Separate output file for every url given in start_urls list of spider in scrapy

问题

I want to create separate output file for every url I have set in start_urls of spider or somehow want to split ouput files start url wise.

Following is the start_urls of my spider

start_urls = ['http://www.dmoz.org/Arts/', 'http://www.dmoz.org/Business/', 'http://www.dmoz.org/Computers/']

I want to create separate output file like

Arts.xml
Business.xml
Computers.xml

I don't know exactly how to do this. I am thinking to achieve this by implementing some thing like following in spider_opened method of item pipeline class,

import re
from scrapy import signals
from scrapy.contrib.exporter import XmlItemExporter

class CleanDataPipeline(object):
    def __init__(self):
        self.cnt = 0
        self.filename = ''

    @classmethod
    def from_crawler(cls, crawler):
        pipeline = cls()
        crawler.signals.connect(pipeline.spider_opened, signals.spider_opened)
        crawler.signals.connect(pipeline.spider_closed, signals.spider_closed)
        return pipeline

    def spider_opened(self, spider):
        referer_url = response.request.headers.get('referer', None)
        if referer_url in spider.start_urls:
            catname = re.search(r'/(.*)$', referer_url, re.I)
            self.filename = catname.group(1)

        file = open('output/' + str(self.cnt) + '_' + self.filename + '.xml', 'w+b')
        self.exporter = XmlItemExporter(file)
        self.exporter.start_exporting()

    def spider_closed(self, spider):
        self.exporter.finish_exporting()
        #file.close()

    def process_item(self, item, spider):
        self.cnt = self.cnt + 1
        self.spider_closed(spider)
        self.spider_opened(spider)
        self.exporter.export_item(item)
        return item

Where I am trying to find the referer url of every scraped item within the start_urls list. If referer url is found in start_urls then file name will be created using that referer url. But problem is how to access response object inside spider_opened() method. If I can access it there, I can create file based on that.

Any help to find a way to perform this? Thanks in advance!

[EDIT]

Solved my problem by changing my pipelines code as followed.

import re
from scrapy import signals
from scrapy.contrib.exporter import XmlItemExporter

class CleanDataPipeline(object):
    def __init__(self):
        self.filename = ''
        self.exporters = {}

    @classmethod
    def from_crawler(cls, crawler):
        pipeline = cls()
        crawler.signals.connect(pipeline.spider_opened, signals.spider_opened)
        crawler.signals.connect(pipeline.spider_closed, signals.spider_closed)
        return pipeline

    def spider_opened(self, spider, fileName = 'default.xml'):
        self.filename = fileName
        file = open('output/' + self.filename, 'w+b')
        exporter = XmlItemExporter(file)
        exporter.start_exporting()
        self.exporters[fileName] = exporter

    def spider_closed(self, spider):
        for exporter in self.exporters.itervalues(): 
            exporter.finish_exporting()

    def process_item(self, item, spider):
        fname = 'default'
        catname = re.search(r'http://www.dmoz.org/(.*?)/', str(item['start_url']), re.I)
        if catname:
            fname = catname.group(1)
        self.curFileName = fname + '.xml'

        if self.filename == 'default.xml':
            if os.path.isfile('output/' + self.filename):
                os.rename('output/' + self.filename, 'output/' + self.curFileName)
            exporter = self.exporters['default.xml']
            del self.exporters['default.xml']
            self.exporters[self.curFileName] = exporter
            self.filename = self.curFileName

        if self.filename != self.curFileName and not self.exporters.get(self.curFileName):
            self.spider_opened(spider, self.curFileName)

        self.exporters[self.curFileName].export_item(item)
        return item

Also Implemented make_requests_from_url in spider to set start_url for every item.

def make_requests_from_url(self, url):
    request = Request(url, dont_filter=True)
    request.meta['start_url'] = url
    return request

回答1:

I'd implement a more explicit approach (not tested):

• configure list of possible categories in settings.py:

  CATEGORIES = ['Arts', 'Business', 'Computers']
  

• define your start_urls based on the setting

  start_urls = ['http://www.dmoz.org/%s' % category for category in settings.CATEGORIES]
  

• add category Field to the Item class

• in the spider's parse method set the category field according to the current response.url, e.g.:

  def parse(self, response):
       ...
       item['category'] = next(category for category in settings.CATEGORIES if category in response.url)
       ...
  

• in the pipeline open up exporters for all categories and choose which exporter to use based on the item['category']:

  def spider_opened(self, spider):
      ...
      self.exporters = {}
      for category in settings.CATEGORIES:
          file = open('output/%s.xml' % category, 'w+b')
          exporter = XmlItemExporter(file)
          exporter.start_exporting()
          self.exporters[category] = exporter
  
  def spider_closed(self, spider):
      for exporter in self.exporters.itervalues(): 
          exporter.finish_exporting()
  
  def process_item(self, item, spider):
      self.exporters[item['category']].export_item(item)
      return item
  

You would probably need to tweak it a bit to make it work but I hope you got the idea - store the category inside the item being processed. Choose a file to export to based on the item category value.

Hope that helps.

回答2:

As long as you don't store it in the item itself, you can't really know the staring url. The following solution should work for you:

• redefine the make_request_from_url to send the starting url with each Request you make. You can store it in meta attribute of your Request. Bypass this starting url with each following Request.

• as soon as you decide to pass the element to pipeline, fill in the starting url for the item from response.meta['start_url']

Hope it helps. Following links may be helpful:

http://doc.scrapy.org/en/latest/topics/spiders.html#scrapy.spider.Spider.make_requests_from_url

http://doc.scrapy.org/en/latest/topics/request-response.html?highlight=meta#passing-additional-data-to-callback-functions

来源：https://stackoverflow.com/questions/23868784/separate-output-file-for-every-url-given-in-start-urls-list-of-spider-in-scrapy