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Celery任务创建了一个新的模型对象，但是django signal不会将其拾取

[英]Celery task creates a new model object, but django signals doesn't pick it up

原文 2019-03-01 06:56:17 4 1 python/ django-models/ celery

I've read around and see this has to do with concurrency, that celery can't edit the app variables because it's in a different thread from a python app. 我已经阅读了一下，发现这与并发有关，因为celery与python应用程序位于不同的线程中，因此它们无法编辑应用程序变量。 So instead I'm modifying the database by creating an object in the hope that django signal would've picked it up, but it's not. 因此，相反，我通过创建一个对象来修改数据库，希望django signal会选择它，但事实并非如此。 I'm doing this since I need to know if an update has occured in the back end so the SPA front end can fetch it automatically. 我这样做是因为我需要知道后端是否发生了更新，以便SPA前端可以自动获取它。 Any tips? 有小费吗？

1 个解决方案

Django signals only work in-process. Django信号仅在进程中工作。 So making a model change in celery will trigger the signal to occur in celery. 因此，在芹菜中进行模型更改会触发信号在芹菜中发生。 If you need to notify a single page app either (1) use a websocket or (2) pass the SPA the task_id so it can poll for when the task is complete and then take the necessary action. 如果您需要通知单个页面应用程序，则（1）使用websocket或（2）通过SPA将task_id传递给SPA，以便它可以在任务完成时进行轮询，然后采取必要的措施。

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