File upload (single file) in FastAPI - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main FastAPI class used to handle file uploads?

FastAPI uses the <code>UploadFile</code> class to handle file uploads efficiently without loading the entire file into memory.

beginner

How do you declare a single file upload parameter in a FastAPI endpoint?

You declare a parameter with type UploadFile and use File() from fastapi to mark it as a file upload, for example: file: UploadFile = File(...).

intermediate

What method do you use to read the contents of an uploaded file in FastAPI?

You use the read() method on the UploadFile object to get the file content as bytes, e.g., contents = await file.read().

intermediate

Why is UploadFile preferred over bytes for file uploads in FastAPI?

UploadFile streams the file and uses temporary storage, which is more memory efficient for large files compared to loading the entire file as bytes.

intermediate

How do you save an uploaded file to disk in FastAPI?

You open a file in write-binary mode and write the contents read from UploadFile, for example:

with open('filename', 'wb') as f:
    f.write(await file.read())

Which FastAPI import is required to declare a file upload parameter?

A FileHandler from fastapi

B File from fastapi

C FileUpload from fastapi

D UploadFile from starlette

You must import File from fastapi to declare a file upload parameter.

What type should a single file upload parameter have in FastAPI?

A UploadFile

B str

C bytes

D File

The parameter should be typed as UploadFile to handle file uploads efficiently.

How do you access the filename of an uploaded file in FastAPI?

A file.filename

B file.name

C file.file_name

D file.get_name()

The UploadFile object has a filename attribute with the original file name.

Which method reads the entire content of an uploaded file asynchronously?

A file.read()

B file.read_bytes()

C await file.read()

D await file.get()

You must use await file.read() because read() is an async method.

What is a benefit of using UploadFile over bytes for uploads?

A Faster upload speed

B Supports multiple files only

C Automatically compresses files

D Streams file to disk, saving memory

UploadFile streams the file and uses temporary storage, which is more memory efficient.

Explain how to create a FastAPI endpoint that accepts a single file upload and saves it to disk.

Think about the parameter type, reading the file, and saving it.

Describe why using UploadFile is better than bytes for handling file uploads in FastAPI.

Consider memory use and file size.

Practice

1. In FastAPI, which parameter type is used to accept a single uploaded file in an endpoint?

easy

A. List[UploadFile]

B. str

C. UploadFile

D. int

Solution

1. Step 1: Understand FastAPI file upload types

   FastAPI uses UploadFile to handle file uploads efficiently.

2. Step 2: Identify single file upload parameter

   For a single file, the parameter type is UploadFile, not a list or primitive type.

3. Final Answer:

   <code>UploadFile</code> -> Option C

4. Quick Check:

   Single file upload uses UploadFile [OK]

Hint: Use UploadFile for single file upload in FastAPI [OK]

Common Mistakes:

• Using str or int instead of UploadFile
• Using List[UploadFile] for single file
• Not importing UploadFile from fastapi

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

easy

A. file: str = File(...)

B. file: bytes = File(...)

C. file: UploadFile

D. file: UploadFile = File(...)

Solution

1. Step 1: Check parameter declaration for file upload

   FastAPI requires File(...) to mark the parameter as a file upload.

2. Step 2: Match type with File marker

   The type must be UploadFile combined with File(...) for single file upload.

3. Final Answer:

   file: UploadFile = File(...) -> Option D

4. Quick Check:

   UploadFile with File(...) is correct syntax [OK]

Hint: Use UploadFile = File(...) to accept single file [OK]

Common Mistakes:

• Omitting File(...) marker
• Using str or bytes instead of UploadFile
• Not assigning default File(...)

3. What will be the output of this FastAPI endpoint when a file named 'test.txt' with content 'hello' is uploaded?

from fastapi import FastAPI, UploadFile, File
app = FastAPI()

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

medium

A. {'filename': 'test.txt', 'content_type': 'text/plain', 'size': 5}

B. {'filename': 'test.txt', 'content_type': 'application/octet-stream', 'size': 0}

C. {'filename': '', 'content_type': '', 'size': 5}

D. RuntimeError due to missing await

Solution

1. Step 1: Read file content asynchronously

   The code uses await file.read() to get the file bytes, so content length is 5 for 'hello'.

2. Step 2: Return file info correctly

   Filename is 'test.txt', content type is 'text/plain' (default for .txt), size is length of content (5).

3. Final Answer:

   {'filename': 'test.txt', 'content_type': 'text/plain', 'size': 5} -> Option A

4. Quick Check:

   File info returned matches uploaded file [OK]

Hint: Await file.read() to get content size correctly [OK]

Common Mistakes:

• Not awaiting file.read() causing empty content
• Assuming wrong content_type
• Returning filename as empty string

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

from fastapi import FastAPI, UploadFile
app = FastAPI()

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

medium

A. Missing import of File from fastapi

B. All of the above

C. file.read() should be awaited in async function

D. Endpoint function must be async to await file.read()

Solution

1. Step 1: Check imports and function signature

   The code misses importing File and the function is not async but calls file.read() which is async.

2. Step 2: Identify async usage and await

   To read file content, function must be async and use await file.read().

3. Final Answer:

   All of the above -> Option B

4. Quick Check:

   Missing import, async, and await cause errors [OK]

Hint: Import File, make function async, await file.read() [OK]

Common Mistakes:

• Forgetting to import File
• Using sync function with async file.read()
• Not awaiting file.read()

5. You want to create a FastAPI endpoint that accepts a single file upload and returns the file's name, content type, and first 10 bytes as a UTF-8 string. Which code snippet correctly implements this?

hard

A. from fastapi import FastAPI, UploadFile, File
app = FastAPI()

@app.post('/upload')
async def upload(file: UploadFile = File(...)):
content = await file.read()
preview = content[:10].decode('utf-8')
return {'filename': file.filename, 'content_type': file.content_type, 'preview': preview}

B. from fastapi import FastAPI, UploadFile
app = FastAPI()

@app.post('/upload')
def upload(file: UploadFile):
content = file.read()
preview = content[:10].decode('utf-8')
return {'filename': file.filename, 'content_type': file.content_type, 'preview': preview}

C. from fastapi import FastAPI, UploadFile, File
app = FastAPI()

@app.post('/upload')
async def upload(file: bytes = File(...)):
preview = file[:10].decode('utf-8')
return {'filename': 'unknown', 'content_type': 'unknown', 'preview': preview}

D. from fastapi import FastAPI, UploadFile, File
app = FastAPI()

@app.post('/upload')
async def upload(file: UploadFile = File(...)):
content = await file.read()
preview = content.decode('utf-8')[:10]
return {'filename': file.filename, 'content_type': file.content_type, 'preview': preview}

Solution

1. Step 1: Use async function with UploadFile and File(...)

   Correct signature is async def with parameter file: UploadFile = File(...).

2. Step 2: Read file content and decode first 10 bytes

   Read content with await file.read(), then slice first 10 bytes and decode to UTF-8 string.

3. Step 3: Return filename, content_type, and preview string

   Return dictionary with correct file info and preview snippet.

4. Final Answer:

   Option A code snippet -> Option A

5. Quick Check:

   Async read, slice, decode first 10 bytes [OK]

Hint: Await read, slice first 10 bytes, decode UTF-8 [OK]

Common Mistakes:

• Using sync function without await
• Decoding entire content before slicing
• Using bytes type instead of UploadFile