芹菜:如何限制队列中的任务数量并在满时停止喂食? [英] Celery: how to limit number of tasks in queue and stop feeding when full?

问题描述

我对 Celery 很陌生，这是我的问题:

I am very new to Celery and here is the question I have:

假设我有一个脚本，它应该不断地从数据库中获取新数据并使用 Celery 将其发送给工作人员.

Suppose I have a script that is constantly supposed to fetch new data from DB and send it to workers using Celery.

tasks.py

# Celery Task
from celery import Celery

app = Celery('tasks', broker='amqp://guest@localhost//')

@app.task
def process_data(x):
    # Do something with x
    pass

fetch_db.py

fetch_db.py

# Fetch new data from DB and dispatch to workers.
from tasks import process_data

while True:
    # Run DB query here to fetch new data from DB fetched_data

    process_data.delay(fetched_data)

    sleep(30);

我担心的是:每 30 秒获取一次数据.据我所知，process_data() 函数可能需要更长的时间，并且取决于工作人员的数量(特别是如果太少)，队列可能会受到限制.

Here is my concern: the data is being fetched every 30 seconds. process_data() function could take much longer and depending on the amount of workers (especially if too few) the queue might get throttled as I understand.

1. 我无法增加工人数量.
2. 我可以修改代码以避免在队列满时喂食.

问题是如何设置队列大小以及如何知道队列已满?一般情况下，如何处理这种情况?

The question is how do I set queue size and how do I know it is full? In general, how to deal with this situation?

推荐答案

可以设置rabbitmq x-max-length 在队列中使用 海带

示例:

import time
from celery import Celery
from kombu import Queue, Exchange

class Config(object):
    BROKER_URL = "amqp://guest@localhost//"

    CELERY_QUEUES = (
        Queue(
            'important',
            exchange=Exchange('important'),
            routing_key="important",
            queue_arguments={'x-max-length': 10}
        ),
    )

app = Celery('tasks')
app.config_from_object(Config)


@app.task(queue='important')
def process_data(x):
    pass

或使用政策

rabbitmqctl set_policy Ten "^one-meg$" '{"max-length-bytes":1000000}' --apply-to queues

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