TypeError：“ModelBase”对象不支持索引。答案

【问题标题】：TypeError: 'ModelBase' object does not support indexing.TypeError：“ModelBase”对象不支持索引。
【发布时间】：2019-04-18 11:18:20
【问题描述】：

我创建了 models.py 文件，当我尝试执行“makemigrations”命令时，出现以下错误：

    return bool(self.related_name) and self.related_name[-1] == '+'
TypeError: 'ModelBase' object does not support indexing

我试图找到一个论坛回复，但主题主要涉及从views.py文件中出现的错误。在我的情况下，该文件尚未创建。

任何帮助将不胜感激，包括数据库中更透明的代码。

我的models.py

from django.db import models


#additives
class TestUser(models.Model):
    name = models.CharField(max_length=50)

class Trade(models.Model):
    all_trade_here = models.TextField()

class ImageGallery(models.Model):
    image = models.ImageField()

class SocialMediaLinks(models.Model):
    link = models.URLField()

# Option for premium user
class Prem_PhotoGallery(models.Model):
    name_photo_gallery = models.CharField(max_length=50)
    image_gallery = models.ForeignKey(ImageGallery,
                                      on_delete=models.CASCADE)

class Prem_PDF(models.Model):
    name_pdf = models.CharField(max_length=50)
    file = models.FileField()

class Prem_YouTube(models.Model):
    name_youtube = models.CharField(max_length=50)
    link = models.URLField()

class Prem_VilmVeo(models.Model):
    name_vilmveo = models.CharField(max_length=50)
    link = models.URLField()

class Prem_Audio(models.Model):
    name_audio = models.CharField(max_length=50)
    link = models.URLField()

class Prem_Website(models.Model):
    name_website = models.CharField(max_length=50)
    link = models.URLField()

class Prem_SocialMedia(models.Model):
    name_social_button_1 = models.CharField(max_length=50)
    links = models.ForeignKey(SocialMediaLinks, on_delete=models.CASCADE)

class Prem_GoogleMaps(models.Model):
    name_google_maps = models.CharField(max_length=50)
    link = models.URLField()

# element write by user
class Person(models.Model):
    #basic information, all write user
    name = models.CharField(max_length=100)
    surname = models.CharField(max_length=100)
    phone = models.CharField(max_length=15)
    e_mail = models.EmailField(max_length=100)
    website = models.URLField(max_length=250)
    photo = models.FileField()
    small_photo = models.FileField()
    description = models.TextField()
    recommended_by = models.ForeignKey(TestUser,
                                       on_delete=models.CASCADE)
    my_projects = models.ManyToManyField(Prem_PhotoGallery,
                                         Prem_PDF,
                                         Prem_YouTube,
                                         Prem_VilmVeo,
                                         Prem_Audio,
                                         Prem_Website,
                                         Prem_SocialMedia,
                                         Prem_GoogleMaps,
                                         blank=True)
    keywords = models.TextField()
    trade = models.ManyToManyField(Trade)

    #premium
    premium_accounts = models.BooleanField(default=False)

    #location
    city = models.CharField(max_length=100)
    street = models.CharField(max_length=100)
    zip_code = models.CharField(max_length=8)

    #data joined
    time_create_account = models.DateTimeField()

【问题讨论】：

标签： django sqlite model

【解决方案1】：

my_projects 的 ManyToManyField 配置不正确。 ManyToManyField 将单个类作为它的第一个位置参数。它还可以接受其他可选参数，例如related_name。

您已将多个类作为参数传递。 ManyToManyField 因此会假设这些其他类对应于可选参数的值，从而导致您看到的错误。

您需要将my_projects 分隔成单独的ManyToManyFields：

my_photogalleries = models.ManyToManyField(Prem_PhotoGallery)
my_pdfs = models.ManyToManyField(Prem_PDF)
... etc ...

【讨论】：