in response.css("a.uni-link"):
            uni_link = response.urljoin(uni.css("::attr(href)").get())
            yield {
                "name": uni.css("::text").get(),
                "link": uni_link
            }

        yield SplashRequest(
            url=response.url,
            callback=self.parse, endpoint="execute",
            args={"lua_source": self.next_page}
        )

python

web-scraping

scrapy

scrapy-splash

云+校园专区

助力在校大学生快速入门云计算，畅游云端

回答 1

Stack Overflow用户

回答已采纳

发布于 2022-05-12 23:40:52

你不需要花这个简单的网站。

尝试加载以下链接：

https://www.topuniversities.com/sites/default/files/qs-rankings-data/en/2057712.txt

这有所有的大学，网站只加载这个文件/json一次，然后显示信息与分页。

下面是简短的代码(不使用scrapy)：

from requests import get
from json import loads, dumps
from lxml.html import fromstring

url = "https://www.topuniversities.com/sites/default/files/qs-rankings-data/en/2057712.txt"
html = get(url, stream=True)

## another approach for loading json
# jdata = loads(html.content.decode())

jdata = html.json()
for x in jdata['data']:
    core_id = x['core_id']
    country = x['country']
    city = x['city']
    guide = x['guide']
    nid = x['nid']
    title = x['title']
    logo = x['logo']
    score = x['score']
    rank_display = x['rank_display']
    region = x['region']
    stars = x['stars']
    recm = x['recm']
    dagger = x['dagger']

    ## convert title to text
    soup = fromstring(title)
    title = soup.xpath(".//a/text()")[0]

    print ( title )

以上代码打印个别大学的“标题”，尝试将其与其他可用列一起保存在CSV/Excel文件中。结果如下：

Massachusetts Institute of Technology (MIT) 
Stanford University
Harvard University
California Institute of Technology (Caltech)
University of Oxford
ETH Zurich - Swiss Federal Institute of Technology
University of Cambridge
Imperial College London