如何将Django访问者文件上载到新的子目录，并且目录名应为会话名？

我正在创建一个django应用程序。它只有一个模型类是文件上传。它没有用户登录和注册系统

我的目标是，当有人访问网页并上传文件时，它会在

“/media”

目录中创建一个新目录。目录名应该是唯一的，也应该是会话名。仔细阅读：

当访问者访问页面并上传文件时，应自动创建一个新目录“a”，即使访问者重新加载页面并上传另一个新文件，也应创建另一个新目录“B”，并将会话上传的文件存储在目录“B”上。

My

model.py

from django.db import models

# Create your models here.
class ImageForExtractText(models.Model):
    image = models.FileField()

from django import forms
from .models import ImageForExtractText

class UploadImage(forms.ModelForm):
    class Meta:
        model = ImageForExtractText
        fields = [
            'image',
        ]

from .forms import UploadImage
from django.shortcuts import render, redirect

def upload(request):
    form = UploadImage(request.POST or None, request.FILES or None)
    if form.is_valid():
        instance = form.save(commit=False)
        instance.save()
    return render(request, 'home.html', {'form': form})

MEDIA_URL = '/media/'
MEDIA_ROOT = os.path.join(os.path.dirname(BASE_DIR), "media")

和我的

form.py

from django.db import models

# Create your models here.
class ImageForExtractText(models.Model):
    image = models.FileField()

from django import forms
from .models import ImageForExtractText

class UploadImage(forms.ModelForm):
    class Meta:
        model = ImageForExtractText
        fields = [
            'image',
        ]

from .forms import UploadImage
from django.shortcuts import render, redirect

def upload(request):
    form = UploadImage(request.POST or None, request.FILES or None)
    if form.is_valid():
        instance = form.save(commit=False)
        instance.save()
    return render(request, 'home.html', {'form': form})

MEDIA_URL = '/media/'
MEDIA_ROOT = os.path.join(os.path.dirname(BASE_DIR), "media")

和

view.py

from django.db import models

# Create your models here.
class ImageForExtractText(models.Model):
    image = models.FileField()

from django import forms
from .models import ImageForExtractText

class UploadImage(forms.ModelForm):
    class Meta:
        model = ImageForExtractText
        fields = [
            'image',
        ]

from .forms import UploadImage
from django.shortcuts import render, redirect

def upload(request):
    form = UploadImage(request.POST or None, request.FILES or None)
    if form.is_valid():
        instance = form.save(commit=False)
        instance.save()
    return render(request, 'home.html', {'form': form})

MEDIA_URL = '/media/'
MEDIA_ROOT = os.path.join(os.path.dirname(BASE_DIR), "media")

和my

setting.py

from django.db import models

# Create your models here.
class ImageForExtractText(models.Model):
    image = models.FileField()

from django import forms
from .models import ImageForExtractText

class UploadImage(forms.ModelForm):
    class Meta:
        model = ImageForExtractText
        fields = [
            'image',
        ]

from .forms import UploadImage
from django.shortcuts import render, redirect

def upload(request):
    form = UploadImage(request.POST or None, request.FILES or None)
    if form.is_valid():
        instance = form.save(commit=False)
        instance.save()
    return render(request, 'home.html', {'form': form})

MEDIA_URL = '/media/'
MEDIA_ROOT = os.path.join(os.path.dirname(BASE_DIR), "media")

---

希望我清楚我的问题是什么，如果我说的任何事情都不清楚，请告诉我

您可以将函数引用传递到文件字段的

upload\u

属性。这个函数应该有一个动态路径（一个随机目录），它将像这样自动创建

import os
import uuid
from django.conf import settings
from django.db import models


def image_path(_, file_name):
    return os.path.join(settings.MEDIA_ROOT, str(uuid.uuid4()), file_name)


# Create your models here.
class ImageForExtractText(models.Model):
    image = models.FileField(upload_to=image_path)

---

您可以将函数引用传递到文件字段的

upload\u

属性。这个函数应该有一个动态路径（一个随机目录），它将像这样自动创建

import os
import uuid
from django.conf import settings
from django.db import models


def image_path(_, file_name):
    return os.path.join(settings.MEDIA_ROOT, str(uuid.uuid4()), file_name)


# Create your models here.
class ImageForExtractText(models.Model):
    image = models.FileField(upload_to=image_path)

---

谢谢你的解决方案，但我需要它应该是季节名称，我怎么做？谢谢你的解决方案，但我需要它应该是季节名称，我怎么做？