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}

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

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}

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

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}

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

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.

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