File upload (single file) in FastAPI - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the correct FastAPI class.

FastAPI

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

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

AUploadFile

BRequest

CFile

DFastAPI

Attempts:

3 left

💡 Hint

Common Mistakes

Importing Request instead of FastAPI.

Using UploadFile or File to create the app.

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]):
    return {"filename": file.filename}

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AUploadFile

BRequest

CFile

Dstr

Attempts:

3 left

💡 Hint

Common Mistakes

Using str instead of UploadFile for the file parameter.

Using File directly as the parameter type.

The UploadFile type is used to receive uploaded files in FastAPI.

3fill in blank

hard

Fix the error in the code to correctly declare the file parameter with File dependency.

FastAPI

from fastapi import UploadFile, File

@app.post("/upload")
async def upload_file(file: UploadFile = [1]):
    return {"filename": file.filename}

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AUploadFile()

BFile()

CFile

DUploadFile

Attempts:

3 left

💡 Hint

Common Mistakes

Using File without parentheses.

Using UploadFile() instead of File().

The File() function is used as a dependency to tell FastAPI this parameter is a file upload.

4fill in blank

hard

Fill both blanks to read the uploaded file content as bytes and return its size.

FastAPI

@app.post("/upload")
async def upload_file(file: UploadFile = File()):
    content = await file.[1]()
    return {"size": len(content) [2] 0}

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Aread

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using read without parentheses.

Using subtraction or multiplication instead of addition.

Use read() to get file bytes asynchronously. Adding 0 returns the size as is.

5fill in blank

hard

Fill all three blanks to save the uploaded file content to disk with the original filename.

FastAPI

import shutil

@app.post("/upload")
async def upload_file(file: UploadFile = File()):
    with open([1], "wb") as buffer:
        shutil.[2](file.[3], buffer)
    return {"filename": file.filename}

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

Afile.filename

Bcopyfileobj

Cfile

Dwrite

Attempts:

3 left

💡 Hint

Common Mistakes

Using write instead of copyfileobj.

Using file.read() instead of file.file.

Using a string literal instead of file.filename.

Open a file with the uploaded filename, then use shutil.copyfileobj(file.file, buffer) to efficiently copy the uploaded file content to disk.