Tag: django

To handle file uploads from an HTML form in your Python Frameworks, you typically handle them differently depending on the web framework you’re using (e.g., Flask, Django, FastAPI, Frappe, etc.).

Note: HTML form must include enctype="multipart/form-data".

In this article, we learn how to handle file uploads in some common Python web frameworks:

Using Flask

Design an HTML form with file field as follows,

<form action="/upload" method="post" enctype="multipart/form-data">
  <input type="file" name="myfile">
  <input type="submit">
</form>

You can get the uploaded file from the myfile field as follows,

from flask import Flask, request

app = Flask(__name__)

@app.route('/upload', methods=['POST'])
def upload_file():
    uploaded_file = request.files['myfile']
    if uploaded_file.filename != '':
        uploaded_file.save(f"./uploads/{uploaded_file.filename}")
    return "File uploaded successfully!"

Using Django

In Django, HTML form is exactly same. But, you have to add csrf_token to each form.

<form method="post" enctype="multipart/form-data">
  {% csrf_token %}
  <input type="file" name="myfile">
  <input type="submit">
</form>

You can get the file from the myfile field as follows,

from django.core.files.storage import FileSystemStorage
from django.http import HttpResponse

def upload_file(request):
    if request.method == 'POST' and request.FILES['myfile']:
        uploaded_file = request.FILES['myfile']
        fs = FileSystemStorage()
        fs.save(uploaded_file.name, uploaded_file)
        return HttpResponse('File uploaded!')
    return HttpResponse('Upload form')

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Using FastAPI

The HTML design file is same as the Django HTML file with csrf_token.

Backend code to get the uploaded file is as follows,

from fastapi import FastAPI, File, UploadFile
from fastapi.responses import HTMLResponse

app = FastAPI()

@app.post("/upload/")
async def upload_file(file: UploadFile = File(...)):
    contents = await file.read()
    with open(f"./uploads/{file.filename}", "wb") as f:
        f.write(contents)
    return {"filename": file.filename}

Using Frappe

In Frappe, when you submit a file from an HTML form, you can handle file uploads using its built-in frappe.utils.file_manager.save_file() method. The HTML design file for Frappe is as follows,

<form method="POST" enctype="multipart/form-data" action="/api/method/my_app.api.upload_file">
  <input type="file" name="file">
  <input type="submit" value="Upload">
</form>

In Frappe backend, you have to create whitelisted method for handling file upload as follows,

import frappe
from frappe.utils.file_manager import save_file
from frappe import _

@frappe.whitelist(allow_guest=True)  # or remove `allow_guest=True` if auth is needed
def upload_file():
    # Access the uploaded file from the request
    uploaded_file = frappe.request.files.get('file')
    
    if not uploaded_file:
        frappe.throw(_("No file uploaded"))

    # Save the file using Frappe's file manager
    saved_file = save_file(
        filename=uploaded_file.filename,
        content=uploaded_file.stream.read(),
        dt=None,  # You can pass doctype here if you want to attach it
        dn=None   # and document name
    )

    return {
        "message": "File uploaded successfully",
        "file_url": saved_file.file_url,
        "file_name": saved_file.file_name
    }

Conclusion

In above post, we learned file uploads using four different Python backed frameworks. We learned that different framework have different ways to handle file uploads.

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