How to make my scrapy read the file which in the same directory?

Question

The target file urls.txt contains all the url to be downloaded.

├─spiders
│  │  stockInfo.py
│  │  urls.txt
│  │  __init__.py

stockInfo.py is my scrapy file.

import scrapy
import os
import re

class QuotesSpider(scrapy.Spider):
    name = "stockInfo"
    projectFile = r"d:/toturial/toturial/spiders/urls.txt"
    with open(projectFile,"r") as f:
        urls = f.readlines()
    start_urls = [url.strip() for url in urls]

    def parse(self, response):
        pass

I have tested that the above stockInfo.py can run successfully in my local pc end with command:

scrapy crawl  stockInfo

Now i deploy the project into remote end scrapy hub with

pip install shub
shub login
API key: xxxxxxxxxxxxxxxxx
shub deploy 380020

It run into trouble:

IOError: [Errno 2] No such file or directory: 'd:/toturial/toturial/spiders/urls.txt'

How to fix it when to deploy my scrapy into the hub ? It is useful to rewrite

projectFile = r"d:/toturial/toturial/spiders/urls.txt"

as

projectFile = "./urls.txt"

when to run it in my local pc end.

Strangely, it is no use to rewrite

projectFile = r"d:/toturial/toturial/spiders/urls.txt"

as

projectFile = "./urls.txt"

when to run it in remote end scrapy hub .

Answer 1

1.add new directory and move urls.txt in it.
To add a new directory resources ,and save urls.txt in it.
My new directory tree is as below.

tutorial
├─tutorial
│  ├─resources
|     |--urls.txt
│  ├─spiders
|     |--stockInfo.py

2.rewrite the setup.py as below.

from setuptools import setup, find_packages

setup(
    name='tutorial',
    version='1.0',
    packages=find_packages(),
    package_data={
        'tutorial': ['resources/*.txt']
    },
    entry_points={
        'scrapy': ['settings = tutorial.settings']
    },
    zip_safe=False,
)

3.rewrite stockInfo.py as below.

import scrapy
import os 
import re
import pkgutil
class QuotesSpider(scrapy.Spider):
    name = "stockInfo"
    data = pkgutil.get_data("tutorial", "resources/urls.txt")
    data = data.decode()
    start_urls = data.split("\r\n")

    def parse(self, response):
        pass