How do I add a url to SgmlLinkExtractor? That is, how do I add an arbitrary url to run the callback on?
To elaborate, using dirbot as an example: https://github.com/scrapy/dirbot/blob/master/dirbot/spiders/googledir.py
parse_category only accesses everything that matches the SgmlLinkExtractor SgmlLinkExtractor(allow='directory.google.com/[A-Z][a-zA-Z_/]+$')
Use BaseSpider instead of CrawlSpider, then set add to start_requests or start_urls []
class MySpider(BaseSpider):
name = "myspider"
def start_requests(self):
return [Request("https://www.example.com",
callback=self.parse)]
def parse(self, response):
hxs = HtmlXPathSelector(response)
...
class ThemenHubSpider(CrawlSpider):
name = 'themenHub'
allowed_domains = ['themen.t-online.de']
start_urls = ["http://themen.t-online.de/themen-a-z/a"]
rules = [Rule(SgmlLinkExtractor(allow=['id_\d+']), 'parse_news')]
Related
I want to get all external links from a given website using Scrapy. Using the following code the spider crawls external links as well:
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors import LinkExtractor
from myproject.items import someItem
class someSpider(CrawlSpider):
name = 'crawltest'
allowed_domains = ['someurl.com']
start_urls = ['http://www.someurl.com/']
rules = (Rule (LinkExtractor(), callback="parse_obj", follow=True),
)
def parse_obj(self,response):
item = someItem()
item['url'] = response.url
return item
What am I missing? Doesn't "allowed_domains" prevent the external links to be crawled? If I set "allow_domains" for LinkExtractor it does not extract the external links. Just to clarify: I wan't to crawl internal links but extract external links. Any help appriciated!
You can also use the link extractor to pull all the links once you are parsing each page.
The link extractor will filter the links for you. In this example the link extractor will deny links in the allowed domain so it only gets outside links.
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors import LxmlLinkExtractor
from myproject.items import someItem
class someSpider(CrawlSpider):
name = 'crawltest'
allowed_domains = ['someurl.com']
start_urls = ['http://www.someurl.com/']
rules = (Rule(LxmlLinkExtractor(allow=()), callback='parse_obj', follow=True),)
def parse_obj(self,response):
for link in LxmlLinkExtractor(allow=(),deny = self.allowed_domains).extract_links(response):
item = someItem()
item['url'] = link.url
An updated code based on 12Ryan12's answer,
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors.lxmlhtml import LxmlLinkExtractor
from scrapy.item import Item, Field
class MyItem(Item):
url= Field()
class someSpider(CrawlSpider):
name = 'crawltest'
allowed_domains = ['someurl.com']
start_urls = ['http://www.someurl.com/']
rules = (Rule(LxmlLinkExtractor(allow=()), callback='parse_obj', follow=True),)
def parse_obj(self,response):
item = MyItem()
item['url'] = []
for link in LxmlLinkExtractor(allow=(),deny = self.allowed_domains).extract_links(response):
item['url'].append(link.url)
return item
A solution would be make usage a process_link function in the SgmlLinkExtractor
Documentation here http://doc.scrapy.org/en/latest/topics/link-extractors.html
class testSpider(CrawlSpider):
name = "test"
bot_name = 'test'
allowed_domains = ["news.google.com"]
start_urls = ["https://news.google.com/"]
rules = (
Rule(SgmlLinkExtractor(allow_domains=()), callback='parse_items',process_links="filter_links",follow= True) ,
)
def filter_links(self, links):
for link in links:
if self.allowed_domains[0] not in link.url:
print link.url
return links
def parse_items(self, response):
### ...
I want to crawl complete website using scrapy but right now its only crawling single page
import scrapy
from scrapy.http import HtmlResponse
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from scrapy.contrib.exporter import JsonItemExporter
class IzodspiderSpider(scrapy.Spider):
name = 'izodspider'
allowed_domains = ['izod.com']
start_urls = ['http://izod.com/']
rules = [Rule(SgmlLinkExtractor(), callback='parse_item', follow=True)]
def parse(self, response):
hxs = scrapy.Selector(response)
meta = hxs.xpath('//meta[#name=\'description\']/#content').extract()
name = hxs.xpath('//div[#id=\'product-details\']/h5').extract()
desc = hxs.xpath('//div[#id=\'product-details\']/p').extract()
is there any way to extract meta tags using portia ?
There is an error in the rule definition and inside the callback.
Since the parse function you use is parse_item you have to call it inside the callback instead of parse
You can find more information about the callback function on the documentation here http://doc.scrapy.org/en/latest/topics/request-response.html?highlight=callback#topics-request-response-ref-request-callback-arguments
class IzodspiderSpider(CrawlSpider):
name = "izod"
depth_limit= 0
bot_name = 'izod'
allowed_domains = ['izod.com']
start_urls = ['http://www.izod.com']
rules = (
Rule(SgmlLinkExtractor(allow=('')), callback='parse_items',follow= True),
)
def parse_items(self, response):
hxs = scrapy.Selector(response)
meta = hxs.xpath('//meta[#name=\'description\']/#content').extract()
name = hxs.xpath('//div[#id=\'product-details\']/h5').extract()
desc = hxs.xpath('//div[#id=\'product-details\']/p').extract()
I am making a crawler to crawl the website recursively but the problem is the spider does not enter the parse_item method.The name of my spider is example.py. The code is given below:
from scrapy.spider import Spider
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from scrapy.selector import Selector
from scrapy.http.request import Request
from scrapy.utils.response import get_base_url
class CrawlSpider(CrawlSpider):
name = "example"
download_delay = 2
allowed_domains = ["dmoz.org"]
print allowed_domains
start_urls = [
"http://www.dmoz.org/Arts/"
]
print start_urls
rules = (
Rule(SgmlLinkExtractor(allow=('/Arts', )), callback='parse_item',follow=True),
)
#The spide is not entering into this parse_item
def parse_item(self, response):
print "hello parse"
sel = Selector(response)
title = sel.xpath('//title/text()').extract()
print title
Why are you trying to define and call a function explicitly?
Try this:
class CrawlSpider(CrawlSpider):
name = "example"
download_delay = 2
allowed_domains = ["dmoz.org"]
print allowed_domains
start_urls = ["http://www.dmoz.org/Arts/"]
def parse(self, response):
print "hello parse"
sel = Selector(response)
title = sel.xpath('//title/text()').extract()
print title
I am very new to scrapy and also i didn't used regular expressions before
The following is my spider.py code
class ExampleSpider(BaseSpider):
name = "test_code
allowed_domains = ["www.example.com"]
start_urls = [
"http://www.example.com/bookstore/new/1?filter=bookstore",
"http://www.example.com/bookstore/new/2?filter=bookstore",
"http://www.example.com/bookstore/new/3?filter=bookstore",
]
def parse(self, response):
hxs = HtmlXPathSelector(response)
Now if we look at start_urls all the three urls are same except they differ at integer value 2?, 3? and so on i mean unlimited according to urls present on the site , i now that we can use crawlspider and we can construct regular expression for the URL like below,
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
import re
class ExampleSpider(CrawlSpider):
name = 'example.com'
allowed_domains = ['example.com']
start_urls = [
"http://www.example.com/bookstore/new/1?filter=bookstore",
"http://www.example.com/bookstore/new/2?filter=bookstore",
"http://www.example.com/bookstore/new/3?filter=bookstore",
]
rules = (
Rule(SgmlLinkExtractor(allow=(........),))),
)
def parse(self, response):
hxs = HtmlXPathSelector(response)
can u please guide me , that how can i construct a crawl spider Rule for the above start_url list.
If i understand you correctly, you want a lot of start URL with a certain pattern.
If so, you can override BaseSpider.start_requests method:
class ExampleSpider(BaseSpider):
name = "test_code"
allowed_domains = ["www.example.com"]
def start_requests(self):
for i in xrange(1000):
yield self.make_requests_from_url("http://www.example.com/bookstore/new/%d?filter=bookstore" % i)
...
If you are using CrawlSpider, it's not usually a good idea to override the parse method.
Rule object can filter the urls you are interesed to the ones you do not care for.
See CrawlSpider in the docs for reference.
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
import re
class ExampleSpider(CrawlSpider):
name = 'example.com'
allowed_domains = ['example.com']
start_urls = ['http://www.example.com/bookstore']
rules = (
Rule(SgmlLinkExtractor(allow=('\/new\/[0-9]\?',)), callback='parse_bookstore'),
)
def parse_boostore(self, response):
hxs = HtmlXPathSelector(response)
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
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']
There's an XMLFeedSpider one can use nowadays.
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...
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