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How to Upload File using FastAPI?
Home Backend Development Python Tutorial Why is my FastAPI file upload always empty, and how can I fix it?

Why is my FastAPI file upload always empty, and how can I fix it?

Dec 09, 2024 am 06:53 AM

Why is my FastAPI file upload always empty, and how can I fix it?

How to Upload File using FastAPI?

Problem:

When attempting to upload a file using FastAPI according to the official documentation, the file2store variable is consistently empty. Rare instances of successful file byte retrieval occur, but this is uncommon.

Solution:

1. Install Python-Multipart:

To enable file uploading, which is transmitted as "form data," install python-multipart if you haven't already:

pip install python-multipart
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2. Use .file Attribute for Single File Upload:

Use the .file attribute of the UploadFile object to obtain the actual Python file (i.e., SpooledTemporaryFile). This allows you to call methods like .read() and .close().

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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3. Handle Large Files:

Use chunks if the file exceeds the 1MB memory limit. Adjust the chunk size as needed.

4. Asynchronous Reading/Writing:

If your endpoint requires async def, use asynchronous methods for reading and writing the file contents.

5. Upload Multiple Files:

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

    return {"message": f"Successfully uploaded {[file.filename for file in files]}."}
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6. HTML Form Example:

Refer to the provided link for an example HTML form to upload files.

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