Uploading A Local File With HTTP POST

noellarkin · Aug 17, 2022

For some reason, the answer to this has been impossible for me to find on StackOverflow.
I want to upload a local file to an API.
The file is located at C:\samplefolder\samplefile.wav

The API documentation tells me to upload it like this:

Python:

import requests
filename = "/path/to/foo.wav"
def read_file(filename, chunk_size=5242880):
    with open(filename, 'rb') as _file:
        while True:
            data = _file.read(chunk_size)
            if not data:
                break
            yield data

headers = {'authorization': "YOUR-API-TOKEN"}
response = requests.post('https://api.assemblyai.com/v2/upload',
                        headers=headers,
                        data=read_file(filename))

print(response.json())

Here's the part I don't get. filename = "/path/to/foo.wav" is how every single file upload example on the internet seems to do it. But filename = "C:\samplefolder\samplefile.wav" doesn't seem to work.
How do I upload a local file from my hard drive, and why is it so effin' hard to find an example of this on the internet?

noellarkin · Aug 17, 2022

Update:
I also tried formatting the file as
file:///C://samplefolder//samplefile.wav - - if I enter this in my browser navigation bar it opens the file.
but yeah doesn't work for the API endpoint either.

codingMinds · Aug 17, 2022

Can you please tell me what error your are getting while uploading the file?

Smackhead · Aug 17, 2022

Need to figure out if you error Is in the request of when finding the file.. are you able to see the http response?

noellarkin · Aug 17, 2022

It's happening with every single code snippet that has to do with uploading a file to an API endpoint, not just the example snippet I used.
I'm using Insomnia (similar to Postman) to test requests now.
If I use the "multipart form" format, with params
Content-Disposition: form-data
name: file
filename: file:///C://samplefolder//samplefile.wav
and with the Header
Content-Type: multipart/form-data

I get errors like:
{
"error": "File not found! Attach a file to your request."
}

What's frustrating is every example I see has something like filename: samplefile.wav
How does one set the working folder?

Smackhead · Aug 17, 2022

I'd have expected C://samplefolder//samplefile.wav to work. Try executing your script from the same folder as the file is located, and set the path to either the file name samplefile.wav or ./samplefile.wav

BeZzzymec · Aug 17, 2022

Try "C:\\samplefolder\\samplefile.wav"

RRX · Aug 17, 2022

Python:

import requests
import os
filename = f"{os.getcwd}/path/to/foo.wav"
def read_file(filename, chunk_size=5242880):
    with open(filename, 'rb') as _file:
        while True:
            data = _file.read(chunk_size)
            if not data:
                break
            yield data

headers = {'authorization': "YOUR-API-TOKEN"}
response = requests.post('https://api.assemblyai.com/v2/upload',
                        headers=headers,
                        data=read_file(filename))

print(response.json()

Maybe?
what i did was added os.getcwd() to get current working directory

or if you want to make an absolute path try

Python:

import requests

import os

filename = f"{os.path.abspath}/path/to/foo.wav"

def read_file(filename, chunk_size=5242880):

    with open(filename, 'rb') as _file:

        while True:

            data = _file.read(chunk_size)

            if not data:

                break

            yield data

headers = {'authorization': "YOUR-API-TOKEN"}

response = requests.post('https://api.assemblyai.com/v2/upload',

                        headers=headers,

                        data=read_file(filename))
print(response.json()

Uploading A Local File With HTTP POST

noellarkin

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noellarkin

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codingMinds

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Smackhead

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noellarkin

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Smackhead

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BeZzzymec

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RRX

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