python - 如何在没有设置文件的情况下更改 scrapy 用户代理

Question

我已经像主要示例一样通过脚本实现了我的蜘蛛:

import scrapy

class BlogSpider(scrapy.Spider):
    name = 'blogspider'
    start_urls = ['https://blog.scrapinghub.com']

    def parse(self, response):
        for title in response.css('h2.entry-title'):
            yield {'title': title.css('a ::text').extract_first()}

        next_page = response.css('div.prev-post > a ::attr(href)').extract_first()
        if next_page:
            yield scrapy.Request(response.urljoin(next_page), callback=self.parse)

我运行:

scrapy runspider myspider.py

如果我没有设置或从 startproject 创建，如何更改用户代理？如此处指定:

https://doc.scrapy.org/en/latest/topics/settings.html

Answer 1

您可以在请求中手动添加 header ，以便您可以指定自定义 User Agent。

在您的蜘蛛文件中，当您请求时:

yield scrapy.Request(self.start_urls, callback=self.parse, headers={"User-Agent": "Your Custom User Agent"})

所以你的蜘蛛会是这样的:

class BlogSpider(scrapy.Spider):
    name = 'blogspider'
    start_urls = ['https://blog.scrapinghub.com']

    def start_requests(self):
        yield scrapy.Request(self.start_urls, callback=self.parse, headers={"User-Agent": "Your Custom User Agent"})

    def parse(self, response):
        for title in response.css('h2.entry-title'):
            yield {'title': title.css('a ::text').extract_first()}

        next_page = response.css('div.prev-post > a ::attr(href)').extract_first()
        if next_page:
            yield scrapy.Request(response.urljoin(next_page), callback=self.parse, headers={"User-Agent": "Your Custom User Agent"})