REST API for IoT device in Raspberry Pi - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the Flask library needed to create a REST API.

Raspberry Pi

from [1] import Flask

Drag options to blanks, or click blank then click option'

Arequests

BFlask

Cflask

Djson

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Flask' instead of 'flask' as the module name.

We import Flask from the flask module to create the REST API.

2fill in blank

medium

Complete the code to create a Flask app instance named 'app'.

Raspberry Pi

app = [1](__name__)

Drag options to blanks, or click blank then click option'

Aflask

BFlask

CApp

Dapplication

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'flask' instead of 'Flask'.

The Flask app is created by calling Flask(__name__).

3fill in blank

hard

Fix the error in the route decorator to handle GET requests at '/status'.

Raspberry Pi

@app.route('/status', methods=[[1]])

Drag options to blanks, or click blank then click option'

A'GET'

B'POST'

C'PUT'

D'DELETE'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'POST' or other HTTP methods instead of 'GET'.

The route should accept GET requests, so the method is 'GET'.

4fill in blank

hard

Fill both blanks to return a JSON response with device status and HTTP status code 200.

Raspberry Pi

return [1]({'status': 'online'}), [2]

Drag options to blanks, or click blank then click option'

Ajsonify

B200

Cjson

D404

Attempts:

3 left

💡 Hint

Common Mistakes

Returning plain dict without jsonify.

Using wrong status code like 404.

Use jsonify to create a JSON response and return HTTP status code 200 for success.

5fill in blank

hard

Fill all three blanks to run the Flask app on host '0.0.0.0' and port 5000 with debug mode on.

Raspberry Pi

if __name__ == '__main__':
    app.run(host=[1], port=[2], debug=[3])

Drag options to blanks, or click blank then click option'

A'127.0.0.1'

B5000

CTrue

D'0.0.0.0'

Attempts:

3 left

💡 Hint

Common Mistakes

Using '127.0.0.1' limits access to localhost only.

Setting debug to False disables helpful error messages.

To make the app accessible on the local network, use host '0.0.0.0', port 5000, and enable debug mode with True.

Practice

1. What is the main purpose of a REST API on a Raspberry Pi IoT device?

easy

A. To connect the Raspberry Pi to a monitor

B. To allow remote communication and control over the network

C. To store large amounts of data locally

D. To increase the Raspberry Pi's processing speed

Solution

1. Step 1: Understand REST API role

   A REST API enables devices to communicate over a network using standard web protocols.

2. Step 2: Identify IoT device needs

   IoT devices use REST APIs to send sensor data or receive commands remotely.

3. Final Answer:

   To allow remote communication and control over the network -> Option B

4. Quick Check:

   REST API = Remote communication [OK]

Hint: REST APIs enable network communication for IoT devices [OK]

Common Mistakes:

• Confusing REST API with hardware speed
• Thinking REST API stores data locally
• Assuming REST API connects to display devices

2. Which Python library is commonly used on Raspberry Pi to create a REST API for IoT devices?

easy

A. Flask

B. NumPy

C. Pandas

D. Matplotlib

Solution

1. Step 1: Identify common Python libraries

   NumPy, Pandas, and Matplotlib are for data processing and visualization, not REST APIs.

2. Step 2: Recognize Flask's role

   Flask is a lightweight web framework used to build REST APIs easily on Raspberry Pi.

3. Final Answer:

   Flask -> Option A

4. Quick Check:

   Flask = REST API framework [OK]

Hint: Flask is the go-to Python web framework for REST APIs [OK]

Common Mistakes:

• Choosing data libraries instead of web frameworks
• Confusing Flask with data analysis tools
• Assuming Matplotlib creates APIs

3. What will be the output of this Flask route code on a Raspberry Pi?

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

@app.route('/sensor')
def sensor():
    data = {'temperature': 22, 'humidity': 60}
    return jsonify(data)

# Assume app is running and /sensor is requested

medium

A. {"temperature": 22, "humidity": 60}

B. temperature=22, humidity=60

C. temperature: 22, humidity: 60

D. Error: jsonify not defined

Solution

1. Step 1: Understand jsonify function

   jsonify converts Python dict to JSON string with correct headers for API response.

2. Step 2: Analyze route output

   The route returns JSON with keys 'temperature' and 'humidity' and their values.

3. Final Answer:

   {"temperature": 22, "humidity": 60} -> Option A

4. Quick Check:

   jsonify(dict) = JSON string [OK]

Hint: jsonify returns JSON string from dict in Flask [OK]

Common Mistakes:

• Expecting plain text instead of JSON
• Confusing jsonify with HTML output
• Missing import of jsonify causing errors

4. Identify the error in this Flask route code for Raspberry Pi IoT device:

from flask import Flask
app = Flask(__name__)

@app.route('/led/on')
def led_on():
    GPIO.output(17, True)
    return 'LED is ON'

medium

A. Return statement must be JSON, not string

B. Route decorator syntax is incorrect

C. Missing import and setup for GPIO module

D. Flask app is not created properly

Solution

1. Step 1: Check GPIO usage

   GPIO module must be imported and configured before use to control pins.

2. Step 2: Verify code completeness

   Code lacks 'import RPi.GPIO as GPIO' and GPIO setup like GPIO.setmode() and GPIO.setup().

3. Final Answer:

   Missing import and setup for GPIO module -> Option C

4. Quick Check:

   GPIO needs import and setup before use [OK]

Hint: Always import and setup GPIO before controlling pins [OK]

Common Mistakes:

• Assuming Flask route syntax is wrong
• Thinking return must be JSON always
• Ignoring GPIO initialization steps

5. You want to create a REST API on Raspberry Pi that returns sensor data only if temperature is above 25°C. Which code snippet correctly implements this logic using Flask?

hard

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

@app.route('/sensor')
def sensor():
temp = 30
if temp >= 25:
return {'temperature': temp}
else:
return {'message': 'Temperature too low'}

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

@app.route('/sensor')
def sensor():
temp = 20
if temp < 25:
return jsonify({'temperature': temp})
else:
return jsonify({'message': 'Temperature too low'})

C. from flask import Flask
app = Flask(__name__)

@app.route('/sensor')
def sensor():
temp = 30
if temp > 25:
return {'temperature': temp}
else:
return {'message': 'Temperature too low'}

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

@app.route('/sensor')
def sensor():
temp = 30
if temp > 25:
return jsonify({'temperature': temp})
else:
return jsonify({'message': 'Temperature too low'})

Solution

1. Step 1: Check temperature condition logic

   We want to return data only if temperature is above 25, so 'temp > 25' is correct.

2. Step 2: Verify response format

   Flask needs jsonify to return proper JSON response with correct headers.

3. Final Answer:

   Code with temp=30, condition temp > 25, and jsonify responses -> Option D

4. Quick Check:

   Use temp > 25 and jsonify for JSON response [OK]

Hint: Use if temp > 25 and jsonify() to return JSON [OK]

Common Mistakes:

• Using wrong temperature comparison operator
• Returning dict without jsonify causing errors
• Setting temperature below threshold incorrectly