Web抓取错误-网站所有者错误:网站密钥无效的域

问题描述 投票:0回答:1

我试图获取此URL的内容-https://www.zillow.com/homedetails/131-Avenida-Dr-Berkeley-CA-94708/24844204_zpid/我好用这是我的代码。

import scrapy
class QuotesSpider(scrapy.Spider):
    name = "quotes"
    start_urls = [
        'https://www.zillow.com/homedetails/131-Avenida-Dr-Berkeley-CA-94708/24844204_zpid/',
    ]
    def parse(self, response):
        filename = 'test.html'
        with open(filename, 'wb') as f:
            f.write(response.body)
        self.log('Saved file %s' % filename)

我打开了抓取的数据(test.html),我得到了这个内容。enter image description here我试图找到解决方案,并且尝试了-ERROR for site owner: Invalid domain for site key但这并不能解决我的问题。

python web-scraping scrapy recaptcha
1个回答
0
投票

首先,尝试这种方法,看看是否可行:

Headerz = {
    "accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9",
    "accept-encoding": "gzip, deflate, br",
    "accept-language": "en-US,en;q=0.9",
    "cache-control": "no-cache",
    "content-type": "application/x-www-form-urlencoded; charset=UTF-8",
    "pragma": "no-cache",
    "sec-fetch-mode": "navigate",
    "sec-fetch-site": "cross-site",
    "sec-fetch-user": "?1",
    "upgrade-insecure-requests": "1",
    "user-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/79.0.3945.130 Safari/537.36",
}

class QuotesSpider(scrapy.Spider):
    name = "quotes"
    start_urls = [
        'https://www.zillow.com/homedetails/131-Avenida-Dr-Berkeley-CA-94708/24844204_zpid/',
    ]

    def start_requests(self):
        yield scrapy.Request(start_urls[0], callback=self.parse, headers=Headerz)

    def parse(self, response):
        filename = 'test.html'
        with open(filename, 'wb') as f:
            f.write(response.body)
        self.log('Saved file %s' % filename)

我们在普通浏览器中看不到输出的原因是,我们没有使用适当的标头,否则标头总是由浏览器发送的。

您需要按照上述代码中的说明添加标头,或者通过在settings.py中对其进行更新。

一种更好的方法是将“旋转代理”存储库与“旋转用户代理”存储库一起使用。

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