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我需要 scrapy 从命令行获取一个参数 (-a FILE_NAME="stuff") 并将其应用于在 pipelines.py 文件中的 CSVWriterPipeLine 中创建的文件。（我使用 pipeline.py 的原因是内置导出器重复数据并重复输出文件中的标头。相同的代码，但在管道中写入修复了它。）

我尝试从 scrapy.utils.project import get_project_settings 中看到

如何从项目管道访问scrapy设置

但我无法从命令行更改文件名。

我也尝试过实现页面上的@avaleske 解决方案，因为它专门解决了这个问题，但我不知道将他谈到的代码放在我的scrapy 文件夹中的什么位置。

帮助？

设置.py：

BOT_NAME = 'internal_links'

SPIDER_MODULES = ['internal_links.spiders']
NEWSPIDER_MODULE = 'internal_links.spiders'
CLOSESPIDER_PAGECOUNT = 100
ITEM_PIPELINES = ['internal_links.pipelines.CsvWriterPipeline']
# Crawl responsibly by identifying yourself (and your website) on the       user-agent
USER_AGENT = 'internal_links (+http://www.mycompany.com)'
FILE_NAME = "mytestfilename"

管道.py：

import csv

class CsvWriterPipeline(object):

    def __init__(self, file_name):
        header = ["URL"]
        self.file_name = file_name
        self.csvwriter = csv.writer(open(self.file_name, 'wb'))
        self.csvwriter.writerow(header)


    def process_item(self, item, internallinkspider):
        # build your row to export, then export the row
        row = [item['url']]
        self.csvwriter.writerow(row)
        return item

蜘蛛.py：

from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.contrib.spiders import CrawlSpider, Rule
from internal_links.items import MyItem



class MySpider(CrawlSpider):
    name = 'internallinkspider'
    allowed_domains = ['angieslist.com']
    start_urls = ['http://www.angieslist.com']

    rules = (Rule(SgmlLinkExtractor(), callback='parse_url', follow=True), )

    def parse_url(self, response):
        item = MyItem()
        item['url'] = response.url

        return item
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1 回答 1

4

您可以使用“设置”概念和-s命令行参数：

scrapy crawl internallinkspider -s FILE_NAME="stuff"

然后，在管道中：

import csv

class CsvWriterPipeline(object):
    @classmethod
    def from_crawler(cls, crawler):
        settings = crawler.settings
        file_name = settings.get("FILE_NAME")
        return cls(file_name)

    def __init__(self, file_name):
        header = ["URL"]
        self.csvwriter = csv.writer(open(file_name, 'wb'))
        self.csvwriter.writerow(header)

    def process_item(self, item, internallinkspider):
        # build your row to export, then export the row
        row = [item['url']]
        self.csvwriter.writerow(row)
        return item
于 2015-07-20T22:13:09.983 回答