Scrapy 从 Link 中提取

Question

我正在尝试从某些链接中提取信息，但我无法访问这些链接，我从 start_url 中提取信息，但我不确定为什么。

这是我的代码：

import scrapy
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from tutorial.items import DmozItem
from scrapy.selector import HtmlXPathSelector

class DmozSpider(scrapy.Spider):
    name = "dmoz"
    allowed_domains = ["dmoz.org"]
    start_urls = [
        "http://www.dmoz.org/Computers/Programming/Languages/Python"
    ]
    rules = [Rule(SgmlLinkExtractor(allow=[r'Books']), callback='parse')] 


    def parse(self, response):
        hxs = HtmlXPathSelector(response)
        item = DmozItem()

        # Extract links
        item['link'] = hxs.select("//li/a/text()").extract()  # Xpath selector for tag(s)

        print item['title']

        for cont, i in enumerate(item['link']):
            print "link: ", cont, i

我没有从“http://www.dmoz.org/Computers/Programming/Languages/Python/Books", instead I get the links from "http://www.dmoz.org/Computers/Programming/Languages/Python”获得链接。

为什么？

Answer 1

要rules正常工作，您需要使用 CrawlSpider 而不是一般的 scrapy Spider。

此外，您需要将第一个解析函数重命名为 parse 以外的名称。否则，您将覆盖 CrawlSpider 的一个重要方法，它将无法工作。请参阅文档中的警告 http://doc.scrapy.org/en/0.24/topics/spiders.html?highlight=rules#crawlspider

您的代码正在抓取来自“http://www.dmoz.org/Computers/Programming/Languages/Python”的链接，因为 rule 命令被通用 Spider 忽略。

此代码应该有效：

import scrapy
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from dmoz.items import DmozItem
from scrapy.selector import HtmlXPathSelector

class DmozSpider(CrawlSpider):
    name = "dmoz"
    allowed_domains = ["dmoz.org"]
    start_urls = [
        "http://www.dmoz.org/Computers/Programming/Languages/Python"
    ]
    rules = [Rule(SgmlLinkExtractor(allow=[r'Books']), callback='parse_item')] 


    def parse_item(self, response):
        hxs = HtmlXPathSelector(response)
        item = DmozItem()

        # Extract links
        item['link'] = hxs.select("//li/a/text()").extract()  # Xpath selector for tag(s)

        print item['link']

        for cont, i in enumerate(item['link']):
            print "link: ", cont, i

Scrapy 从 Link 中提取

Scrapy extracting from Link

python

scrapy

scrapy-spider