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Scrapy/Python yield and continue processing possible?

I am trying this sample code
from scrapy.spiders import Spider, Request
import scrapy
class MySpider(Spider):
name = 'toscrapecom'
start_urls = ['http://books.toscrape.com/catalogue/page-1.html']
urls = (
'http://books.toscrape.com/catalogue/page-{}.html'.format(i + 1) for i in range(50)
)
def parse(self, response):
for url in self.urls:
return Request(url)
It crawls all the pages fine. However if I yield an item before the for loop then it crawls only the first page. (as shown below)
from scrapy.spiders import Spider, Request
import scrapy
class MySpider(Spider):
name = 'toscrapecom'
start_urls = ['http://books.toscrape.com/catalogue/page-1.html']
urls = (
'http://books.toscrape.com/catalogue/page-{}.html'.format(i + 1) for i in range(50)
)
def parse(self, response):
yield scrapy.item.Item()
for url in self.urls:
return Request(url)
But I can use yield Request(url) instead of return... and it scrapes the pages backwards from last page to first.
I would like to understand why return does not work anymore once an item is yielded? Can somebody explain this in a simple way?

You ask why the second code does not work, but I don’t think you fully understand why the first code works :)
The for loop of your first code only loops once.
What is happening is:
self.parse() is called for the URL in self.start_urls.
self.parse() gets the first (and only the first!) URL from self.urls, and returns it, exiting self.parse().
When a response for that first URL arrives, self.parse() gets called again, and this time it returns a request (only 1 request!) for the second URL from self.urls, because the previous call to self.parse() already consumed the first URL from it (self.urls is an iterator).
The last step repeats in a loop, but it is not the for loop that does it.
You can change your original code to this and it will work the same way:
def parse(self, response):
try:
return next(self.urls)
except StopIteration:
pass

Because to call items/requests it should be generator function.
You even cannot use yield and return in the same function with the same "meaning", it will raise SyntaxError: 'return' with argument inside generator.
The return is (almost) equivalent to raising StopIteration. In this topic Return and yield in the same function you can find very detailed explanation, with links specification.

Scrapy get pre-redirect url

I've a crawler running without troubles but i need to get the start_url and not the redirected one.
The problem is i'm using rules to pass parameters to the URL ( like field-keywords=xxxxx ) and finally get the correct url.
The parse function starts getting the item attributes without any troubles but when i want the start URL ( the true one ) it stores the redirected one ...
I've tryed:
response.url
response.request.meta.get('redirect_urls')
Both returns the final url ( the redirected one ) and not the start_url.
Some one know why, or has any clue ?
Thanks in advance.

use a Spider Middleware to keep track of the start url from every response:
from scrapy import Request
class StartRequestsMiddleware(object):
start_urls = {}
def process_start_requests(self, start_requests, spider):
for i, request in enumerate(start_requests):
request.meta.update(start_url=request.url)
yield request
def process_spider_output(self, response, result, spider):
for output in result:
if isinstance(output, Request):
output.meta.update(
start_url=response.meta['start_url'],
)
yield output
keep track of the start_url every response comes from with:
response.meta['start_url']

Have you tried response.request.url? I personally would override the start_requests method adding the original url in the meta, something like:
yield Request(url, meta={'original_request': url})
And then extract it using response.meta['original_request']

Scrapy yield request from one spider to another

I have the following code:
#FirstSpider.py
class FirstSpider(scrapy.Spider):
name = 'first'
start_urls = ['https://www.basesite.com']
next_urls = []
def parse(self, response):
for url in response.css('bunch > of > css > here'):
self.next_urls.append(url.css('more > css > here'))
l = Loader(item=Item(), selector=url.css('more > css'))
l.add_css('add', 'more > css')
...
...
yield l.load_item()
for url in self.next_urls:
new_urls = self.start_urls[0] + url
yield scrapy.Request(new_urls, callback=SecondSpider.parse_url)
#SecondSpider.py
class SecondSpider(scrapy.Spider):
name = 'second'
start_urls = ['https://www.basesite.com']
def parse_url(self):
"""Parse team data."""
return self
# self is a HtmlResponse not a 'response' object
def parse(self, response):
"""Parse all."""
summary = self.parse_url(response)
return summary
#ThirdSpider.py
class ThirdSpider(scrapy.Spider):
# take links from second spider, continue:
I want to be able to pass the url scraped in Spider 1 to Spider 2 (in a different script). I'm curious as to why when I do, the 'response' is a HtmlResponse and not a response object ( When doing something similar to a method in the same class as Spider 1; I don't have this issue )
What am i missing here? How do i just pass the original response(s) to the second spider? ( and from the second onto the third, etc..?)

You could use Redis as shared resource between all spiders https://github.com/rmax/scrapy-redis
Run all N spiders (don't close on idle state), so each of them will be connected to same Redis and waiting tasks(url, request headers) from there;
As the side-effect push task data to Redis from X_spider with specific key (Y_spider name).

What about using inheritance? "parse" function names should be different.
If your first spider inherits from the second, it will be able to set the callback to self.parse_function_spider2

Scrapy Start_request parse

I am writing a scrapy script to search and scrape result from a website. I need to search items from website and parse each url from the search results. I started with Scrapy's start_requests where i'd pass the search query and redirect to another function parse which will retrieve the urls from the search result. Finally i called another function parse_item to parse the results. I'm able to extract the all the search results url but i'm not being able to parse the results ( parse_item is not working). Here is the code:
# -*- coding: utf-8 -*-
from scrapy.http.request import Request
from scrapy.spider import BaseSpider
class xyzspider(BaseSpider):
name = 'dspider'
allowed_domains = ["www.example.com"]
mylist = ['Search item 1','Search item 2']
url = 'https://example.com/search?q='
def start_requests(self):
for i in self.mylist:
i = i.replace(' ','+')
starturl = self.url+ i
yield Request(starturl,self.parse)
def parse(self,response):
itemurl = response.xpath(".//section[contains(#class, 'search-results')]/a/#href").extract()
for j in itemurl:
print j
yield Request(j,self.parse_item)
def parse_item(self,response):
print "hello"
'''rating = response.xpath(".//ul(#class = 'ratings')/li[1]/span[1]/text()").extract()
print rating'''
Could anyone please help me. Thank you.

I was getting a Filtered offsite request error. I changed the allowed domain from allowed_domains = www.xyz.com to
xyz.com and it worked perfectly.

Your code looks good. So you might need to use the Request attribute dont_filter set to True:
yield Request(j,self.parse_item, dont_filter=True)
From the docs:
dont_filter (boolean) – indicates that this request should not be filtered by the scheduler. This is used when you want to perform an identical request multiple times, to ignore the duplicates filter. Use it with care, or you will get into crawling loops. Default to False.
Anyway I recommend you to have a look at the item Pipelines.
Those are used to process scraped items using the command:
yield my_object
Item pipelines are used to post-process everything yielded by the spider.

How to avoid duplication in a crawler

I wrote a crawler using the scrapy framework in python to select some links and meta tags.It then crawls the start urls and write the data in a JSON encoded format onto a file.The problem is that when the crawler is run two or three times with the same start urls the data in the file gets duplicated .To avoid this I used a downloader middleware in scrapy which is this : http://snippets.scrapy.org/snippets/1/
What I did was copy and paste the above code in a file inside my scrapy project and I enabled it in the settings.py file by adding the following line:
SPIDER_MIDDLEWARES = {'a11ypi.removeDuplicates.IgnoreVisitedItems':560}
where "a11ypi.removeDuplicates.IgnoreVisitedItems" is the class path name and finally I went in and modified my items.py file and included the following fields
visit_id = Field()
visit_status = Field()
But this doesn't work and still the crawler produces the same result appending it to the file when run twice
I did the writing to the file in my pipelines.py file as follows:
import json
class AYpiPipeline(object):
def __init__(self):
self.file = open("a11ypi_dict.json","ab+")
# this method is called to process an item after it has been scraped.
def process_item(self, item, spider):
d = {}
i = 0
# Here we are iterating over the scraped items and creating a dictionary of dictionaries.
try:
while i<len(item["foruri"]):
d.setdefault(item["foruri"][i],{}).setdefault(item["rec"][i],{})[item["foruri_id"][i]] = item['thisurl'] + ":" +item["thisid"][i]
i+=1
except IndexError:
print "Index out of range"
json.dump(d,self.file)
return item
And my spider code is as follows:
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from a11ypi.items import AYpiItem
class AYpiSpider(CrawlSpider):
name = "a11y.in"
allowed_domains = ["a11y.in"]
# This is the list of seed URLs to begin crawling with.
start_urls = ["http://www.a11y.in/a11ypi/idea/fire-hi.html"]
# This is the callback method, which is used for scraping specific data
def parse(self,response):
temp = []
hxs = HtmlXPathSelector(response)
item = AYpiItem()
wholeforuri = hxs.select("//#foruri").extract() # XPath to extract the foruri, which contains both the URL and id in foruri
for i in wholeforuri:
temp.append(i.rpartition(":"))
item["foruri"] = [i[0] for i in temp] # This contains the URL in foruri
item["foruri_id"] = [i.split(":")[-1] for i in wholeforuri] # This contains the id in foruri
item['thisurl'] = response.url
item["thisid"] = hxs.select("//#foruri/../#id").extract()
item["rec"] = hxs.select("//#foruri/../#rec").extract()
return item
Kindly suggest what to do.

try to understand why the snippet is written as it is:
if isinstance(x, Request):
if self.FILTER_VISITED in x.meta:
visit_id = self._visited_id(x)
if visit_id in visited_ids:
log.msg("Ignoring already visited: %s" % x.url,
level=log.INFO, spider=spider)
visited = True
Notice in line 2, you actually require a key in in Request.meta called FILTER_VISITED in order for the middleware to drop the request. The reason is well-intended because every single url you have visited will be skipped and you will not have urls to tranverse at all if you do not do so. So, FILTER_VISITED actually allows you to choose what url patterns you want to skip. If you want links extracted with a particular rule skipped, just do
Rule(SgmlLinkExtractor(allow=('url_regex1', 'url_regex2' )), callback='my_callback', process_request = setVisitFilter)
def setVisitFilter(request):
request.meta['filter_visited'] = True
return request
P.S I do not know if it works for 0.14 and above as some of the code has changed for storing spider context in the sqlite db.