Scrapy feed output contains the expected output several times instead of just once

Question

I've written a spider of which the sole purpose is to extract one number from http://www.funda.nl/koop/amsterdam/, namely, the maximum number of pages from the pager at the bottom (e.g., the number 255 in the example below).

I managed to do this using the LinkExtractor based on the regular expression that URLs of these pages match. The spider is shown below:

import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor
from scrapy.crawler import CrawlerProcess
from Funda.items import MaxPageItem

class FundaMaxPagesSpider(CrawlSpider):
    name = "Funda_max_pages"
    allowed_domains = ["funda.nl"]
    start_urls = ["http://www.funda.nl/koop/amsterdam/"]

    le_maxpage = LinkExtractor(allow=r'%s+p\d+' % start_urls[0])   # Link to a page containing thumbnails of several houses, such as http://www.funda.nl/koop/amsterdam/p10/

    rules = (
    Rule(le_maxpage, callback='get_max_page_number'),
    )

    def get_max_page_number(self, response):
        links = self.le_maxpage.extract_links(response)
        max_page_number = 0                                                 # Initialize the maximum page number
        page_numbers=[]
        for link in links:
            if link.url.count('/') == 6 and link.url.endswith('/'):         # Select only pages with a link depth of 3
                page_number = int(link.url.split("/")[-2].strip('p'))       # For example, get the number 10 out of the string 'http://www.funda.nl/koop/amsterdam/p10/'
                page_numbers.append(page_number)
                # if page_number > max_page_number:
                #     max_page_number = page_number                           # Update the maximum page number if the current value is larger than its previous value
        max_page_number = max(page_numbers)
        print("The maximum page number is %s" % max_page_number)
        yield {'max_page_number': max_page_number}

If I run this with feed output by entering scrapy crawl Funda_max_pages -o funda_max_pages.json at the command line, the resulting JSON file looks like this:

[
{"max_page_number": 257},
{"max_page_number": 257},
{"max_page_number": 257},
{"max_page_number": 257},
{"max_page_number": 257},
{"max_page_number": 257},
{"max_page_number": 257}
]

I find it strange that the dict is outputted 7 times instead of just once. After all, the yield statement is outside of the for loop. Can anyone explain this behavior?

Answer 1

As a workaround, I've written the output to a text file to be used instead of the JSON feed output:

import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor
from scrapy.crawler import CrawlerProcess

class FundaMaxPagesSpider(CrawlSpider):
    name = "Funda_max_pages"
    allowed_domains = ["funda.nl"]
    start_urls = ["http://www.funda.nl/koop/amsterdam/"]

    le_maxpage = LinkExtractor(allow=r'%s+p\d+' % start_urls[0])   # Link to a page containing thumbnails of several houses, such as http://www.funda.nl/koop/amsterdam/p10/

    rules = (
    Rule(le_maxpage, callback='get_max_page_number'),
    )

    def get_max_page_number(self, response):
        links = self.le_maxpage.extract_links(response)
        max_page_number = 0                                                 # Initialize the maximum page number
        for link in links:
            if link.url.count('/') == 6 and link.url.endswith('/'):         # Select only pages with a link depth of 3
                print("The link is %s" % link.url)
                page_number = int(link.url.split("/")[-2].strip('p'))       # For example, get the number 10 out of the string 'http://www.funda.nl/koop/amsterdam/p10/'
                if page_number > max_page_number:
                    max_page_number = page_number                           # Update the maximum page number if the current value is larger than its previous value
        print("The maximum page number is %s" % max_page_number)
        place_name = link.url.split("/")[-3]                                # For example, "amsterdam" in 'http://www.funda.nl/koop/amsterdam/p10/'
        print("The place name is %s" % place_name)
        filename = str(place_name)+"_max_pages.txt"                         # File name with as prefix the place name
        with open(filename,'wb') as f:
            f.write('max_page_number = %s' % max_page_number)               # Write the maximum page number to a text file
        yield {'max_page_number': max_page_number}

process = CrawlerProcess({
    'USER_AGENT': 'Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1)'
})

process.crawl(FundaMaxPagesSpider)
process.start() # the script will block here until the crawling is finished

I've also adapted the spider to run it as a script. The script will generate a text file amsterdam_max_pages.txt with a single line max_page_number: 257.

Answer 2

1. Your spider goes to first start_url.
2. Uses LinkExtractor to extract 7 urls.
3. Downloads every one of those 7 urls and calls get_max_page_number on every one of those.
4. For every url get_max_page_number returns a dictionary.