Python：如何通过RQ获得工作结果

我有这样的代码

from rq import Queue, use_connection
from task import add
from redis import Redis
import time
import logging

# use redis by default
# create work queue
redis_conn = Redis()
q = Queue(connection=redis_conn) 

#notice: cann't run a task function in __main__ module
#because rq save module and function name in redis
#when rqworker running, __main__ is another module
# enqueue tasks,function enqueue returns the job instance
job = q.enqueue(add, 3, 9)
job = q.enqueue(add, 4, 9)
job = q.enqueue(add, 5, 9)
job = q.enqueue(add, 6, 9)

time.sleep(3)
#get the job result by job.result
logging.warn("result is %s", job.result)

当我运行应用程序时，我得到这样的输出

WARNING:root:result is 15

如何获得上一个结果？

我只是想办法。

def recommend(request):
    user_id = int(request.GET.get('user_id','287'))
    job = django_rq.enqueue(recommended,user_id) # Enqueue Job
    logger.debug(job.key)
    response_data = {'status':'processing', 'job': job.key}
    return HttpResponse(json.dumps(response_data), content_type="application/json")

def get_result(request):
    job_id = request.GET.get('job_id')
    redis_conn = django_rq.get_connection()
    job_id=job_id.split(':')[2] 
    logger.info(job_id)
    job = Job.fetch(job_id,redis_conn) #fetch Job from redis
    if job.is_finished:
        ret = job.return_value
    elif job.is_queued:
        ret = {'status':'in-queue'}
    elif job.is_started:
        ret = {'status':'waiting'}
    elif job.is_failed:
        ret = {'status': 'failed'}
    return HttpResponse(json.dumps(ret), content_type="application/json")

这是我使用django_rq库的Django项目中的一些代码。只要记住Job.key看起来像这样的“ job：id：job_key”。除以：并获得第二个元素