Django：如何将字节 object 保存到 models.FileField？

Question

My web application has the following structure:

1. backend with Django
2. frontend with React.

I have a form with React. I send a file from client form and I receive the file in my Django application with an APIView.

I receive a m3u file as bytes object.

b'------WebKitFormBoundaryIaAPDyj9Qrx8DrWA\r\nContent-Disposition: 
form-data; name="upload"; 
filename="test.m3u"\r\nContent-Type: audio/x- 
mpegurl\r\n\r\n#EXTM3U\n#EXTINF:-1 tvg-ID="" tvg-name="...

I would save the file in a Django model to a models.FileField and convert bytes object to m3u file. How you do it?

Answer 1

1. models.FileField ( models.ImageField ) needs django.core.files.base.File like objects ex)

   • django.core.files.images.ImageFile

   • django.core.files.base.ContentFile

2. ImageFile or ContentFile needs two args.

   1. IO object : which has seek() method (ex) io.BytesIO ).

   2. name : str . (important! without name, it will not works).

3. bytes object doesn't have IO's methods(ex) seek() ). it should be converted to IO object.

---

models.py

class Message(models.Model):
    image = models.ImageField(upload_to='message_image/%Y/%m', null=True)

---

views.py or consumers.py or some-where.py

import io
from django.core.files.images import ImageFile
from myapp.models import Message

def create_image_message(image_bytes):
    image = ImageFile(io.BytesIO(image_bytes), name='foo.jpg')  # << the answer!
    new_message = Message.objects.create(image=image)

Answer 2

You can try:

from django.core.files.base import ContentFile
import base64

file_data = ContentFile(base64.b64decode(fileData))
object.file.save(file_name, file_data)

You can use your file_name with an .m3u extension, and you shall have it.

Answer 3

I solved using temporary file. I used this code:

extM3u = str.encode(body.decode('utf8').split('EXTM3U\n#')[1].split('------WebKitFormBoundary')[0])

fileTemp = NamedTemporaryFile(delete=True, dir='media/tmp')
fileTemp.write(extM3u)
filenameRe = re.compile('.*?filename=[\'"](.*?)[\'"]')
filename = regParse(filenameRe, body.decode('utf8'))
file = File(fileTemp, name=filename)
m3u = M3u(titleField=filename, fileField=file)
m3u.save()

Answer 4

You can use a ContentFile . To do so:

from django.core.files.base import ContentFile

content_file = ContentFile(file_bytes, name=file_name)
# Assumes MyModel has a FileField called `file`
MyModel.objects.create(file=content_file)

Answer 5

convert string to bytes

bytes_data = ... # bytes

string_data = bytes_data.hex() # this is a string that you can save in TextField in django model

then, to get bytes from the string:

bytes_data_2 = bytes.fromhex(string_data)

I apologize for the crooked translation, I'm still only learning English.