Keluaran umpan tergores berisi keluaran yang diharapkan beberapa kali, bukan hanya sekali

1. Laba-laba Anda menuju ke start_url pertama.
2. Menggunakan LinkExtractor untuk mengekstrak 7 url.
3. Unduh masing-masing dari 7 url tersebut dan panggil get_max_page_number pada setiap url tersebut.
4. Untuk setiap url get_max_page_number mengembalikan kamus.

Sebagai solusinya, saya telah menulis output ke file teks untuk digunakan sebagai pengganti output feed JSON:

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

Saya juga mengadaptasi laba-laba untuk menjalankannya sebagai skrip. Script akan menghasilkan file teks amsterdam_max_pages.txt dengan satu baris max_page_number: 257.