具有多个页面的 Scrapy

Question

我创建了一个简单的 scrapy 项目，在其中，我从初始站点示例中获取了总页码。com/full。现在我需要从 example.com/page-2 开始抓取所有页面到 100（如果总页数是 100）。我该怎么做？

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：

我尝试使用那个 urls = ('https://example.com/page-{}'.format(i) for i in range(1,total_pages)) 但它不起作用，可能是我做错了什么。

更新 #2：我已经像这样更改了我的代码

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)

但仍然循环只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出来。我该怎么做？

Answer 1

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 2

如 tutorial 所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。

Answer 3

您必须搜索 'next_page' 对象并在其位于页面上时继续循环。

# -*- 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)

如果你运行使用“-o sanet.json”键你会得到下面的结果。

scrapy runspider 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"}
]

具有多个页面的 Scrapy

Scrapy with multiple pages

scrapy

web-scraping

python-3.x

scrapy-spider