在rq队列中放置类方法时出现Python RQ错误

Question

我需要将类方法放到RQ队列中。但它给出了一个错误

这里是worker.py

import os

import redis
from rq import Worker, Queue, Connection

listen = ['high', 'default', 'low']

redis_url = os.getenv('REDISTOGO_URL', 'redis://localhost:6379')

conn = redis.from_url(redis_url)

if __name__ == '__main__':
    with Connection(conn):
        worker = Worker(map(Queue, listen))
        worker.work()

这里是单独文件中的类

import requests
import re

class Robots():
    def __init__(self, url):
        self.url = url


    def get_url(self):
        if self.url.endswith('/'):
            return self.url + "robots.txt" 
        else:
            return self.url + "/robots.txt"


    def get_status_code(self):
        return requests.get(self.get_url()).status_code

这里是app.py

import time
from rq import Queue
from worker import conn
from rob import Robots
url = 'http://heroku.com'
q = Queue(connection=conn)

task = q.enqueue(Robots(url).get_status_code())
print (task.result)
time.sleep(4)
print (task.result)

当我运行代码时，它会出现一个错误。

File "app.py", line 8, in <module>
    task = q.enqueue(Robots(url).get_status_code())
  File "/home/atom/Desktop/Python/Tests/herokurq/lib/python3.8/site-packages/rq/queue.py", line 381, in enqueue
    depends_on, job_id, at_front, meta, args, kwargs) = Queue.parse_args(f, *args, **kwargs)
  File "/home/atom/Desktop/Python/Tests/herokurq/lib/python3.8/site-packages/rq/queue.py", line 353, in parse_args
    if not isinstance(f, string_types) and f.__module__ == '__main__':

当我把简单的函数而不是类方法放在--一切都很好的时候。但我得把这个方法。它给出了一个错误。

Answer 1

我发现了一个问题，我应该使用q.enqueue=.get_status_code代替q.enqueue(Robots(url).get_status_code())