Question

我需要scrapy从命令行获取参数（-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在输出文件中重复标题。相同的代码，但在管道中写入修复它。）

我尝试从scrapy.utils.project导入get_project_settings，如

中所示

How to access scrapy settings from item Pipeline

但我无法从命令行更改文件名。

我也尝试在页面上实现@ avaleske的解决方案，因为它专门解决了这个问题，但我不知道在scrapy文件夹中将他所说的代码放在哪里。

帮助？

settings.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"

pipelines.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

spider.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

Answer 1

您可以使用＆＃34;设置＆＃34;概念和scrapy crawl internallinkspider -s FILE_NAME="stuff"命令行参数：

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

然后，在管道中：

body {
        top: 5rem;
        right: 5rem;
        bottom: 5rem;
        left: 5rem;
        position: absolute;
}

将file_name参数传递给scrapy中用于csv导出的管道

1 个答案: