Using python script to upload a file to django server localhost

Question

I build an django rest framework that can help me to upload a file

in view.py

class FileResultViewSet(viewsets.ModelViewSet):
    queryset = FileResult.objects.order_by('-timestamp')
    serializer_class = FileResultSerializer
    parser_classes = (MultiPartParser, FormParser)

    def perform_create(self, serializer):
        serializer.save()

The script work well without any error, and the file is stored in media folder

Then, I use an python script that can help me to upload file to django

import requests
import openpyxl

url = "http://localhost:8000/upload/"
headers = {'Content-Type': 'multipart/form-data'}
file_path = "H:/MEGA_H/Data/uploaded/Book1.xlsx"
'''
with open(file_path, 'rb') as f:
    file = {'file': f}
    response = requests.get(url, headers=headers, files=file)
'''

filename = "Book1.xlsx"

# Open the file and read its contents

with open(file_path, 'rb') as f:
    file_contents = f.read()

# Create a dictionary to store the file information
#file_data = {'file': (filename, file_contents,result_file)}
file_data = {
    "result_file": file_contents
}


# Make a POST request to the URL with the file information
response = requests.post(url, files=file_data)

# Check if the request was successful
if response.status_code == 201:
    print("File successfully uploaded")
else:
    print("Failed to upload file")

The code still works with an issue, after upload the file to media folder, the file doesnt include the file type, so it can not read. I have to change the name of the file , and typing file type ".xlsx"

The file name is only Book1, without Book1.xlsx

John Gordon · Accepted Answer · 2023-02-12 17:31:05Z

1

with open(file_path, 'rb') as f:
    file_contents = f.read()

file_data = {
    "result_file": file_contents
}

This is the wrong way to do it. The dict should contain the file object itself, not the file contents.

with open(file_path, 'rb') as f:
    file_data = {"result_file": f}
    response = requests.post(url, files=file_data)

answered Feb 12, 2023 at 17:31

John Gordon

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