CODEWITHSUNDEEP
pythondjango
NachoMiguel
NachoMiguel

Reputation: 993

Python/Django - using a view method on an upload file

I have this view: 

def index(request):
    file = open("SK ✌😜✌.txt", encoding="UTF-8")
    data = file.read()
    file.close()

    lines = data.split("\n")
    ...More code...

In this view i open a file from the very first moment the app starts and i do some work on the file, is a story, and when i start the server and go to http://127.0.0.1:8000/(Name Of The App), i see all the work that i have done on that file.

What i want to do is to do that same work, starting with the reading of the file, BUT i want to do that with the file that the user uploads in that moment. I have this that i took from bootstrap:

<div class="form-group">
    <label for="exampleInputFile">File input</label>
    <input type="file" id="exampleInputFile">
</div>

I guess i have to use in some way the id of the input but i`m not really sure how to pass this file that the user uploads in the ui to the method that i have in my views.py

Any help will be really appreciated

Upvotes: 3

Views: 1254

Answers (1)

Rahul Gupta
Rahul Gupta

Reputation: 47846

You need to have a name attribute in your <input> template code.

<input type="file" id="exampleInputFile" name="some_file">

Then to access the file in your view, you need to use request.FILES attribute.

As per the Django docs on HttpRequest.FILES attribute:

A dictionary-like object containing all uploaded files. Each key in FILES is the name from the <input type="file" name="" />. Each value in FILES is an UploadedFile.

Your code should be something like:

def index(request):
    if request.method=="POST":
        uploaded_file = request.FILES['some_file'] # get the uploaded file
        # do something with the file

Note: request.FILES will only contain data if the request method was POST and the <form> that posted to the request had enctype="multipart/form-data. Otherwise, FILES will be a blank dictionary-like object.

Upvotes: 3

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