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How to use Scrapy for URL crawling

I want to crawl the link https://www.aparat.com/.
I crawl it correctly and get all the video links with header tag;like this :
import scrapy
class BlogSpider(scrapy.Spider):
name = 'aparatspider'
start_urls = ['https://www.aparat.com/']
def parse(self, response):
print '=' * 80 , 'latest-trend :'
ul5 = response.css('.block-grid.xsmall-block-grid-2.small-block-grid-3.medium-block-grid-4.large-block-grid-5.is-not-center')
ul5 = ul5.css('ul').css('li')
latesttrend = []
for li5 in ul5:
latesttrend.append(li5.xpath('div/div[1]/a').xpath('#onmousedown').extract_first().encode('utf8'))
print(latesttrend)
now my question is this:
How can I get all the links from the داغ ترین ها tag, more than 1000? Currently, I get only 60, more or less.

I did this with the following code :
import scrapy
from scrapy.linkextractors import LinkExtractor
from scrapy.spiders import CrawlSpider, Rule
from scrapy.http import Request
class aparat_hotnewsItem(scrapy.Item):
videourl = scrapy.Field()
class aparat_hotnewsSpider(CrawlSpider):
name = 'aparat_hotnews'
allowed_domains = ['www.aparat.com']
start_urls = ['http://www.aparat.com/']
# Xpath for selecting links to follow
xp = 'your xpath'
rules = (
Rule(LinkExtractor(restrict_xpaths=xp), callback='parse_item', follow=True),
)
def parse_item(self, response):
item = aparat_hotnewsItem()
item['videourl'] = response.xpath('your xpath').extract()
yield item

Scrapy crawlspider rule not working for next pages

I am trying to crawl next page links using crawlspider but I am not able to get any result if I change parse function to something else. My Rule is not working.I am able to fetch only current page with parse function.Where am I going wrong.
This is my naukri_spider.py file
import scrapy
from scrapy import Spider
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from naukri.items import NaukriItem
class NaukriSpider(Spider):
name = "naukri"
allowed_domains = ["naukri.com"]
start_urls = ["http://www.naukri.com/information-technology-jobs?xt=catsrch&qf[]=24"]
rules = (
Rule(SgmlLinkExtractor(allow=(), restrict_xpaths=('//div[#class="pagination"]/a/button[#class="grayBtn"]',)), callback="parse", follow= True),
)
def parse(self,response):
for sel in response.xpath('//*[#class="content"]'):
item = NaukriItem()
item['title'] = sel.xpath('span[#class="desig"]/text()').extract()
item['location'] = sel.xpath('span[#class="loc"]/span/text()').extract()
item['organization'] = sel.xpath('span[#class="org"]/text()').extract()
yield item

The parse method implemeted by CrawlSpider used to follow links. Change your rule callback to parse_start_url and override it.
This code works fine.
import scrapy
from scrapy import Spider
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from stackoverflow.items import NaukriItem
class NaukriSpider(CrawlSpider):
name = "naukri"
allowed_domains = ["naukri.com"]
start_urls = ["http://www.naukri.com/information-technology-jobs?xt=catsrch&qf[]=24"]
rules = (
Rule(SgmlLinkExtractor(allow=('information-technology-jobs.*', )), callback="parse_start_url", follow= True),
)
def parse_start_url(self,response):
for sel in response.xpath('//*[#class="content"]'):
item = NaukriItem()
item['title'] = sel.xpath('span[#class="desig"]/text()').extract()
item['location'] = sel.xpath('span[#class="loc"]/span/text()').extract()
item['organization'] = sel.xpath('span[#class="org"]/text()').extract()
yield item

scrapy spider code check

so im' trying to scrape the website in the SgmlLinkExtractor parameters below website with scrapy, and this is what my spider looks like:
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from desidime_sample.items import DesidimeItem
import string
class DesidimeSpider(CrawlSpider):
name = "desidime"
allowed_domains = ["desidime.com"]
start_urls = ["http://www.desidime.com/forums/hot-deals-online"]
rules = (
Rule(SgmlLinkExtractor(allow=(), restrict_xpaths=('''//td[not(#*)]/div
[not(#*)]/a[not(#class)]/#href''')), callback="parse_items", follow=True),
)
def parse_items(self, response):
hxs = HtmlXPathSelector(response)
deals = hxs.select('''//div[#class='user-comment-text'][1]''')
items = []
for deals in deals:
item = DesidimeItem()
item["deal"] = deals.select("//div[#class='user-comment-text'][1]/p/text()").extract()
item["link"] = deals.select("//div[#class='user-comment-text'][1]/p[1]/a[1]/#href").extract()
items.append(item)
return items
It should be quite obvious what I'm trying to do, but for some reason when I tell the spider to crawl and export the text and links to the CVS file, I end up with:
link,deal http://wwww.facebook.com/desidime,
http://wwww.facebook.com/desidime,
(same thing for many more lines, then:)
",,"
, " same url" ,
(same thing for many more lines, then:)
"link,deals"
So, can anyone tell me what the problem is? If you run each of my above xpaths as reponse.xpath("xpath").extract() after scrapy shell "//corresponingcrawlruleurl", you'll get the right results.

The problem is inside the parse_items callback. When you iterate over the deals, the deal context-specific locators have to be relative. In other words, start your XPath expressions inside the loop with a dot:
def parse_items(self, response):
for deal in response.xpath("//div[#class='user-comment-text'][1]"):
item = DesidimeItem()
item["deal"] = deal.xpath(".//p/text()").extract()
item["link"] = deal.xpath(".//p[1]/a[1]/#href").extract()
yield item
(note that I've also simplified the code).
Here is the complete spider, I'm executing (it does scrape the text and links, though I don't know what is your desired output):
import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor
class DesidimeItem(scrapy.Item):
deal = scrapy.Field()
link = scrapy.Field()
class DesidimeSpider(CrawlSpider):
name = "desidime"
allowed_domains = ["desidime.com"]
start_urls = ["http://www.desidime.com/forums/hot-deals-online"]
rules = [
Rule(LinkExtractor(restrict_xpaths="//td[not(#*)]/div[not(#*)]/a[not(#class)]"),
callback="parse_items",
follow=True),
]
def parse_items(self, response):
for deal in response.xpath("//div[#class='user-comment-text'][1]"):
item = DesidimeItem()
item["deal"] = deal.xpath(".//p/text()").extract()
item["link"] = deal.xpath(".//p[1]/a[1]/#href").extract()
yield item

Scrapy, only follow internal URLS but extract all links found

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):
### ...

Constructing a regular expression for url in start_urls list in scrapy framework python

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)