Separate Output File For Every Url Given In Start_urls List Of Spider In Scrapy

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

Solution 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 categoryField to the Item class

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

  defparse(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']:

  defspider_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
  
  defspider_closed(self, spider):
      for exporter inself.exporters.itervalues(): 
          exporter.finish_exporting()
  
  defprocess_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.

Solution 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

Solution 3:

Here's how I did it for my project without setting the category in item:

Pass the argument from the command line like this:

scrapy crawl reviews_spider -a brand_name=apple

Receive the argument and set to spider args in the my_spider.py

def__init__(self, brand_name, *args, **kwargs):
    self.brand_name = brand_name
    super(ReviewsSpider, self).__init__(*args, **kwargs)

    # i am reading start_urls from an external file depending on the passed argumentwithopen('make_urls.json') as f:
        self.start_urls = json.loads(f.read())[self.brand_name]

In pipelines.py:

classReviewSummaryItemPipeline(object):
    @classmethoddeffrom_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

    defspider_opened(self, spider):
        # change the output file name based on argument
        self.file = open(f'reviews_summary_{spider.brand_name}.csv', 'w+b')
        self.exporter = CsvItemExporter(self.file)
        self.exporter.start_exporting()

    defspider_closed(self, spider):
        self.exporter.finish_exporting()
        self.file.close()

    defprocess_item(self, item, spider):
        self.exporter.export_item(item)
        return item