在scrapy的spider的start_urls列表中给定的每个URL单独的输出文件

FMoridhara 发表于 Dev

Fmoridhara

我想为我在spider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，以明智的方式启动url。

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，如

Arts.xml
Business.xml
Computers.xml

我不知道该怎么做。我正在考虑通过在item管道类的spider_opened方法中实现以下操作来实现这一点，

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

我在哪里尝试在start_urls列表中找到每个已抓取项目的引用网址。如果在start_urls中找到引用网址，则将使用该引用网址创建文件名。但是问题是如何在spider_opened（）方法内访问响应对象。如果可以在那里访问它，则可以基于该文件创建文件。

有帮助找到执行此操作的方法吗？提前致谢！

[编辑]

通过更改管道代码如下解决了我的问题。

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

同样make_requests_from_url在Spider中实现，可以为每个项目设置start_url。

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

我将采用更明确的方法（未经测试）：

在以下位置配置可能类别的列表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类

在Spider的parse方法中，category根据current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，然后根据以下条件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