API : specify filename when uploading a file

Technical Help
,
1

Hi,

When uploading a file using the upload-file API, can you specifiy the original name of the file ?
Something similar to this (see the original_name part I added):

requests.post(
        url = "http://vision.axe.lan:3001/api/user-files/upload-file/",
        headers = {"Authorization": f"Token {token}"},
        files =  {'file': data, 'original_name': 'toto.jpg'}
)

I’m asking because, the file I’m trying to upload comes from a portal in base64 format. I’m just trying to transfer it to baserow without having to save it localy. Just passing the binary data…

2

Hi,

You cannot specify the original file name, but if you pass a file through the API with the following code, the response should be a json object containing the original file name as one of its properties.

# Upload the file to the baserow instance
file_url = 'http://vision.axe.lan:3001/api/user-files/upload-file/'  
# Select the file
csv_file_path = 'demo.csv'  

# Prepare the file for multipart upload
with open(csv_file_path, 'rb') as file:
    file_response = requests.post(file_url, headers=auth_header, files={'file': ('demo.csv', file, 'text/csv')})

# Store the uploaded file name
original_filename = file_response.json().get('original_name')
print(original_filename)