File upload (single file) in FastAPI - Mini Project: Build & Apply

File upload (single file)

📖 Scenario: You are building a simple web API that allows users to upload a single file. This is common in many web applications where users need to send images, documents, or other files to the server.

🎯 Goal: Create a FastAPI app that accepts a single file upload through a POST request and returns the filename and content type.

📋 What You'll Learn

Use FastAPI to create the web app

Create an endpoint to accept a single file upload

Use the correct type hint for the uploaded file

Return a JSON response with the filename and content type

💡 Why This Matters

🌍 Real World

Uploading files is common in web apps for user profile pictures, documents, or media content.

💼 Career

Backend developers often implement file upload endpoints using FastAPI or similar frameworks.

1

Create FastAPI app instance

Import FastAPI from fastapi and create an app instance called app.

FastAPI

from fastapi import FastAPI

# Create FastAPI app instance
# Your code here

Need a hint?

Use app = FastAPI() to create the app instance.

2

Import UploadFile and File

Import UploadFile and File from fastapi to handle file uploads.

FastAPI

from fastapi import FastAPI, UploadFile, File

app = FastAPI()
# Your code here

Need a hint?

Import both UploadFile and File from fastapi in the same line.

3

Create POST endpoint for file upload

Create a POST endpoint at /uploadfile/ using @app.post. Define an async function called upload_file that accepts a parameter file of type UploadFile using File(...).

FastAPI

from fastapi import FastAPI, UploadFile, File

app = FastAPI()

# Create POST endpoint for file upload
# Your code here

Need a hint?

Use @app.post('/uploadfile/') decorator and define an async function with the correct parameter type.

4

Return filename and content type

Inside the upload_file function, return a dictionary with keys filename and content_type using file.filename and file.content_type.

FastAPI

from fastapi import FastAPI, UploadFile, File

app = FastAPI()

@app.post('/uploadfile/')
async def upload_file(file: UploadFile = File(...)):
    # Return filename and content type
    # Your code here

Need a hint?

Return a dictionary with the file's filename and content_type.