CODEWITHSUNDEEP
pythonweb-crawlerscrapy
Zoran
Zoran

Reputation: 1026

Scrapy - Follow RSS links

I was wondering if anyone ever tried to extract/follow RSS item links using SgmlLinkExtractor/CrawlSpider. I can't get it to work...

I am using the following rule:


   rules = (
       Rule(SgmlLinkExtractor(tags=('link',), attrs=False),
           follow=True,
           callback='parse_article'),
       )

(having in mind that rss links are located in the link tag).

I am not sure how to tell SgmlLinkExtractor to extract the text() of the link and not to search the attributes ...

Any help is welcome, Thanks in advance

Upvotes: 7

Views: 5609

Answers (4)

NGloom
NGloom

Reputation: 408

XML Example From scrapy doc XMLFeedSpider

from scrapy.spiders import XMLFeedSpider
from myproject.items import TestItem

class MySpider(XMLFeedSpider):
    name = 'example.com'
    allowed_domains = ['example.com']
    start_urls = ['http://www.example.com/feed.xml']
    iterator = 'iternodes'  # This is actually unnecessary, since it's the default value
    itertag = 'item'

    def parse_node(self, response, node):
        self.logger.info('Hi, this is a <%s> node!: %s', self.itertag, ''.join(node.extract()))

        #item = TestItem() 
        item = {} # change to dict for removing the class not found error
        item['id'] = node.xpath('@id').extract()
        item['name'] = node.xpath('name').extract()
        item['description'] = node.xpath('description').extract()
        return item

Upvotes: -2

opyate
opyate

Reputation: 5428

There's an XMLFeedSpider one can use nowadays.

Upvotes: 6

Zoran
Zoran

Reputation: 1026

I have done it using CrawlSpider:

class MySpider(CrawlSpider):
   domain_name = "xml.example.com"

   def parse(self, response):
       xxs = XmlXPathSelector(response)
       items = xxs.select('//channel/item')
       for i in items: 
           urli = i.select('link/text()').extract()
           request = Request(url=urli[0], callback=self.parse1)
           yield request

   def parse1(self, response):
       hxs = HtmlXPathSelector(response)
       # ...
       yield(MyItem())

but I am not sure that is a very proper solution...

Upvotes: 0

Pablo Hoffman
Pablo Hoffman

Reputation: 1540

CrawlSpider rules don't work that way. You'll probably need to subclass BaseSpider and implement your own link extraction in your spider callback. For example:

from scrapy.spider import BaseSpider
from scrapy.http import Request
from scrapy.selector import XmlXPathSelector

class MySpider(BaseSpider):
    name = 'myspider'

    def parse(self, response):
        xxs = XmlXPathSelector(response)
        links = xxs.select("//link/text()").extract()
        return [Request(x, callback=self.parse_link) for x in links]

You can also try the XPath in the shell, by running for example:

scrapy shell http://blog.scrapy.org/rss.xml

And then typing in the shell:

>>> xxs.select("//link/text()").extract()
[u'http://blog.scrapy.org',
 u'http://blog.scrapy.org/new-bugfix-release-0101',
 u'http://blog.scrapy.org/new-scrapy-blog-and-scrapy-010-release']

Upvotes: 7

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