Scrapy + splash:无法选择元素

Question

我正在学习如何使用scrapy.作为练习,我正在尝试访问https://www.ubereats.com/stores/,单击地址文本框,输入位置,然后按Enter按钮移动到包含可用于此的餐馆的下一页地点.我有以下lua代码:

function main(splash)
  local url = splash.args.url
  assert(splash:go(url))
  assert(splash:wait(5))

  local element = splash:select('.base_29SQWm')
  local bounds = element:bounds()
  assert(element:mouseclick{x = bounds.width/2, y = bounds.height/2})
    assert(element:send_text("Wall Street"))
  assert(splash:send_keys("<Return>"))
  assert(splash:wait(5))

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

当我点击"渲染!" 在splash API中,我收到以下错误消息:

  {
      "info": {
          "message": "Lua error: [string \"function main(splash)\r...\"]:7: attempt to index local 'element' (a nil value)",
          "type": "LUA_ERROR",
          "error": "attempt to index local 'element' (a nil value)",
          "source": "[string \"function main(splash)\r...\"]",
          "line_number": 7
      },
      "error": 400,
      "type": "ScriptError",
      "description": "Error happened while executing Lua script"
  }

不知怎的,我的css表达式是假的,导致试图访问未定义的元素/ nil!我尝试了其他表达方式,但我似乎无法弄明白!

问:有谁知道如何解决这个问题？

编辑:即使我仍然想知道如何实际点击元素,我想出了如何通过使用键获得相同的结果:

function main(splash)
    local url = splash.args.url
    assert(splash:go(url))
    assert(splash:wait(5))
    splash:send_keys("<Tab>")
    splash:send_keys("<Tab>")
    splash:send_text("Wall Street, New York")
    splash:send_keys("<Return>")
    assert(splash:wait(10))

    return {
    html = splash:html(),
    png = splash:png(),
    }
  end

但是,splash API中返回的html/images来自您输入地址的页面,而不是您输入地址后单击输入后看到的页面.

Q2:如何成功加载第二页？

Answer 1

不是一个完整的解决方案,但这是我到目前为止所拥有的:

import json
import re

import scrapy
from scrapy_splash import SplashRequest


class UberEatsSpider(scrapy.Spider):
    name = "ubereatspider"
    allowed_domains = ["ubereats.com"]

    def start_requests(self):
        script = """
        function main(splash)
            local url = splash.args.url
            assert(splash:go(url))
            assert(splash:wait(10))

            splash:set_viewport_full()

            local search_input = splash:select('#address-selection-input')
            search_input:send_text("Wall Street, New York")
            assert(splash:wait(5))

            local submit_button = splash:select('button[class^=submitButton_]')
            submit_button:click()

            assert(splash:wait(10))

            return {
                html = splash:html(),
                png = splash:png(),
            }
          end
        """
        headers = {
            'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_2) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/55.0.2883.95 Safari/537.36'
        }
        yield SplashRequest('https://www.ubereats.com/new_york/', self.parse, endpoint='execute', args={
            'lua_source': script,
            'wait': 5
        }, splash_headers=headers, headers=headers)

    def parse(self, response):
        script = response.xpath("//script[contains(., 'cityName')]/text()").extract_first()
        pattern = re.compile(r"window.INITIAL_STATE = (\{.*?\});", re.MULTILINE | re.DOTALL)

        match = pattern.search(script)
        if match:
            data = match.group(1)
            data = json.loads(data)
            for place in data["marketplace"]["marketplaceStores"]["data"]["entity"]:
                print(place["title"])

请注意Lua脚本中的更改:我找到了搜索输入,将搜索文本发送给它,然后找到"查找"按钮并单击它.在屏幕截图中,无论我设置的时间延迟,我都没有看到加载的搜索结果,但我设法从script内容中获取了餐馆名称.该place对象包含所有必要的信息来筛选所需的餐馆.

另请注意,我导航到的URL是"纽约"(不是一般的"商店").

我不完全确定为什么搜索结果页面没有加载,但希望它对你来说是一个好的开始,你可以进一步改进这个解决方案.