python - 如何从模型中获取加扰的图像文件名以查看django

Question

我正在尝试获取将要上传的图像的乱序名称并将其发送到 View 。在 View 中，我知道如何获取图像的真实名称，但如何从模型中获取乱序名称？

如果有人知道，我会朝这个方向尝试，希望您能告诉我，我是否朝正确的方向前进:

opts = UploadImage._meta
    print ([f.name for f in opts.fields])
    print ( [(field.name, getattr(UploadImage,field.name)) for field in UploadImage._meta.fields])
    print (UploadImage._meta.get_field('image'))
    print([f.name for f in opts.many_to_many])

模型.py

from django.db import models
import uuid

def scramble_uploaded_filename(instance, filename):
    extension = filename.split(".")[-1]
    return "{}.{}".format(uuid.uuid4(), extension)

def filename(instance, filename):
    return filename

# Create your models here.
class UploadImage(models.Model):
#    print (scramble_uploaded_filename)
    image = models.ImageField("Uploaded image",upload_to=scramble_uploaded_filename)
    img_type = models.IntegerField("Image type")
    created = models.DateTimeField(auto_now_add=True)
    user = models.CharField(default='1',max_length= 100,editable=False)

views.py

import sys,os
sys.path.append(os.path.join(os.path.dirname(__file__), '..'))
from rest_framework import viewsets
from api.models import UploadImage
from api.serializers import UploadedImageSerializer
from rest_framework import authentication, permissions
from rest_framework.parsers import MultiPartParser,FormParser
import MySQLdb


class FileUploadViewSet(viewsets.ModelViewSet):
    #create queryset view
    permission_classes = (permissions.IsAuthenticated,)
    queryset = UploadImage.objects.filter(id=1,user='auth.User')
    serializer_class = UploadedImageSerializer
    parser_classes = (MultiPartParser, FormParser,)

    #after post action get this
    def perform_create(self, serializer):
        #grab request
        image_name = self.request.data['image']
        username = self.request.user
        #grab scrambled filename

Answer 1

如果这是您要求的原因，图像处理应该在模型中完成。 同样在 fillen 函数中，您可以找到图像的名称。

将此添加到 model.py

# Create your models here.

class UploadImage(models.Model):
#    print (scramble_uploaded_filename)
    image = models.ImageField("Uploaded image",upload_to=scramble_uploaded_filename)
    img_type = models.IntegerField("Image type")
    created = models.DateTimeField(auto_now_add=True)
    user = models.CharField(default='1',max_length= 100,editable=False)
    def filen(self):
        path = self.image.path
        print('image path',path)
        tip = self.captcha_type
        text = **here is your function that returns result of image process**
        return text 
    result = filen