Multiple file uploads in FastAPI - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the correct class for handling file uploads in FastAPI.

FastAPI

from fastapi import FastAPI, [1]

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

AUploadFile

BRequest

CFile

DForm

Attempts:

3 left

💡 Hint

Common Mistakes

Using File instead of UploadFile

Importing Form instead of UploadFile

FastAPI uses UploadFile to handle file uploads efficiently.

2fill in blank

medium

Complete the code to accept multiple files in the endpoint function parameter.

FastAPI

async def upload_files(files: list[[1]]):

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

Abytes

BFile

CUploadFile

Dstr

Attempts:

3 left

💡 Hint

Common Mistakes

Using list[File] instead of list[UploadFile]

Using list[str] which is incorrect for files

To accept multiple files, use a list of UploadFile objects.

3fill in blank

hard

Fix the error in the endpoint decorator to accept multiple files with the correct parameter.

FastAPI

@app.post("/upload")
async def upload(files: list[UploadFile] = [1]):

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

AFile()

BFile(...)

CForm(...)

DDepends()

Attempts:

3 left

💡 Hint

Common Mistakes

Using File() which makes the parameter optional

Using Form(...) which is for form fields, not files

Use File(...) to mark the parameter as required for multiple file uploads.

4fill in blank

hard

Fill both blanks to read the filename and content of each uploaded file.

FastAPI

for file in files:
    filename = file.[1]
    content = await file.[2]()

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

Afilename

Bread

Ccontent

Dwrite

Attempts:

3 left

💡 Hint

Common Mistakes

Using file.content instead of file.filename

Calling file.write() instead of file.read()

Use file.filename to get the file name and await file.read() to get its content.

5fill in blank

hard

Fill all three blanks to define the FastAPI app, import UploadFile and File, and create the upload endpoint.

FastAPI

from fastapi import [1], UploadFile, File

app = [2]()

@app.post("/upload")
async def upload(files: list[UploadFile] = File(...)):
    return {file.filename: await file.read() for file in [3]

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

AFastAPI

Bapp

Cfiles

DRequest

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'app' instead of 'FastAPI' in import

Using 'app' instead of 'FastAPI()' to create the app

Using wrong variable name instead of 'files' in the comprehension

Import FastAPI, create the app with FastAPI(), and iterate over files in the endpoint.

Practice

1. In FastAPI, which type hint should you use to accept multiple uploaded files in an endpoint?

easy

A. str with UploadFile

B. List[UploadFile] with File(...)

C. List[str] with File(...)

D. UploadFile without List

Solution

1. Step 1: Understand FastAPI file upload types

   FastAPI uses UploadFile to handle file uploads efficiently.

2. Step 2: Use List for multiple files

   To accept multiple files, you wrap UploadFile in a List and use File(...) to declare it as a file input.

3. Final Answer:

   List[UploadFile] with File(...) -> Option B

4. Quick Check:

   Multiple files = List[UploadFile] [OK]

Hint: Use List[UploadFile] with File(...) for multiple files [OK]

Common Mistakes:

• Using UploadFile without List for multiple files
• Using List[str] which expects strings, not files
• Confusing str type with file uploads

2. Which of the following is the correct syntax to declare a FastAPI endpoint that accepts multiple files named 'files'?

easy

A. files: List[UploadFile] = File(...)

B. files: UploadFile = File(...)

C. files: List[str] = File(...)

D. files: str = UploadFile(...)

Solution

1. Step 1: Check parameter type for multiple files

   Multiple files require List[UploadFile] type hint.

2. Step 2: Use File(...) to mark as file input

   File(...) is needed to tell FastAPI this is a file upload field.

3. Final Answer:

   files: List[UploadFile] = File(...) -> Option A

4. Quick Check:

   Multiple files syntax = List[UploadFile] = File(...) [OK]

Hint: Use List[UploadFile] = File(...) for multiple files param [OK]

Common Mistakes:

• Using UploadFile without List for multiple files
• Using List[str] which is incorrect for files
• Assigning UploadFile(...) instead of File(...)

3. Given this FastAPI endpoint code, what will be the output if two files named 'a.txt' and 'b.txt' are uploaded?

from fastapi import FastAPI, File, UploadFile
from typing import List

app = FastAPI()

@app.post('/upload')
async def upload(files: List[UploadFile] = File(...)):
    return {"filenames": [file.filename for file in files]}

medium

A. {"filenames": ["a.txt", "b.txt"]}

B. {"filenames": ["files", "files"]}

C. {"filenames": []}

D. Runtime error due to wrong type

Solution

1. Step 1: Understand the endpoint logic

   The endpoint returns a dictionary with a list of filenames extracted from the uploaded files.

2. Step 2: Check the uploaded files names

   Two files named 'a.txt' and 'b.txt' are uploaded, so their filenames will be returned in the list.

3. Final Answer:

   {"filenames": ["a.txt", "b.txt"]} -> Option A

4. Quick Check:

   Returned filenames list matches uploaded files [OK]

Hint: Returned filenames list matches uploaded files names [OK]

Common Mistakes:

• Expecting file content instead of filenames
• Confusing parameter name with file names
• Assuming empty list if files uploaded

4. What is wrong with this FastAPI endpoint for multiple file uploads?

from fastapi import FastAPI, File, UploadFile
from typing import List

app = FastAPI()

@app.post('/upload')
async def upload(files: UploadFile = File(...)):
    return {"count": len(files)}

medium

A. File(...) should be replaced with UploadFile(...)

B. Missing async keyword in function definition

C. len(files) is invalid because files is a string

D. files should be List[UploadFile] to accept multiple files

Solution

1. Step 1: Check parameter type for multiple files

   The parameter 'files' is typed as UploadFile, which accepts only one file.

2. Step 2: Correct type for multiple files

   To accept multiple files, it must be List[UploadFile].

3. Final Answer:

   files should be List[UploadFile] to accept multiple files -> Option D

4. Quick Check:

   Multiple files need List[UploadFile] type [OK]

Hint: Use List[UploadFile] for multiple files, not UploadFile alone [OK]

Common Mistakes:

• Using UploadFile instead of List[UploadFile] for multiple files
• Confusing File(...) with UploadFile(...)
• Forgetting async keyword (though present here)

5. You want to create a FastAPI endpoint that accepts multiple files and returns a JSON with each file's name and size in bytes. Which code snippet correctly implements this?

hard

A. async def upload(files: List[str] = File(...)):
return [{"name": f, "size": 0} for f in files]

B. async def upload(files: UploadFile = File(...)):
return [{"name": files.filename, "size": len(await files.read())}]

C. async def upload(files: List[UploadFile] = File(...)):
results = []
for file in files:
content = await file.read()
results.append({"name": file.filename, "size": len(content)})
return results

D. async def upload(files: List[UploadFile] = File(...)):
return [{"name": file.name, "size": file.size} for file in files]

Solution

1. Step 1: Accept multiple files with correct type

   Use List[UploadFile] with File(...) to accept multiple files.

2. Step 2: Read each file content to get size

   Use await file.read() to get bytes, then len() to get size.

3. Step 3: Return list of dicts with name and size

   Build a list of dictionaries with filename and size for each file.

4. Final Answer:

   Code snippet that correctly reads files and returns name and size -> Option C

5. Quick Check:

   Read files async, return name and size list [OK]

Hint: Read files async, use len(content) for size, return list [OK]

Common Mistakes:

• Using UploadFile without List for multiple files
• Trying to get size without reading file content
• Using wrong attribute names like file.name or file.size