Чтобы программно сохранить изображение в Django ImageField, вы можете использовать различные методы. Вот несколько разных подходов:

Метод 1: использование класса Fileиз модуля django.core.filesDjango

from django.core.files import File
# Assume you have an instance of your model with the `ImageField`
my_model_instance = MyModel()
# Open the image file
with open('path/to/image.jpg', 'rb') as image_file:
    # Create a Django `File` object
    django_file = File(image_file)
    # Assign the `File` object to the `ImageField` of the model instance
    my_model_instance.image_field.save('image.jpg', django_file)

Метод 2: непосредственное использование функции Python open()

import os
from django.conf import settings
# Assume you have an instance of your model with the `ImageField`
my_model_instance = MyModel()
# Open the image file
with open('path/to/image.jpg', 'rb') as image_file:
    # Get the file name and path
    file_name = os.path.basename(image_file.name)
    file_path = os.path.join(settings.MEDIA_ROOT, file_name)
    # Save the image file to the desired location
    with open(file_path, 'wb') as destination:
        for chunk in image_file.chunks():
            destination.write(chunk)
    # Assign the file path to the `ImageField` of the model instance
    my_model_instance.image_field = file_path
    my_model_instance.save()

Метод 3: использование класса ContentFileиз модуля django.core.files.baseDjango

from django.core.files.base import ContentFile
# Assume you have an instance of your model with the `ImageField`
my_model_instance = MyModel()
# Open the image file
with open('path/to/image.jpg', 'rb') as image_file:
    # Read the image content
    image_content = image_file.read()
    # Create a Django `ContentFile` object
    content_file = ContentFile(image_content)
    # Assign the `ContentFile` object to the `ImageField` of the model instance
    my_model_instance.image_field.save('image.jpg', content_file)

Обязательно замените 'path/to/image.jpg'фактическим путем к файлу изображения. Кроме того, 'image_field'следует заменить именем вашего ImageFieldв модели.