Custom status codes in Flask - Mini Project: Build & Apply

Custom status codes in Flask

📖 Scenario: You are building a simple web API using Flask. You want to send custom HTTP status codes with your responses to indicate different situations clearly.

🎯 Goal: Create a Flask app that returns a JSON message with a custom status code when a specific route is accessed.

📋 What You'll Learn

Create a Flask app instance named app

Define a route /status that returns a JSON response

Use a custom status code 299 with the response

Return a JSON dictionary with a key message and value "Custom status code response"

💡 Why This Matters

🌍 Real World

Custom status codes help APIs communicate specific states or errors beyond standard HTTP codes, improving client-server communication.

💼 Career

Understanding how to send custom status codes is useful for backend developers building APIs that need precise control over response statuses.

1

Set up the Flask app

Import Flask from flask and create a Flask app instance called app.

Flask

from flask import Flask

# Create the Flask app instance called app
# Your code here

Hint

Use app = Flask(__name__) to create the app instance.

2

Create a route for /status

Define a route /status using @app.route('/status') and create a function called status that will handle requests to this route.

Flask

from flask import Flask

app = Flask(__name__)

# Define a route /status with a function named status
# Your code here

Hint

Use the @app.route decorator to define the route and a function named status.

3

Return JSON with custom status code

Inside the status function, import jsonify from flask and return a JSON response with {"message": "Custom status code response"} and a custom status code 299.

Flask

from flask import Flask

app = Flask(__name__)

@app.route('/status')
def status():
    # Import jsonify and return JSON with custom status code 299
    # Your code here

Hint

Use return jsonify({...}), 299 to send JSON with a custom status code.

4

Add the app run block

Add the standard Flask app run block: if __name__ == '__main__': and inside it call app.run() to start the server.

Flask

from flask import Flask, jsonify

app = Flask(__name__)

@app.route('/status')
def status():
    return jsonify({"message": "Custom status code response"}), 299

# Add the app run block below
# Your code here

Hint

This block runs the Flask app when the script is executed directly.

Practice

1. What is the correct way to return a custom status code 450 from a Flask route?

easy

A. return "Error", 450

B. return 450, "Error"

C. return Response(status=450)

D. return make_response(450, "Error")

Solution

1. Step 1: Understand Flask return tuple format

   Flask routes return a tuple with (response_body, status_code) to set custom status codes.

2. Step 2: Match the tuple with custom code 450

   Returning "Error", 450 matches the format (body, status_code) correctly.

3. Final Answer:

   return "Error", 450 -> Option A

4. Quick Check:

   Tuple (body, status) = return "Error", 450 [OK]

Hint: Return (body, code) tuple to set custom status [OK]

Common Mistakes:

• Swapping order of status and body in return tuple
• Using Response without import or wrong parameters
• Using make_response incorrectly

2. Which of the following is the correct syntax to return a JSON response with a custom status code 499 in Flask?

easy

A. return jsonify(message="Fail"), 499

B. return jsonify(message="Fail", status=499)

C. return jsonify(message="Fail", 499)

D. return jsonify(message="Fail").status_code(499)

Solution

1. Step 1: Use jsonify to create JSON response

   jsonify returns a Response object with JSON content.

2. Step 2: Return tuple with jsonify and status code

   To set custom status code, return (jsonify(...), status_code) tuple.

3. Final Answer:

   return jsonify(message="Fail"), 499 -> Option A

4. Quick Check:

   jsonify + tuple (response, code) = return jsonify(message="Fail"), 499 [OK]

Hint: Return (jsonify(...), code) tuple for JSON + custom code [OK]

Common Mistakes:

• Passing status code inside jsonify arguments
• Trying to call status_code as method
• Incorrect tuple order or syntax

3. What will be the HTTP status code returned by this Flask route?

from flask import Flask
app = Flask(__name__)

@app.route('/test')
def test():
    return "Custom", 499

medium

A. 404

B. 200

C. 499

D. 500

Solution

1. Step 1: Check the return statement in route

   The route returns a tuple ("Custom", 499), where 499 is the status code.

2. Step 2: Flask uses the second tuple element as status code

   Flask sets HTTP status code to 499 as specified in the return tuple.

3. Final Answer:

   499 -> Option C

4. Quick Check:

   Return tuple second value = status code 499 [OK]

Hint: Second tuple value in return sets status code [OK]

Common Mistakes:

• Assuming default 200 status code
• Confusing return order
• Thinking 499 is invalid and defaults to 500

4. Identify the error in this Flask route that tries to return a custom status code 499:

from flask import Flask
app = Flask(__name__)

@app.route('/error')
def error():
    return "Oops", status=499

medium

A. Route function must return a Response object

B. Custom status code 499 is not allowed in Flask

C. Missing jsonify for JSON response

D. Using 'status' keyword argument in return is invalid

Solution

1. Step 1: Check return statement syntax

   Flask does not accept 'status' keyword in return statement directly.

2. Step 2: Correct way is to return tuple (body, status_code)

   Use return "Oops", 499 instead of return "Oops", status=499.

3. Final Answer:

   Using 'status' keyword argument in return is invalid -> Option D

4. Quick Check:

   Return tuple (body, code) not keyword args [OK]

Hint: Return tuple (body, code), not keyword args [OK]

Common Mistakes:

• Using status=code in return instead of tuple
• Assuming Flask auto-detects status from keywords
• Confusing with Response object parameters

5. You want to create a Flask route that returns a JSON message with a custom status code 460 only if a query parameter 'valid' is 'no'. Otherwise, return status 200 with a success message. Which code correctly implements this?

hard

A. from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/check')
def check():
if request.args.get('valid') == 'no':
return jsonify(error="Invalid", status=460)
return jsonify(message="OK")

B. from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/check')
def check():
if request.args.get('valid') == 'no':
return jsonify(error="Invalid"), 460
return jsonify(message="OK"), 200

C. from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/check')
def check():
if request.args.get('valid') == 'no':
return jsonify(error="Invalid").status_code = 460
return jsonify(message="OK")

D. from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/check')
def check():
if request.args.get('valid') == 'no':
return jsonify(error="Invalid")
return jsonify(message="OK"), 200

Solution

1. Step 1: Check conditional logic for query parameter

   Use request.args.get('valid') to check if it equals 'no'.

2. Step 2: Return JSON with custom status code 460 if invalid

   Return (jsonify(...), 460) tuple to set custom status code correctly.

3. Step 3: Return success JSON with status 200 otherwise

   Return (jsonify(...), 200) tuple for success response.

4. Final Answer:

   Correct conditional returns with tuples for custom and normal codes -> Option B

5. Quick Check:

   Return (jsonify, code) tuples for all cases [OK]

Hint: Return (jsonify(...), code) tuples for all responses [OK]

Common Mistakes:

• Passing status inside jsonify arguments
• Trying to set status_code attribute directly
• Missing status code tuple on error response