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Code Block
languagepy
# models.py
import uuid


from django.db import models

# 저장할 파일 이름을 반환합니다.
def getFileName(instance, filename):
    ext = filename.split('.')[-1]
    return "%s.%s" % (uuid.uuid4(), ext)

class UploadFile(models.Model):
    file = models.FileField(upload_to=getFileName)

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