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I am using Gradio to build the UI. I am using the following code to upload a file.

import gradio as gr

def upload_file(files):
    file_paths = [file.name for file in files]
    return file_paths

with gr.Blocks() as demo:
    file_output = gr.File()
    opt = gr.Label()
    upload_button = gr.UploadButton("Click to Upload a File", file_types=["image", "video"], file_count="multiple")
    upload_button.upload(upload_file, upload_button, file_output)

demo.launch()

How to get the file location in a variable?

Nithin
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2 Answers2

1

Not sure if you're still looking for an answer, but for anyone else who is dealing with file uploads when using Gradio I have found the following function (and variations of it) that makes use of the shutil library infinitely useful!

def process_file(fileobj):
    path = "/home/ubuntu/temps/" + os.path.basename(fileobj)
    shutil.copyfile(fileobj.name, path)
    # now you can process the file at path as needed, e.g:
    # do_something_to_file(path)

Without doing this, I found I had a lot of issues treating the file initially uploaded to Gradio as a tempfile._TemporaryFileWrapper object. I found I was encountering a lot of permission issues and it was hard to do everything I wanted with the object and the tempfile library.

This new method with shutil gives you complete control over a permanent file object, and if you need to delete it afterwards just add in the code to do so when you're done.

In the context of using this with Gradio, it would work in a simple example as follows:

import gradio as gr
import os
import shutil

def process_file(fileobj):
    path = "/home/ubuntu/temps/" + os.path.basename(fileobj)  #NB*
    shutil.copyfile(fileobj.name, path)
    return do_something_to_file(path)

demo = gr.Interface(
    fn=process_file,
    inputs=[
        "file",
    ],
    outputs="text"
)
demo.launch(server_name='0.0.0.0')

NB: I am doing this on an Ubuntu instance, obviously please modify path name accordingly for your operating system and needs. If you have issues with this method check that you and your python script have permission to write to whichever directory you specify in your path.

KingRanTheMan
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You could use the following code:

import gradio as gr

def process_file(fileobj): 
    file_path = fileobj.name 
    return do_something(file_path)

demo = gr.Interface(
    fn=process_file,
    inputs=[
        "file",
    ],
    outputs="text"
)

demo.launch(server_name='0.0.0.0')
Vedant Desai
  • 1
  • Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Aug 04 '23 at 04:51