Development server with debug mode in Flask

Introduction

Debug mode helps you find and fix errors quickly while you build your web app. The development server restarts automatically when you change your code.

When you are building a new Flask web app and want to see changes immediately.

When you want detailed error messages to understand what went wrong.

When you want the server to reload automatically after you edit your code.

When testing new features before making your app live.

When you want to avoid restarting the server manually after every change.

Syntax

Flask

app.run(debug=True)

Use debug=True inside app.run() to turn on debug mode.

Do not use debug mode in production because it can show sensitive information.

Examples

This example starts the Flask app with debug mode on. The server reloads on code changes and shows detailed errors.

Flask

from flask import Flask
app = Flask(__name__)

@app.route('/')
def home():
    return 'Hello, Flask!'

if __name__ == '__main__':
    app.run(debug=True)

This runs the server without debug mode. Changes won't reload automatically and errors are less detailed.

Flask

app.run(debug=False)

Sample Program

This program creates a simple Flask app that shows a welcome message. Running it with debug=True means you get live reloads and helpful error messages.

Flask

from flask import Flask
app = Flask(__name__)

@app.route('/')
def home():
    return 'Welcome to debug mode!'

if __name__ == '__main__':
    app.run(debug=True)

Important Notes

Debug mode shows detailed error pages with helpful info.

Auto-reloading saves time by restarting the server when you save files.

Never use debug mode on a public or production server for security reasons.

Summary

Debug mode helps you develop faster by showing errors and reloading automatically.

Enable it by passing debug=True to app.run().

Turn it off before deploying your app to keep it safe.

Practice

1. What does setting debug=True in app.run() do in a Flask app?

easy

A. It hides all error messages from the user.

B. It disables the server from running.

C. It makes the app run faster in production.

D. It shows detailed error messages and reloads the server on code changes.

Solution

1. Step 1: Understand debug mode purpose

   Debug mode helps developers by showing detailed error messages and automatically restarting the server when code changes.

2. Step 2: Check what debug=True does in app.run()

   Passing debug=True enables this helpful behavior during development.

3. Final Answer:

   It shows detailed error messages and reloads the server on code changes. -> Option D

4. Quick Check:

   Debug mode = detailed errors + auto reload [OK]

Hint: Debug mode shows errors and reloads automatically [OK]

Common Mistakes:

• Thinking debug mode speeds up production
• Believing debug mode hides errors
• Confusing debug mode with disabling the server

2. Which of the following is the correct way to start a Flask development server with debug mode enabled?

easy

A. app.run(debug=True)

B. app.run(debug=1)

C. app.run(mode='debug')

D. app.run(enable_debug=True)

Solution

1. Step 1: Recall Flask run method parameters

   The app.run() method accepts a debug parameter as a boolean to enable debug mode.

2. Step 2: Identify correct syntax

   Using app.run(debug=True) is the standard and correct way to enable debug mode.

3. Final Answer:

   app.run(debug=True) -> Option A

4. Quick Check:

   Correct debug syntax = debug=True [OK]

Hint: Use debug=True exactly in app.run() [OK]

Common Mistakes:

• Using incorrect parameter names like mode or enable_debug
• Passing debug as a string or number instead of boolean
• Confusing debug with other run parameters

3. What will happen when you run this Flask code?

from flask import Flask
app = Flask(__name__)

@app.route('/')
def home():
    return 'Hello!'

if __name__ == '__main__':
    app.run(debug=True)

medium

A. The server will not start because the route is missing a template.

B. The server starts but debug mode is off, so no error details are shown.

C. The server starts with debug mode, showing errors and auto-reloading on code changes.

D. The code will raise a syntax error because debug=True is invalid here.

Solution

1. Step 1: Analyze the code structure

   The code defines a Flask app with a simple route returning 'Hello!'. The server is started with debug=True.

2. Step 2: Understand effect of debug=True

   Passing debug=True enables debug mode, so errors show detailed info and the server reloads on code changes.

3. Final Answer:

   The server starts with debug mode, showing errors and auto-reloading on code changes. -> Option C

4. Quick Check:

   app.run(debug=True) = debug mode active [OK]

Hint: debug=True means errors show and server reloads [OK]

Common Mistakes:

• Thinking debug=True causes syntax errors
• Assuming debug mode is off by default
• Believing routes need templates to start server

4. You wrote this code but the Flask server does not reload on code changes:

from flask import Flask
app = Flask(__name__)

@app.route('/')
def index():
    return 'Hi'

if __name__ == '__main__':
    app.run()

What is the likely fix?

medium

A. Change app = Flask(__name__) to app = Flask().

B. Add debug=True to app.run() to enable auto-reload.

C. Add reload=True parameter to app.run().

D. Remove the if __name__ == '__main__' guard.

Solution

1. Step 1: Identify missing debug mode

   The code calls app.run() without debug=True, so debug mode and auto-reload are off.

2. Step 2: Enable debug mode for auto-reload

   Adding debug=True to app.run() activates debug mode, enabling error messages and auto-reload on code changes.

3. Final Answer:

   Add debug=True to app.run() to enable auto-reload. -> Option B

4. Quick Check:

   Missing debug=True disables auto-reload [OK]

Hint: Add debug=True to app.run() for auto-reload [OK]

Common Mistakes:

• Trying to use reload=True which is invalid
• Removing the main guard which is unrelated
• Changing Flask app initialization unnecessarily

5. You want to run your Flask app with debug mode but only on your local machine, not in production. Which code snippet correctly achieves this?

hard

A. if __name__ == '__main__':
app.run(debug=True)

B. app.run(debug=True) # runs debug mode always

C. if __name__ == '__main__':
app.run() # debug mode off

D. app.run() # debug mode on by default

Solution

1. Step 1: Understand the main guard usage

   The if __name__ == '__main__' block ensures the server runs only when the script is executed directly, typically local development.

2. Step 2: Enable debug mode only inside main guard

   Placing app.run(debug=True) inside this block enables debug mode only locally, avoiding debug mode in production.

3. Step 3: Check other options

   Running app.run(debug=True) outside the guard runs debug mode always, which is unsafe for production.

4. Final Answer:

   if __name__ == '__main__': app.run(debug=True) -> Option A

5. Quick Check:

   Main guard + debug=True = safe local debug [OK]

Hint: Use main guard with debug=True for local only [OK]

Common Mistakes:

• Running debug mode outside main guard risks production
• Assuming debug mode is off by default
• Not using main guard and causing unwanted server start