6

我有这样的问题：

如何在scrapy中根据url过滤重复请求

因此，我不希望网站被多次抓取。我调整了中间件并编写了一个打印语句来测试它是否正确分类了已经看到的网站。确实如此。

尽管如此，解析似乎被执行了多次，因为我收到的 json-File 包含双重条目。

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from scrapy.item import Item

from crawlspider.items import KickstarterItem

from HTMLParser import HTMLParser

### code for stripping off HTML tags:
class MLStripper(HTMLParser):
    def __init__(self):
        self.reset()
        self.fed = []
    def handle_data(self, d):
        self.fed.append(d)
    def get_data(self):
        return str(''.join(self.fed))

def strip_tags(html):
    s = MLStripper()
    s.feed(html)
    return s.get_data()
###

items = []

class MySpider(CrawlSpider):
    name = 'kickstarter'
    allowed_domains = ['kickstarter.com']
    start_urls = ['http://www.kickstarter.com']

    rules = (
        # Extract links matching 'category.php' (but not matching 'subsection.php')
        # and follow links from them (since no callback means follow=True by default).
        Rule(SgmlLinkExtractor(allow=('discover/categories/comics', ))),

        # Extract links matching 'item.php' and parse them with the spider's method parse_item
        Rule(SgmlLinkExtractor(allow=('projects/', )), callback='parse_item'),
    )

    def parse_item(self, response):
        self.log('Hi, this is an item page! %s' % response.url)

        hxs = HtmlXPathSelector(response)
        item = KickstarterItem()

        item['date'] = hxs.select('//*[@id="about"]/div[2]/ul/li[1]/text()').extract()
        item['projname'] = hxs.select('//*[@id="title"]/a').extract()
        item['projname'] = strip_tags(str(item['projname']))

        item['projauthor'] = hxs.select('//*[@id="name"]')
        item['projauthor'] = item['projauthor'].select('string()').extract()[0]

        item['backers'] = hxs.select('//*[@id="backers_count"]/data').extract()
        item['backers'] = strip_tags(str(item['backers']))

        item['collmoney'] = hxs.select('//*[@id="pledged"]/data').extract()
        item['collmoney'] = strip_tags(str(item['collmoney']))

        item['goalmoney'] = hxs.select('//*[@id="stats"]/h5[2]/text()').extract()
        items.append(item)
        return items

我的 items.py 看起来像这样：

# Define here the models for your scraped items
#
# See documentation in:
# http://doc.scrapy.org/topics/items.html

from scrapy.item import Item, Field

class KickstarterItem(Item):
    # define the fields for your item here like:
    date = Field()
    projname = Field()
    projauthor = Field()
    backers = Field()
    collmoney = Field()
    goalmoney = Field()
    pass

我的中间件如下所示：

import os

from scrapy.dupefilter import RFPDupeFilter
from scrapy.utils.request import request_fingerprint

class CustomFilter(RFPDupeFilter):
def __getid(self, url):
    mm = url.split("/")[4] #extracts project-id (is a number) from project-URL
    print "_____________", mm
    return mm

def request_seen(self, request):
    fp = self.__getid(request.url)
    self.fingerprints.add(fp)
    if fp in self.fingerprints and fp.isdigit(): # .isdigit() checks wether fp comes from a project ID
        print "______fp is a number (therefore a project-id) and has been encountered before______"
        return True
    if self.file:
        self.file.write(fp + os.linesep)

我将此行添加到 settings.py：

DUPEFILTER_CLASS = 'crawlspider.duplicate_filter.CustomFilter'

我使用“scrapy crawl kickstarter -o items.json -t json”调用脚本。然后我从中间件代码中看到了正确的打印语句。关于为什么 json 包含包含相同数据的多个条目的任何评论？

4

1 回答 1

1

所以现在这些是删除重复项的三个修改：

我将此添加到settings.py： ITEM_PIPELINES = ['crawlspider.pipelines.DuplicatesPipeline',]

让 scrapy 知道我在 pipelines.py 中添加了一个 DuplicatesPipeline 函数：

from scrapy import signals
from scrapy.exceptions import DropItem

class DuplicatesPipeline(object):

    def __init__(self):
        self.ids_seen = set()

    def process_item(self, item, spider):
        if item['projname'] in self.ids_seen:
            raise DropItem("Duplicate item found: %s" % item)
        else:
            self.ids_seen.add(item['projname'])
            return item

你不需要调整蜘蛛，也不要使用我之前发布的 dupefilter/middleware 东西。

但是我觉得我的解决方案不会减少通信，因为必须先创建 Item-object，然后才能对其进行评估并可能删除它。但我对此没意见。

（提问者找到的解决方案，移到答案中）

于 2014-08-02T17:31:18.343 回答