python - 如何按照链接列表从 scrapy 的页面获取数据？

Question

我有一个网页要抓取。在页面上，是 <table> 中的链接列表。 .我正在尝试使用规则部分要求 Scrapy 通过链接，并获取链接目标页面上的数据。下面是我的代码:

class ToScrapeSpiderXPath(scrapy.Spider):
    name = 'coinmarketcap'
    start_urls = [
        'https://coinmarketcap.com/currencies/views/all/'
    ]

    rules = (
        Rule(LinkExtractor(allow=(), restrict_xpaths=('//tr/td[2]/a/@href',)), callback="parse", follow= True),
    )

    def parse(self, response):
        print("TEST TEST TEST")
        BTC = BTCItem()
        BTC['source'] = str(response.request.url).split("/")[2]
        BTC['asset'] = str(response.request.url).split("/")[4],
        BTC['asset_price'] = response.xpath('//*[@id="quote_price"]/text()').extract(),
        BTC['asset_price_change'] = response.xpath('/html/body/div[2]/div/div[1]/div[3]/div[2]/span[2]/text()').extract(),
        BTC['BTC_price'] = response.xpath('/html/body/div[2]/div/div[1]/div[3]/div[2]/small[1]/text()').extract(),
        BTC['Prct_change'] = response.xpath('/html/body/div[2]/div/div[1]/div[3]/div[2]/small[2]/text()').extract()
        yield (BTC)

我的问题是 Scrapy 没有跟踪链接。它只是在尝试从该链接中提取数据时获取链接。我错过了什么？

更新 #1: 为什么要抓取而不是抓取？

2017-03-28 23:10:33 [scrapy.core.engine] DEBUG: Crawled (200) <GET https://coinmarketcap.com/currencies/pivx/> (referer: None)
2017-03-28 23:10:33 [scrapy.core.engine] DEBUG: Crawled (200) <GET https://coinmarketcap.com/currencies/zcash/> (referer: None)
2017-03-28 23:10:33 [scrapy.core.engine] DEBUG: Crawled (200) <GET https://coinmarketcap.com/currencies/bitcoin/> (referer: None)
2017-03-28 23:10:33 [scrapy.core.scraper] DEBUG: Scraped from <200 https://coinmarketcap.com/currencies/nem/>

Answer 1

你需要继承一个CrawlSpider链接提取器工作的类:

from scrapy.spiders import CrawlSpider
from scrapy.spiders import Rule
from scrapy.contrib.linkextractors import LinkExtractor


class ToScrapeSpiderXPath(CrawlSpider):
    name = 'coinmarketcap'
    start_urls = [
        'https://coinmarketcap.com/currencies/views/all/'
    ]

    rules = (
        Rule(LinkExtractor(restrict_xpaths='//tr/td[2]/a'), callback="parse_table_links", follow= True),
    )

    def parse_table_links(self, response):
        print(response.url)

请注意，您需要修复 restrict_xpaths 值 - 它应该指向 a 元素而不是元素的 @href 属性.而且，您可以将其定义为字符串而不是元组。

此外，allow 参数是可选的。