Question

这是我的models.py

from django.db import models
from django.urls import reverse
from django.contrib.auth.models import User
from userprofiles.models import UserProfile
# Create your models here.

import os

class Category (models.Model):
    name = models.CharField(max_length=100)
    def __str__(self):
        return self.name

class Image(models.Model):
    owner = models.ForeignKey(UserProfile, on_delete=models.CASCADE)
    category = models.ForeignKey(Category, on_delete=models.CASCADE)
    in_gallery = models.ManyToManyField(Gallery)
    title = models.CharField(max_length=100)
    def __str__(self):
        return self.title

class Gallery(models.Model):
    name = models.CharField(max_length=100)
    have_images = models.ManyToManyField(Image)

    def __str__(self):
        return self.name

class ImageInGallery (models.Model):
    image = models.ForeignKey(Image)
    gallery = models.ForeignKey(Gallery)


line 22, in Image     in_gallery = models.ManyToManyField(Gallery)
NameError: name 'Gallery' is not defined

我希望图像和图库以及类ImageInGallery有很多关系来存储Image~Gallery关系集。

另外，我希望图库中的图片可以从图库中访问，反之亦然，因此我将models.ManyToManyField(Gallery)和models.ManyToManyField(Image)放在两个类中

出现错误的原因：

line 22, in Image     in_gallery = models.ManyToManyField(Gallery)
NameError: name 'Gallery' is not defined

Answer 1

将Gallery模型放在Image

之上

<强>实施例

from django.db import models
from django.urls import reverse
from django.contrib.auth.models import User
from userprofiles.models import UserProfile
# Create your models here.

import os

class Category (models.Model):
    name = models.CharField(max_length=100)
    def __str__(self):
        return self.name

class Gallery(models.Model):
    name = models.CharField(max_length=100)
    have_images = models.ManyToManyField(Image)

    def __str__(self):
        return self.name

class Image(models.Model):
    owner = models.ForeignKey(UserProfile, on_delete=models.CASCADE)
    category = models.ForeignKey(Category, on_delete=models.CASCADE)
    in_gallery = models.ManyToManyField(Gallery)
    title = models.CharField(max_length=100)
    def __str__(self):
        return self.title



class ImageInGallery (models.Model):
    image = models.ForeignKey(Image)
    gallery = models.ForeignKey(Gallery)

Answer 2

Image和Gallery之间存在重复关系。

class Image(models.Model):
    in_gallery = models.ManyToManyField(Gallery)

class Gallery(models.Model):
    have_images = models.ManyToManyField(Image)

您只需要定义其中一个字段，然后就可以像docs一样对其进行描述。

如果删除字段Gallery.have_images，则可以：

# list all galleries a image is part of
image_instance.in_gallery.all()

# list all images from a gallery
gallery_instance.image_set.all()

Answer 3

只需按照您想要的方式保留您的结构，并将模型名称放在引号内像：

have_images = models.ManyToManyField("Image")

如果模特在另一个应用程序中，请使用该应用程序的名称实施例

have_images = models.ManyToManyField("app_name.Image")

Django：ManyToMany关联名称未定义

3 个答案: