REST API for IoT device in Raspberry Pi - Mini Project: Build & Apply

REST API for IoT Device

📖 Scenario: You have a Raspberry Pi connected to a temperature sensor. You want to create a simple REST API so other devices or apps can get the current temperature reading.This project will guide you step-by-step to build a basic REST API using Python and Flask on your Raspberry Pi.

🎯 Goal: Build a REST API on your Raspberry Pi that returns the current temperature reading as JSON when accessed.

📋 What You'll Learn

Create a Python dictionary to simulate sensor data

Add a configuration variable for the API endpoint path

Use Flask to create a REST API endpoint that returns the sensor data

Print the API URL to the console

💡 Why This Matters

🌍 Real World

IoT devices often need to share sensor data with other devices or apps. A REST API is a common way to do this over a network.

💼 Career

Understanding how to build REST APIs on devices like Raspberry Pi is useful for IoT developers, embedded systems engineers, and backend programmers working with connected devices.

1

Create sensor data dictionary

Create a Python dictionary called sensor_data with these exact entries: 'temperature': 22.5, 'unit': 'Celsius'.

Raspberry Pi

# Create the sensor_data dictionary with temperature and unit
# Your code here

Hint

Use curly braces {} to create a dictionary with keys and values.

2

Add API endpoint path configuration

Create a variable called api_path and set it to the string "/temperature".

Raspberry Pi

sensor_data = {'temperature': 22.5, 'unit': 'Celsius'}
# Create the api_path variable with the value '/temperature'
# Your code here

Hint

Assign the string "/temperature" to the variable api_path.

3

Create Flask REST API endpoint

Import Flask and jsonify from flask. Create a Flask app called app. Use @app.route(api_path) to define a function get_temperature() that returns jsonify(sensor_data).

Raspberry Pi

sensor_data = {'temperature': 22.5, 'unit': 'Celsius'}
api_path = "/temperature"
# Import Flask and jsonify, create app, and define route
# Your code here

Hint

Use @app.route(api_path) to create the API endpoint and return the sensor data as JSON.

4

Run the Flask app and print API URL

Add the code to run the Flask app with app.run(host='0.0.0.0', port=5000) inside if __name__ == '__main__':. Before running, print the string "API running at http://localhost:5000".

Raspberry Pi

from flask import Flask, jsonify

sensor_data = {'temperature': 22.5, 'unit': 'Celsius'}
api_path = "/temperature"

app = Flask(__name__)

@app.route(api_path)
def get_temperature():
    return jsonify(sensor_data)

# Print API URL and run the Flask app
# Your code here

Hint

Use if __name__ == '__main__': to run the app and print the API URL before starting.

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