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Why is my FastAPI file upload variable always empty?

Susan Sarandon
Release: 2024-12-08 08:20:10
Original
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Why is my FastAPI file upload variable always empty?

How to Upload File using FastAPI?

Background

FastAPI is a modern, high-performance web framework that provides a clean and efficient way to build APIs. File uploading is a fundamental feature required in many web applications. To facilitate this, FastAPI offers multiple options for file upload handling.

Issue

In certain scenarios, when uploading a file according to the FastAPI documentation, the file2store variable consistently remains empty. This issue persists despite attempts to utilize Python's UploadFile or bytes approach.

Solution

The following measures can be taken to resolve this issue:

1. Install Python-Multipart

FastAPI uses form data to transmit uploaded files. Installing "python-multipart" ensures that uploaded files are correctly handled.

pip install python-multipart
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2. Use Memory-Friendly File Handling

FastAPI uses a SpooledTemporaryFile object to store uploaded files in memory. If the file size exceeds 1 MB, it is written to a temporary file on disk. For large files, consider the following options:

  • Read the File in Chunks: Break the file into smaller chunks and process them sequentially. This reduces memory consumption but may prolong processing time.
  • Use shutil.copyfileobj(): Utilize this function to copy the file contents in manageable chunks, minimizing memory usage.

3. Avoid Blocking Endpoints

When handling file uploads, endpoint functions should be defined with def instead of async def. This prevents the server from blocking while waiting for file operations to complete.

4. Allow Additional Data Transfer

If you need to transmit data along with file uploads, review the provided answers for guidance.

5. Understand Endpoint Differences

Become familiar with the distinctions between def and async def endpoints and their implications for file handling.

Example Code for File Upload

Consider the following code snippet as an example:

from fastapi import File, UploadFile

@app.post("/upload")
def upload(file: UploadFile = File(...)):
    try:
        contents = file.file.read()
        with open(file.filename, 'wb') as f:
            f.write(contents)
    except Exception:
        return {"message": "Error uploading file"}
    finally:
        file.file.close()

    return {"message": f"Successfully uploaded {file.filename}"}
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This code handles file uploads and stores the contents in a file with the same name as the uploaded file. If you encounter any issues, consult the provided resources for further assistance.

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