在scrapy的start_urls列表中给出的每个url的单独输出文件

Question

我想为我在蜘蛛的start_urls中设置的每个url创建单独的输出文件,或者以某种方式想要分割输出文件启动URL明智.

以下是我的蜘蛛的start_urls

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

我想创建单独的输出文件

Arts.xml
Business.xml
Computers.xml

我不知道该怎么做.我想通过在项目管道类的spider_opened方法中实现一些类似的东西来实现这一点,

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

我试图找到start_urls列表中每个被删除项目的引用URL.如果在start_urls中找到referer url,则将使用该referer url创建文件名.但问题是如何在spider_opened()方法中访问响应对象.如果我可以在那里访问它,我可以根据它创建文件.

有什么帮助找到一种方法来执行此操作？提前致谢!

[编辑]

通过更改我的管道代码解决了我的问题.

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

也在make_requests_from_urlspider中实现为每个项设置start_url.

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

Answer 1

我实现了一种更明确的方法(未经测试):

• 配置可能的类别列表settings.py:

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

• start_urls根据设置定义您的

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

• 添加category Field到Item班级

• 在蜘蛛的解析方法中category根据当前设置字段response.url,例如:

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

• 在管道中打开所有类别的出口商,并根据以下内容选择要使用的出口商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
  

您可能需要稍微调整一下以使其工作,但我希望您有这个想法 - 将类别存储在item正在处理的内部.根据项目类别值选择要导出的文件.

希望有所帮助.