👀 Views: 65 💬 Answers: 1 📅 Created: 2025-06-21

python flask dictionary Python

I'm learning this framework and I've been working on this all day and I'm sure I'm missing something obvious here, but I'm developing a Flask application and using a dictionary to store configuration settings for different environments. I've set up a dictionary like this: ```python config = { 'development': { 'DEBUG': True, 'DATABASE_URI': 'sqlite:///dev.db' }, 'production': { 'DEBUG': False, 'DATABASE_URI': 'mysql://user:pass@localhost/prod' } } ``` I then try to access settings based on the environment variable like this: ```python import os current_env = os.getenv('FLASK_ENV', 'development') settings = config[current_env] ``` However, I'm getting a `KeyError` when the `FLASK_ENV` variable is set to 'staging', which I haven't defined in the dictionary. The behavior message is: ``` KeyError: 'staging' ``` I've tried wrapping the access in a `try-except` block, but I want to ensure that if an undefined environment is accessed, it defaults to the 'development' settings instead. Here's what I've attempted: ```python try: settings = config[current_env] except KeyError: settings = config['development'] ``` This works, but it feels like there should be a more elegant way to handle this. Is there a cleaner approach to ensure I get a default value if a key doesn't exist without resorting to exception handling? Also, is there a best practice for managing configuration settings in Flask applications? I appreciate any guidance on this scenario! I'm working on a service that needs to handle this. My team is using Python for this service. Am I missing something obvious?