Why file operations are common in FastAPI - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the FastAPI class.

FastAPI

from fastapi import [1]
app = [1]()

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

AFastAPI

BRequest

CResponse

DFile

Attempts:

3 left

💡 Hint

Common Mistakes

Importing Request or Response instead of FastAPI.

Trying to import File here.

The FastAPI class is imported to create the app instance.

2fill in blank

medium

Complete the code to define a POST endpoint that accepts a file upload.

FastAPI

@app.post("/upload")
async def upload_file(file: [1]):
    content = await file.read()
    return {"size": len(content)}

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

Astr

BUploadFile

CFile

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using File instead of UploadFile as the parameter type.

Using str or int which are not file types.

UploadFile is used in FastAPI to handle file uploads asynchronously.

3fill in blank

hard

Fix the error in the import statement to handle file uploads.

FastAPI

from fastapi import [1]
from fastapi import UploadFile, File

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

ARequest

BFile

CFastAPI

DDepends

Attempts:

3 left

💡 Hint

Common Mistakes

Importing File twice or missing FastAPI import.

Confusing Request with FastAPI.

FastAPI must be imported to create the app instance; UploadFile and File are imported separately.

4fill in blank

hard

Fill both blanks to save the uploaded file content to disk.

FastAPI

async def save_file(file: UploadFile):
    content = await file.[1]()
    with open("uploaded_file", [2]) as f:
        f.write(content)

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

Aread

Bwrite

C"wb"

D"r"

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'write' method on UploadFile which does not exist.

Opening file in read mode instead of write mode.

We read the file content asynchronously, then open a file in binary write mode to save it.

5fill in blank

hard

Fill all three blanks to create a FastAPI endpoint that saves an uploaded file and returns its size.

FastAPI

@app.post("/save")
async def save_upload(file: [1] = [2](...)):
    content = await file.[3]()
    with open("saved_file", "wb") as f:
        f.write(content)
    return {"size": len(content)}

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

AUploadFile

BFile

Cread

Dstr

Attempts:

3 left

💡 Hint

Common Mistakes

Using str instead of UploadFile for file parameter.

Missing File() dependency declaration.

Using write() instead of read() on UploadFile.

The endpoint uses UploadFile type with File() to accept uploads, reads content asynchronously, saves it, and returns size.

Practice

1. Why are file operations common in FastAPI applications?

easy

A. Because file operations replace all API calls

B. Because FastAPI does not support databases

C. Because they allow handling user uploads and downloads easily

D. Because FastAPI only works with local files

Solution

1. Step 1: Understand FastAPI's purpose

   FastAPI is used to build web APIs that often need to accept or send files like images or documents.

2. Step 2: Recognize file operation role

   File operations let apps handle user uploads and downloads, which are common web app features.

3. Final Answer:

   Because they allow handling user uploads and downloads easily -> Option C

4. Quick Check:

   File handling = user uploads/downloads [OK]

Hint: File ops = user file handling in web apps [OK]

Common Mistakes:

• Thinking FastAPI can't use databases
• Believing file ops replace API calls
• Assuming FastAPI only works with local files

2. Which of the following is the correct way to declare a file upload parameter in a FastAPI endpoint?

easy

A. def upload(file: UploadFile):

B. def upload(file: int):

C. def upload(file: str):

D. def upload(file: list):

Solution

1. Step 1: Recall FastAPI file upload type

   FastAPI uses the UploadFile type to handle uploaded files efficiently.

2. Step 2: Match parameter type

   The parameter must be typed as UploadFile to receive file data properly.

3. Final Answer:

   def upload(file: UploadFile): -> Option A

4. Quick Check:

   UploadFile type for file uploads [OK]

Hint: Use UploadFile type for file uploads [OK]

Common Mistakes:

• Using str instead of UploadFile
• Using int or list which are invalid for files
• Omitting type annotation

3. Given this FastAPI code snippet, what will be the output when a file is uploaded?

from fastapi import FastAPI, UploadFile
app = FastAPI()

@app.post('/upload')
async def upload(file: UploadFile):
    content = await file.read()
    return {'filename': file.filename, 'size': len(content)}

medium

A. {'filename': 'example.txt', 'size': 1000} if a 1000-byte file named example.txt is uploaded

B. A syntax error because async functions cannot read files

C. An empty dictionary because file.read() returns None

D. A runtime error because file.filename is not accessible

Solution

1. Step 1: Understand async file reading

   The code uses await file.read() to read the uploaded file content asynchronously.

2. Step 2: Check returned dictionary

   The function returns the filename and the size of the content in bytes, so for a 1000-byte file named example.txt, it returns that info.

3. Final Answer:

   {'filename': 'example.txt', 'size': 1000} if a 1000-byte file named example.txt is uploaded -> Option A

4. Quick Check:

   Async read + filename = correct output [OK]

Hint: Async read returns bytes; filename is accessible [OK]

Common Mistakes:

• Thinking async can't read files
• Assuming file.read() returns None
• Believing file.filename is inaccessible

4. Identify the error in this FastAPI file upload endpoint:

from fastapi import FastAPI, UploadFile
app = FastAPI()

@app.post('/upload')
def upload(file: UploadFile):
    content = file.read()
    return {'size': len(content)}

medium

A. The endpoint path '/upload' is invalid

B. UploadFile should be replaced with str

C. The return statement should return content directly

D. Missing async and await for reading the file

Solution

1. Step 1: Check file reading method

   UploadFile.read() is an async method and must be awaited inside an async function.

2. Step 2: Identify missing async keywords

   The function is not async and does not await file.read(), causing a runtime error.

3. Final Answer:

   Missing async and await for reading the file -> Option D

4. Quick Check:

   Async read requires async def and await [OK]

Hint: Use async def and await for UploadFile.read() [OK]

Common Mistakes:

• Using synchronous def with async read
• Replacing UploadFile with str incorrectly
• Returning content instead of size is not an error here

5. You want to save an uploaded file to disk in FastAPI. Which approach correctly handles this while keeping the app responsive?

from fastapi import FastAPI, UploadFile
app = FastAPI()

@app.post('/save')
async def save_file(file: UploadFile):
    contents = await file.read()
    with open(file.filename, 'wb') as f:
        f.write(contents)
    return {'filename': file.filename}

hard

A. This code is correct and efficient for saving files asynchronously

B. Opening and writing files synchronously blocks the event loop; use async file libraries

C. You must convert contents to string before writing

D. You should not read the file content; just save file.filename directly

Solution

1. Step 1: Analyze file reading and writing

   The file is read asynchronously, but writing with open() is synchronous and blocks the event loop.

2. Step 2: Identify best practice for responsiveness

   To keep FastAPI responsive, use async file writing libraries like aiofiles instead of blocking open().

3. Final Answer:

   Opening and writing files synchronously blocks the event loop; use async file libraries -> Option B

4. Quick Check:

   Async read + sync write blocks event loop [OK]

Hint: Use async file write libs to avoid blocking [OK]

Common Mistakes:

• Assuming sync write is fine in async endpoint
• Converting bytes to string before writing binary
• Trying to save filename without content