How to get file path from UploadFile in FastAPI?

Question

Basically, I'm trying to create an endpoint to Upload files to Amazon S3.

async def upload_files(filepath: str, upload_file_list: List[UploadFile] = File(...)):
    for upload_file in upload_file_list:
        abs_file_path = "/manual/path/works" + upload_file.path
        # Replace above line to get absolute file path from UploadFile
        response = s3_client.upload_file(abs_file_path,bucket_name,
                                                   os.path.join(dest_path, upload_file.filename))

Above is my code to upload multiple files to the S3 bucket. s3_client.upload_file() accepts an absolute file path of the file to upload. It is working when I manually put the full path.

This, however, didn't work:

response = s3_client.upload_file(upload_file.filename, bucket_name,
                                                   os.path.join(dest_path, upload_file.filename))

Is there a way to get this absolute path in FastAPI? Or, any alternative with temp_path without copying or writing the file?

If not, then any alternative with boto3 to upload files to S3 using FastAPI?

Answer 1

UploadFile uses Python's SpooledTemporaryFile, which is a "file stored in memory", and "is destroyed as soon as it is closed". You can either read the file contents (i.e., contents = await file.read()) and then upload these bytes to your server (if it permits), or copy the contents of the uploaded file into a NamedTemporaryFile, as explained here. Unlike SpooledTemporaryFile, NamedTemporaryFile "is guaranteed to have a visible name in the file system" that "can be used to open the file". The temp file's path can be accessed through file_copy.name

contents = await file.read()
file_copy = NamedTemporaryFile('wb', delete=False)
f = None
try:
    # Write the contents to the temp file
    with file_copy as f:
        f.write(contents);

    # Here, upload the file to your S3 service
    f = open(file_copy.name, 'rb')
    print(f.read(10))

finally:
    if f is not None:
        f.close() # Remember to close the file
    os.unlink(file_copy.name)  # delete the file

Update

Additionally, one can access the actual Python file using the file attribute. As per the documentation:

file: A SpooledTemporaryFile (a file-like object). This is the actual Python file that you can pass directly to other functions or libraries that expect a "file-like" object.

Thus, you could also try using upload_fileobj function and passing upload_file.file:

 response = s3_client.upload_fileobj(upload_file.file, bucket_name, os.path.join(dest_path, upload_file.filename))

or, passing a file-like object using the _file attribute of the SpooledTemporaryFile, which returns either an io.BytesIO or io.TextIOWrapper object (depending on whether binary or text mode was specified).

 response = s3_client.upload_fileobj(upload_file.file._file, bucket_name, os.path.join(dest_path, upload_file.filename))

Update 2

You could even keep the bytes in an in-memory buffer BytesIO, use it to upload the contents to the S3 bucket, and finally close it ("The buffer is discarded when the close() method is called."). Remember to call seek(0) method to reset the cursor back to the beginning of the file after you finish writing to the BytesIO stream.

contents = await file.read()
temp_file = io.BytesIO()
temp_file.write(contents)
temp_file.seek(0)
s3_client.upload_fileobj(temp_file, bucket_name, os.path.join(dest_path, upload_file.filename))
temp_file.close()