Scrapy怎么循环生成要爬取的页面url?

Question

Scrapy怎么循环生成要爬取的页面url?
比如下面这个demo的start_requests方法，它是手动写的page1,page2：

import scrapy


class QuotesSpider(scrapy.Spider):
    name = "quotes"

    def start_requests(self):
        urls = [
            'http://quotes.toscrape.com/page/1/',
            'http://quotes.toscrape.com/page/2/',
        ]
        for url in urls:
            yield scrapy.Request(url=url, callback=self.parse)

    def parse(self, response):
        page = response.url.split("/")[-2]
        filename = 'quotes-%s.html' % page
        with open(filename, 'wb') as f:
            f.write(response.body)
        self.log('Saved file %s' % filename)

如果有50页，url分别是：

http://www.example.com/page/1
http://www.example.com/page/2
...
http://www.example.com/page/50

怎么生成这个url，
for循环的语法应该怎么写？

Answer 1

urls = ["http://www.example.com/page/" + str(x) for x in range(1, 50, 1) + "/"]

其实你这demo不用yield，直接用内建的start_urls怼就行：

start_urls = ["http://www.example.com/page/" + str(x) for x in range(1, 50, 1)]

Answer 2

格式化字符串

for i in range(1,51):
    yield scrapy.Request('http://quotes.toscrape.com/page/%d/'%i)