FastAPI Tutorial: How to upload any file to server using FastAPI #1

Welcome to our FastAPI tutorial! In this tutorial, we will be learning how to upload any file to a server using FastAPI. FastAPI is a modern, fast (high-performance), web framework for building APIs with Python 3.6+.

Prerequisites

Before we begin, you will need to have the following installed on your machine:

• Python 3.6+
• FastAPI
• Uvicorn (ASGI server)

Getting Started

First, let’s create a new Python file for our FastAPI application. You can name it app.py. Here’s a simple example of a FastAPI application that allows file upload:

      from fastapi import FastAPI, File, UploadFile

      app = FastAPI()

      @app.post("/uploadfile/")
      async def upload_file(file: UploadFile = File(...)):
          contents = await file.read()
          # Save the file to a specific location
          with open("uploaded_" + file.filename, "wb") as f:
              f.write(contents)
          return {"filename": file.filename}
    

In this example, we define a route /uploadfile/ that accepts a POST request and a file upload. The file parameter is of type UploadFile, which represents a file being uploaded.

Testing the File Upload

To test the file upload, we can use the curl command-line tool or any HTTP client like Postman. Here’s an example using curl:

      $ curl -X POST "http://localhost:8000/uploadfile/" -H "accept: application/json" -H "Content-Type: multipart/form-data" -F "file=@/path/to/file.txt"
    

Replace /path/to/file.txt with the path to the file you want to upload. After executing the curl command, you should receive a JSON response with the filename of the uploaded file.

Conclusion

Congratulations! You have learned how to upload any file to a server using FastAPI. In this tutorial, we covered the basics of file upload with FastAPI and created a simple file upload endpoint. FastAPI provides a convenient way to handle file uploads and many other features for building modern web APIs.