Multiple file uploads in FastAPI - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the purpose of using List[UploadFile] in FastAPI?

It allows the API endpoint to accept multiple files uploaded at once by the client, handling each file as an UploadFile object.

beginner

How do you declare a FastAPI endpoint to accept multiple files?

Use a parameter with type List[UploadFile] and add File(...) to tell FastAPI to expect multiple files in the request.

intermediate

What is the difference between UploadFile and bytes in FastAPI file uploads?

UploadFile provides a file-like interface with metadata and async methods, while bytes reads the entire file content into memory as raw bytes.

intermediate

Why is it better to use UploadFile for multiple file uploads instead of reading files as bytes?

Because UploadFile streams files efficiently without loading everything into memory, which is important when handling many or large files.

beginner

How can you access the filename of each uploaded file in FastAPI?

Each UploadFile object has a filename attribute that contains the original name of the uploaded file.

Which type hint allows a FastAPI endpoint to accept multiple uploaded files?

A UploadFile

B List[UploadFile]

C bytes

D str

Using List[UploadFile] tells FastAPI to expect multiple files uploaded as a list.

What decorator parameter is used to mark a parameter as a file upload in FastAPI?

A Query(...)

B Body(...)

C File(...)

D Path(...)

File(...) marks the parameter as a file upload.

What attribute of UploadFile gives you the original filename?

A filename

B file_name

C name

D title

The filename attribute contains the original file name.

Why is UploadFile preferred over bytes for large files?

A It streams the file efficiently without loading all at once

B It compresses files

C It converts files to text automatically

D It loads the entire file into memory

UploadFile streams files efficiently, saving memory.

How do you declare a FastAPI endpoint parameter to accept multiple files?

A files: UploadFile = File(...)

B files: bytes = File(...)

C files: str = File(...)

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

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

Explain how to create a FastAPI endpoint that accepts multiple file uploads and processes each file.

Think about how to receive and loop through files in the endpoint function.

Describe the advantages of using UploadFile over bytes when handling multiple file uploads in FastAPI.

Consider memory usage and file handling features.

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