多页抓取

Question

我创建了一个简单的scrapy项目，在其中，我从初始网站example.com/full获得了总页数。现在，我需要将所有页面从example.com/page-2刮到100（如果总页面数是100）。我该怎么办？

任何建议都会有所帮助。

代码：

import scrapy


class AllSpider(scrapy.Spider):
    name = 'all'
    allowed_domains = ['example.com']
    start_urls = ['https://example.com/full/']
    total_pages = 0

def parse(self, response):
    total_pages = response.xpath("//body/section/div/section/div/div/ul/li[6]/a/text()").extract_first()
    #urls = ('https://example.com/page-{}'.format(i) for i in range(1,total_pages))
    print(total_pages)

更新＃1：

我尝试使用该urls = ('https://example.com/page-{}'.format(i) for i in range(1,total_pages))，但无法正常工作，可能是我做错了。

更新＃2： 我已经更改了这样的代码

class AllSpider(scrapy.Spider):
name = 'all'
allowed_domains = ['sanet.st']
start_urls = ['https://sanet.st/full/']
total_pages = 0

def parse(self, response):
    total_pages = response.xpath("//body/section/div/section/div/div/ul/li[6]/a/text()").extract_first()
    for page in range(2, int(total_pages)):
        url = 'https://sanet.st/page-'+str(page)
        yield scrapy.Request(url)
        title =  response.xpath('//*[@class="list_item_title"]/h2/a/span/text()').extract()
        print(title)

但是仍然循环仅重复显示第一页标题。 我需要从不同的页面中提取标题并在提示中打印它。 我该怎么办？

Answer 1

您必须搜索“ next_page”对象，并在其在页面上时继续循环。

# -*- coding: utf-8 -*-
import scrapy
from scrapy.http import Request


class SanetSpider(scrapy.Spider):
    name = 'sanet'
    allowed_domains = ['sanet.st']
    start_urls = ['https://sanet.st/full/']

    def parse(self, response):
        yield {
            # Do something.
            'result': response.xpath('//h3[@class="posts-results"]/text()').extract_first()
        }

        # next_page = /page-{}/ where {} number of page.
        next_page = response.xpath('//a[@data-tip="Next page"]/@href').extract_first()

        # next_page = https://sanet.st/page-{}/ where {} number of page.
        next_page = response.urljoin(next_page)

        # If next_page have value
        if next_page:
            # Recall parse with url https://sanet.st/page-{}/ where {} number of page.
            yield scrapy.Request(url=next_page, callback=self.parse)

如果使用“ -o sanet.json”键运行此代码，则会得到以下结果。

  

scrapy runpider sanet.py -o sanet.json

[
{"result": "results 1 - 15 from 651"},
{"result": "results 16 - 30 from 651"},
{"result": "results 31 - 45 from 651"},
...
etc.
...
{"result": "results 631 - 645 from 651"},
{"result": "results 646 - 651 from 651"}
]

Answer 2

from scrapy.http import Request


def parse(self, response):
    total_pages = response.xpath("//body/section/div/section/div/div/ul/li[6]/a/text()").extract_first()
    urls = ('https://example.com/page-{}'.format(i) for i in range(1,total_pages))
    for url in urls:
        yield Request(url, callback=self.parse_page)

def parse_page(self, response):
    # do the stuff

Answer 3

tutorial中显示的另一种方法是使用yield response.follow(url, callback=self.parse_page)，它直接支持相对URL。