尝试使用Scrapy获取少数网站的内容,但它们都返回403(禁止)响应代码。尽管当我使用“获取”功能发出请求时,相同的网站工作正常,如下所示:
import requests
url = "https://www.name_of_website.com/"
headers = {
"User-Agent": "Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.1; WOW64; Trident/5.0)",
}
response = requests.get(url, headers=headers)
print(response.status_code)
此外,网站可以像往常一样通过 Chrome 访问。我尝试通过 DEFAULT_REQUEST_HEADERS 与 Scrapy 中的 chrome 使用相同的标头,但仍然失败。
不知道为什么 scrapy 会失败,而常规 requests.get() 却可以工作。许多网站都观察到这种行为。我尝试使用 scrapy-fake-useragent 以及中间件,但是没有成功
任何线索或解决方案将不胜感激。
我在here看到类似的问题,但是没有帮助,因此需要向该领域的专家寻求新的想法。
谢谢
编辑(回答@ewoks和@Lakshmanrao Simhadri):
我正在尝试以下网址用于研究目的,正如我提到的我收到的响应代码:
https://www.fastcompany.com/ - 403
https://www.ft.com/ - 200
https://www.theinformation.com/ - 200
https://www.pcmag.com/ - 403
https://www.thestreet.com/ - 403
他们都没有使用过 scrapy。
我的Scrapy代码很简单如下:
class TheinformationSpider(scrapy.Spider):
name = "theinformation"
allowed_domains = ["www.theinformation.com"]
start_urls = ["https://www.theinformation.com/"]
def parse(self, response):
print(response)
我现在只是在寻找响应代码。
我更新的设置如下:
DEFAULT_REQUEST_HEADERS = {
"User-Agent": "Mozilla/5.0 (Linux; x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/121.0.0.0 Safari/537.36",
"Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7",
"Accept-Encoding": "gzip, deflate, br, zstd",
"Accept-Language": "en-US,en;q=0.9",
"Referer": "http://www.google.com",
}
我在爬行时收到以下响应:
2024-03-08 15:15:54 [scrapy.core.engine] DEBUG: Crawled (403) <GET https://www.theinformation.com/> (referer: http://www.google.com)
2024-03-08 15:15:54 [scrapy.spidermiddlewares.httperror] INFO: Ignoring response <403 https://www.theinformation.com/>: HTTP status code is not handled or not allowed
2024-03-08 15:15:54 [scrapy.core.engine] INFO: Closing spider (finished)
Total articles scrapped by "theinformation" = 0, null data = 0
我尝试从 chrome 传递精确的标头来发出 scrapy 请求,但仍然失败。通过使用一些代理,我可以获得响应。请看看下面的解决方案并让我知道您的想法
from urllib.parse import urlencode
import scrapy
# Get your own api_key from scrapeops or some other proxy vendor
API_KEY = "api_key"
def get_scrapeops_url(url):
payload = {'api_key': API_KEY, 'url': url}
proxy_url = 'https://proxy.scrapeops.io/v1/?' + urlencode(payload)
return proxy_url
class FastCompany(scrapy.Spider):
name = "fastcompany"
def start_requests(self):
urls = ["https://www.fastcompany.com/"]
for url in urls:
proxy_url = get_scrapeops_url(url)
yield scrapy.Request(url=proxy_url, callback=self.parse)
def parse(self, response):
print(response)