Question

我正在尝试从scrapy中的url列表中为每个抓取的url生成一个csv文件。我明白我会修改pipeline.py，但到目前为止我所有的尝试都失败了。我不明白如何将被抓取的URL传递给管道并将其用作输出的名称并相应地拆分输出。

任何帮助？

由于

这里是蜘蛛和管道

from scrapy import signals
from scrapy.contrib.exporter import CsvItemExporter

class VappPipeline(object):
    def __init__(self):
        self.files = {}

@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):
        file = open('results/%s.csv' % spider.name, 'w+b')
        self.files[spider] = file
        self.exporter = CsvItemExporter(file)
        self.exporter.fields_to_export = ['item']
        self.exporter.start_exporting()

    def spider_closed(self, spider):
        self.exporter.finish_exporting()
        file = self.files.pop(spider)
        file.close()

    def process_item(self, item, spider):
        self.exporter.export_item(item)
        return item

管道：

{{1}}

Answer 1

我认为当你的抓取完成而不是按项目时，你应该批量处理所有这些事情作为后处理步骤，但这里有关于你如何做你想做的草案：

from scrapy import Spider
from scrapy.selector import Selector 
from vApp.items import fItem


class VappSpider(Spider):

    name = "vApp"
    allowed_domains = ["google.co.uk"]
    start_urls = [l.strip() for l in open('data/listOfUrls.txt').readlines()]


def parse(self, response):

    trs = Selector(response).xpath('//[@id="incdiv"]/table/tbody/tr')
    for tr in trs:
        item = fItem()

        try:
            item['item'] = tr.xpath('td/text()').extract()[0]
        except IndexError:
            item['item'] = 'null'
        item['url'] = response.url
        yield item


from scrapy import signals
from scrapy.contrib.exporter import CsvItemExporter
from urlparse import urlparse

class VappPipeline(object):
    def __init__(self):
        self.files = {}
        self.exporter = {}

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

    def process_item(self, item, spider):
        url = item['url']
        parsed_uri = urlparse(url)
        domain = parsed_uri.netloc
        if domain not in self.exporter:
            file = open('results/%s.csv' % domain, 'w+b')
            self.files[domain] = file
            self.exporter[domain] = CsvItemExporter(file)
            self.exporter[domain].fields_to_export = ['item']
            self.exporter[domain].start_exporting()

        assert domain in self.exporter

        self.exporter[domain].export_item(item)

        return item

    def spider_closed(self, spider):
        for domain, exporter in self.exporter.iteritems():
            exporter.finish_exporting()
            self.files[domain].close()

如何在scrapy中分割url列表中的输出

1 个答案: