Interview question

How do you handle file uploads with Django forms? Django forms के साथ file uploads कैसे handle करें?

Answer

File uploads require the form's enctype to be multipart/form-data and the view must access request.FILES separately from request.POST.

from django import forms
from .models import Book

class BookForm(forms.ModelForm):
    class Meta:
        model = Book
        fields = ['title', 'cover']  # cover is an ImageField/FileField on the model

def upload_book(request):
    if request.method == 'POST':
        # BOTH request.POST (text fields) and request.FILES (file fields) needed
        form = BookForm(request.POST, request.FILES)
        if form.is_valid():
            form.save()
            return redirect('book-list')
    else:
        form = BookForm()
    return render(request, 'upload.html', {'form': form})

<!-- Template - enctype is REQUIRED for file uploads -->
<form method='post' enctype='multipart/form-data'>
    {% csrf_token %}
    {{ form.as_p }}
    <button type='submit'>Upload</button>
</form>
<!-- Without enctype='multipart/form-data', file data is silently NOT sent -->

# Model with a file field
class Book(models.Model):
    title = models.CharField(max_length=200)
    cover = models.ImageField(upload_to='covers/%Y/%m/')  # organized by year/month

# Validating uploaded file size/type manually
def validate_file_size(value):
    max_size_mb = 5
    if value.size > max_size_mb * 1024 * 1024:
        raise forms.ValidationError(f'File size cannot exceed {max_size_mb}MB')

class BookForm(forms.ModelForm):
    cover = forms.ImageField(validators=[validate_file_size])
    class Meta:
        model = Book
        fields = ['title', 'cover']

# Accessing an uploaded file's properties directly in a view
def check_upload(request):
    if request.method == 'POST':
        uploaded_file = request.FILES.get('cover')
        if uploaded_file:
            print(uploaded_file.name, uploaded_file.size, uploaded_file.content_type)

File uploads के लिए form का enctype multipart/form-data होना चाहिए और view को request.FILES को request.POST से अलग access करना चाहिए।

from django import forms
from .models import Book

class BookForm(forms.ModelForm):
    class Meta:
        model = Book
        fields = ['title', 'cover']

def upload_book(request):
    if request.method == 'POST':
        form = BookForm(request.POST, request.FILES)
        if form.is_valid():
            form.save()
            return redirect('book-list')
    else:
        form = BookForm()
    return render(request, 'upload.html', {'form': form})

<!-- enctype ज़रूरी है -->
<form method='post' enctype='multipart/form-data'>
    {% csrf_token %}
    {{ form.as_p }}
    <button type='submit'>Upload</button>
</form>

class Book(models.Model):
    title = models.CharField(max_length=200)
    cover = models.ImageField(upload_to='covers/%Y/%m/')

def validate_file_size(value):
    max_size_mb = 5
    if value.size > max_size_mb * 1024 * 1024:
        raise forms.ValidationError(f'File size {max_size_mb}MB से ज़्यादा नहीं हो सकता')

class BookForm(forms.ModelForm):
    cover = forms.ImageField(validators=[validate_file_size])
    class Meta:
        model = Book
        fields = ['title', 'cover']

Was this answer clear?

Subjects

Related questions