Accepting connections in FastAPI - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import FastAPI and create an app instance.

FastAPI

from fastapi import [1]

app = [1]()

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

AFastAPI

BRequest

CResponse

DDepends

Attempts:

3 left

💡 Hint

Common Mistakes

Importing Request or Response instead of FastAPI

Not creating an instance of FastAPI

You need to import FastAPI and create an instance of it to accept connections.

2fill in blank

medium

Complete the code to define a GET route at the root path that returns a welcome message.

FastAPI

@app.[1]("/")
async def read_root():
    return {"message": "Welcome!"}

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

Aget

Bpost

Cput

Ddelete

Attempts:

3 left

💡 Hint

Common Mistakes

Using post or put instead of get for a simple retrieval

Forgetting the decorator

The @app.get decorator defines a GET route to accept connections at the root path.

3fill in blank

hard

Fix the error in the code to run the FastAPI app with uvicorn on port 8000.

FastAPI

import uvicorn

if __name__ == "__main__":
    uvicorn.run("main:[1]", host="127.0.0.1", port=8000, reload=True)

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

Aapi

Bapp

Capplication

Dserver

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong instance name like server or api

Forgetting the colon separator

The string "main:app" tells uvicorn to find the app instance in the main module.

4fill in blank

hard

Fill both blanks to create a POST route at '/submit' that accepts JSON data and returns it.

FastAPI

@app.[1]("/submit")
async def submit_data(data: [2]):
    return data

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

Apost

Bget

Cdict

Dlist

Attempts:

3 left

💡 Hint

Common Mistakes

Using GET instead of POST for data submission

Using list instead of dict for JSON data

Use @app.post to define a POST route and dict to accept JSON data as a Python dictionary.

5fill in blank

hard

Fill all three blanks to create a FastAPI app that accepts connections on all interfaces at port 8080 with reload enabled.

FastAPI

import uvicorn

if __name__ == "__main__":
    uvicorn.run("main:[1]", host="[2]", port=[3], reload=True)

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

Aapp

B0.0.0.0

C8080

D127.0.0.1

Attempts:

3 left

💡 Hint

Common Mistakes

Using localhost (127.0.0.1) instead of 0.0.0.0 to accept external connections

Using wrong port number

Use app as the FastAPI instance name, 0.0.0.0 to accept connections from all interfaces, and port 8080 as specified.

Practice

1. What does it mean when a FastAPI app is "accepting connections"?

easy

A. The app is listening for incoming network requests on a specific address and port.

B. The app is sending data to a database.

C. The app is shutting down all network activity.

D. The app is compiling Python code.

Solution

1. Step 1: Understand the term "accepting connections" in networking

   Accepting connections means the app waits and listens for requests from users or other programs on a network address and port.

2. Step 2: Relate to FastAPI app behavior

   FastAPI apps accept connections by running a server that listens on a host and port for incoming HTTP requests.

3. Final Answer:

   The app is listening for incoming network requests on a specific address and port. -> Option A

4. Quick Check:

   Accepting connections = listening on address and port [OK]

Hint: Accepting connections means listening for requests [OK]

Common Mistakes:

• Confusing accepting connections with sending data
• Thinking it means shutting down the app
• Mixing it with code compilation

2. Which of the following is the correct way to start a FastAPI app to accept connections on all network interfaces at port 8000?

easy

A. uvicorn.run(app, host="127.0.0.1", port=8080)

B. uvicorn.run(app, host="0.0.0.0", port=8000)

C. uvicorn.run(app, host="localhost", port=80)

D. uvicorn.run(app, host="192.168.1.1", port=5000)

Solution

1. Step 1: Identify the host for all network interfaces

   The host "0.0.0.0" means the app listens on all network interfaces, allowing external access.

2. Step 2: Check the port number

   Port 8000 is the default port commonly used for FastAPI apps.

3. Final Answer:

   uvicorn.run(app, host="0.0.0.0", port=8000) -> Option B

4. Quick Check:

   Host 0.0.0.0 + port 8000 = correct start command [OK]

Hint: Use host 0.0.0.0 for all interfaces [OK]

Common Mistakes:

• Using localhost instead of 0.0.0.0 for external access
• Choosing wrong port numbers
• Confusing IP addresses with hostnames

3. Given this code snippet, what will happen when you run it?

import uvicorn
from fastapi import FastAPI
app = FastAPI()

@app.get("/")
async def read_root():
    return {"message": "Hello World"}

uvicorn.run(app, host="127.0.0.1", port=9000)

medium

A. The app will listen on all interfaces at port 9000 but return an error on root URL.

B. The app will fail to start due to syntax error.

C. The app will listen on localhost at port 9000 and respond with JSON {"message": "Hello World"} on root URL.

D. The app will listen on localhost but on default port 8000.

Solution

1. Step 1: Analyze the uvicorn.run parameters

   The app runs on host "127.0.0.1" (localhost) and port 9000, so it listens only on local machine at port 9000.

2. Step 2: Check the route and response

   The root path "/" returns JSON {"message": "Hello World"} asynchronously, so a GET request to localhost:9000/ returns this JSON.

3. Final Answer:

   The app will listen on localhost at port 9000 and respond with JSON {"message": "Hello World"} on root URL. -> Option C

4. Quick Check:

   Host 127.0.0.1 + port 9000 + root JSON response = B [OK]

Hint: Check host, port, and route response carefully [OK]

Common Mistakes:

• Assuming app listens on all interfaces
• Expecting default port 8000 instead of 9000
• Thinking code has syntax errors

4. You wrote this code to start your FastAPI app but it crashes with an error:

import uvicorn
from fastapi import FastAPI
app = FastAPI()

uvicorn.run(app, host="localhost", port="8000")

What is the likely cause of the error?

medium

A. Missing async keyword before uvicorn.run call.

B. The host "localhost" is invalid and causes the crash.

C. FastAPI app must be run with app.run(), not uvicorn.run().

D. The port number should be an integer, not a string.

Solution

1. Step 1: Check the port argument type

   The port argument must be an integer, but here it is passed as a string "8000" which causes a type error.

2. Step 2: Verify host and method usage

   Host "localhost" is valid, and uvicorn.run is the correct method to start FastAPI apps.

3. Final Answer:

   The port number should be an integer, not a string. -> Option D

4. Quick Check:

   Port must be int, not str [OK]

Hint: Port must be a number, not text [OK]

Common Mistakes:

• Passing port as string instead of int
• Thinking host "localhost" is invalid
• Using wrong method to start app

5. You want your FastAPI app to accept connections only from your local machine and run on port 8080. Which code snippet correctly achieves this and why?

Option A:
uvicorn.run(app, host="0.0.0.0", port=8080)

Option B:
uvicorn.run(app, host="127.0.0.1", port=8080)

Option C:
uvicorn.run(app, host="localhost", port=80)

Option D:
uvicorn.run(app, host="192.168.1.100", port=8080)

hard

A. Option B, because 127.0.0.1 restricts connections to local machine only on port 8080.

B. Option A, because 0.0.0.0 listens on all interfaces including local machine.

C. Option C, because localhost always means local machine and port 80 is standard.

D. Option D, because 192.168.1.100 is the local IP and port 8080 is correct.

Solution

1. Step 1: Understand host IP meanings

   Host "127.0.0.1" means the app listens only on the local machine (loopback interface), blocking external connections.

2. Step 2: Check port number

   Port 8080 is specified correctly and is a valid port for local testing.

3. Step 3: Evaluate other options

   0.0.0.0 listens on all interfaces, allowing external access; localhost on port 80 uses the wrong port; 192.168.1.100 binds to a specific LAN IP which may not restrict to local machine only.

4. Final Answer:

   Option A, because 127.0.0.1 restricts connections to local machine only on port 8080. -> Option A

5. Quick Check:

   Local only = 127.0.0.1 + port 8080 [OK]

Hint: Use 127.0.0.1 for local-only access [OK]

Common Mistakes:

• Choosing 0.0.0.0 for local-only access
• Confusing localhost and 127.0.0.1
• Using wrong port number