将类方法放在RQ队列上时出现Python rq错误

问题描述 投票:0回答:1

我需要在RQ队列上放置一个类方法。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,它给出一个错误。

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__':

当我使用简单函数而不是类方法时,一切正常。但是我需要提出方法。它给出了一个错误。

python python-rq
1个回答
0
投票
我发现了问题我应该使用task = q.enqueue(Robots(url).get_status_code)而不是任务= q.enqueue(Robots(url).get_status_code())
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