Multiple file uploads in FastAPI

Introduction

Uploading multiple files lets users send many files at once. This saves time and makes apps easier to use.

When users need to upload several photos to a profile or gallery.

When a form requires multiple documents, like resumes and cover letters.

When users want to upload many files for backup or sharing.

When collecting multiple images for a product listing.

When users submit multiple attachments in a support ticket.

Syntax

FastAPI

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

app = FastAPI()

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

Use List[UploadFile] to accept multiple files.

The File(...) tells FastAPI this is a file upload field.

Examples

Basic example to receive multiple files and return their names.

FastAPI

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

app = FastAPI()

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

This example reads each file's content and returns filename and size.

FastAPI

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

app = FastAPI()

@app.post("/uploadfiles/")
async def upload_files(files: List[UploadFile] = File(...)):
    contents = []
    for file in files:
        content = await file.read()
        contents.append({"filename": file.filename, "size": len(content)})
    return contents

Sample Program

This FastAPI app defines a POST endpoint /uploadfiles/ that accepts multiple files. It returns a JSON with the list of uploaded file names.

FastAPI

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

app = FastAPI()

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

Important Notes

Remember to use async and await when reading file contents.

Use List[UploadFile] to handle multiple files cleanly.

FastAPI automatically handles the multipart form data for file uploads.

Summary

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

Return file info like names or sizes after upload.

FastAPI makes handling multiple file uploads simple and efficient.

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