[FIXED] Use Scrapyd with arguments

Issue

I'm using Scrapyd to use scrapy as webservice.

I would like to use the curl command with parameters like this :

curl http://myip:6800/schedule.json -d project=default -d spider=myspider -d domain=www.google.fr

But I don't know how to get the parameter domain in the Crawler.

import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor

class MyItem(Item):
    url = Field()

class HttpbinSpider(CrawlSpider):

    name = "expired"
    start_urls = [domain]

I need to pass sometimes one domain or multiples in arguments.

Thank's !

Solution

It's not possible due to a missing feature in scrapy.

Users typically workaround this by serializing the arguments to curl and then unserializing in the spider's init().

curl http://myip:6800/schedule.json -d project=default -d spider=myspider -d domains='["www1.example.com", "www2.example.com"]'

Code :

class MySpider(Spider):
    def __init__(self, domains=None):
        domains = json.loads(domains)
        # do something with domains...

Answered By - Pixel