Flask route as API endpoint - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What is the output of this Flask API endpoint?

Consider this Flask route that returns JSON data. What will the client receive when accessing /api/data?

Flask

from flask import Flask, jsonify
app = Flask(__name__)

@app.route('/api/data')
def data():
    return jsonify({'name': 'Alice', 'age': 30})

A {'name': 'Alice', 'age': 30}

B <Response 200 OK>

C name: Alice, age: 30

D {"name": "Alice", "age": 30}

💡 Hint

Think about what jsonify does and what format the client receives.

The jsonify function converts the dictionary into a JSON string with correct content-type. The client receives a JSON string, not a Python dict or a plain string.

📝 Syntax

intermediate

Which Flask route definition will cause a syntax error?

Identify the Flask route decorator that will cause a syntax error when defining an API endpoint.

A

@app.route('/api/user')
def user():
return 'User data'

B

@app.route('/api/user')
def user():
    return 'User data'

C

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D

@app.route('/api/user')
def user():
    return {'user': 'Bob'}

💡 Hint

Check the indentation of the function body.

Option A lacks indentation for the return statement, causing an IndentationError in Python.

❓ state_output

advanced

What is the response status code of this Flask API endpoint?

Given this Flask route, what status code will the client receive when accessing /api/status?

Flask

from flask import Flask, jsonify
app = Flask(__name__)

@app.route('/api/status')
def status():
    return jsonify({'status': 'ok'}), 201

A 201

B 500

C 404

D 200

💡 Hint

Look at the second value returned by the function.

Returning a tuple with (response, status_code) sets the HTTP status code. Here it is 201 Created.

🔧 Debug

advanced

Why does this Flask API endpoint raise a TypeError?

Examine the code below. Why will calling /api/error cause a TypeError?

Flask

from flask import Flask
app = Flask(__name__)

@app.route('/api/error')
def error():
    return {'message': 'fail'}

A Route decorator syntax is invalid

B Missing return statement causes NoneType error

C Flask cannot convert dict to response automatically without jsonify

D Function name 'error' is reserved and causes conflict

💡 Hint

Flask needs a proper response object or string, not a raw dict.

Returning a dict directly without jsonify causes Flask to raise a TypeError because it cannot convert dict to a valid response.

🧠 Conceptual

expert

Which option correctly implements a Flask API endpoint that accepts a URL parameter and returns it in JSON?

You want to create an API endpoint /api/item/<id> that returns the id parameter as JSON. Which code snippet does this correctly?

A

from flask import Flask, jsonify
app = Flask(__name__)

@app.route('/api/item/<id>')
def item():
    return jsonify({'id': id})

B

from flask import Flask, jsonify
app = Flask(__name__)

@app.route('/api/item/<id>')
def item(id):
    return jsonify({'id': id})

C

from flask import Flask, jsonify
app = Flask(__name__)

@app.route('/api/item/<int:id>')
def item(id):
    return {'id': id}

D

from flask import Flask, jsonify
app = Flask(__name__)

@app.route('/api/item/<int:id>')
def item():
    return jsonify({'id': id})

💡 Hint

Check the function parameters and route variable types carefully.

Option B correctly defines the route with a string parameter and passes it to the function. It returns JSON using jsonify. Option B misses the id parameter in the function. Option B misses the id parameter and will cause a NameError. Option B returns a dict without jsonify, causing a TypeError.