How to process a file uploaded using a form in Django?

Question

I built a form with a FileField in order to upload file to be processed. Strangely it seem that file is closed before I can do anything with in the view. I encounter a ValueError after validating the form:

"I/O operation on closed file"

I can see that file is properly detected but closed when attempting to read it.

Note: Django version 2.2.25.

forms.py

class FileImportForm(forms.Form):
    headers = ["lastname","firstname","gender","title","entity","email","company","address","phone"]
    file = forms.FileField(label='CSV file',validators=[CsvFileValidator(headers)])
    
    def clean_file(self):
        file = self.cleaned_data['file']
        return file   

views.py

@login_required
def file_import(request):
    if request.method == 'POST': 
        form = FileImportForm(request.POST,request.FILES)

        if form.is_valid():
            if request.FILES['file']:
                file_post = request.FILES['file']
                # Offending line below (I/O operation on closed file)
                file_content = file_post.read().decode('UTF-8')

                return redirect("/foo")
    else:
        form = FileImportForm()
    
    return render(request,"file_import.html", { 'form': form })

How to properly process uploaded file (read, etc..)?

Answer 1

The error occurs because you didn't open your file properly. Here's the working way:

file_post = request.FILES['file']
f = open(file_post, "w")

As python's file operation also comes with a read-only method, you cannot do file_content = file_post.read().decode('UTF-8') directly because you didn't make this file writable.

Reference.

EDIT

As the upload file type may be incorrect, more conversions are needed. Something like:

def my_view(request):
    uploaded_file = request.FILES['file']
    str_text = ''
    for line in uploaded_file:
        str_text = str_text + line.decode()  # "str_text" will be of `str` type
    # do something
    return something

Reference2