How to Execute Code Once at Django Startup?

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Release: 2024-11-01 20:13:02
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How to Execute Code Once at Django Startup?

Getting Single-Execution Code to Run at Django Startup

When developing Django applications, there may be instances where you need to execute specific code only once during startup, typically for initialization purposes. However, using a Middleware class for this goal can result in multiple executions, as the Django middleware system calls the __init__() method of each middleware at every request.

Django's Solution for Django < 1.7

To address this issue in Django versions prior to 1.7, one effective approach is to place your startup code within the __init__.py file of one of your INSTALLED_APPS. For example, in the myapp application, you can include the following code in myapp/__init__.py:

<code class="python">def startup():
    pass # Place your startup code here

startup()</code>
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This approach ensures that the startup() function is executed during Django startup and is not invoked again upon subsequent requests, providing the desired one-time execution.

Improved Solution for Django 1.7 and Later

Django version 1.7 introduced a more elegant solution for this requirement. By implementing an AppConfig class with a ready() method within the apps.py file of your application, you can specify code to be executed when the application is marked as ready. Here's an example using the myapp application:

<code class="python"># myapp/apps.py
from django.apps import AppConfig
class MyAppConfig(AppConfig):
    name = 'myapp'
    verbose_name = "My Application"
    def ready(self):
        pass # Place your startup code here</code>
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This approach allows you to define and execute your startup code more naturally and is the recommended solution for Django 1.7 and later.

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