【发布时间】:2016-08-20 13:06:20
【问题描述】:
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。我明白我将修改 pipeline.py,但是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。
有什么帮助吗?
谢谢
这里是蜘蛛和管道
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'
yield item
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
【问题讨论】: