CODEWITHSUNDEEP
pythonscrapy
Raheel
Raheel

Reputation: 9024

Cannot Access Parent instance variable in child

I am working on scrapy framework.

I have some common properties which i want in all of my spiders so i made a BaseSpider.

BaseSpider

import scrapy
from src.LoggerFactory import get_logger
import ConfigParser
from redis import Redis


class BaseSpider(scrapy.Spider):
    logger = get_logger()

    def __init__(self, *args, **kwargs):
        super(scrapy.Spider, self).__init__(*args, **kwargs)
        config = ConfigParser.RawConfigParser()
        config.read('../../config.cfg')
        self.config = config
        self.redis = Redis(host=config.get('redis', 'host'), port=config.get('redis', 'port'))

    def parse(self, response):
        pass

And my EbaySpider is as follows

EbaySpider

import scrapy
import json
from scrapper.items import Product
from BaseSpider import BaseSpider


class EbaySpider(BaseSpider):
    name = "ebay"
    allowed_domains = ["ebay.com"]

    def __init__(self, *args, **kwargs):
        super(BaseSpider, self).__init__(*args, **kwargs)
        print self.redis # Throws AttributeError: 'EbaySpider' object has no attribute 'redis'
        exit()

Strangest part is i can still access scrapy.Spider properts in my EbaySpider although it is not inherited from scrapy.Spider.

Also, if there is any scrapy way to extend Spiders please suggest as i wasn't able to find that in their documentations.

Upvotes: 0

Views: 78

Answers (1)

Granitosaurus
Granitosaurus

Reputation: 21436

Your super usage is wrong.

a typical superclass call looks like this:

class C(B):
    def method(self, arg):
        super(C, self).method(arg)

As per official 2.7 docs

Where's your looks like:

class C(B):
    def method(self, arg):
        super(B, self).method(arg)
              ^

Upvotes: 4

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