从多对多字段中选择项目到列表中的 link 个特定项目到管理面板中的模型

Question

我正在开发食谱书应用程序，目前我的模型以这种方式连接：

class Tool(models.Model):
    name = models.CharField(max_length=50)
    description = models.CharField(max_length=200)

    def __str__(self):
        return self.name

class Recipe(models.Model):
    name = models.CharField(max_length=50)
    description = models.CharField(max_length=200)
    servings = models.IntegerField(default=1, blank=False)
    tools = models.ManyToManyField(Tool)

    def __str__(self):
        return self.name

管理面板输入当前如下所示：

如截图所示保存数据后，出现以下错误：

OperationalError at /admin/recipeBook/recipe/add/
no such table: recipeBook_recipe_tools

这是我的admin.py，以防有用：

from django.contrib import admin

from .models import Recipe, Ingredient, Instruction, Tool

# Register your models here.
class IngredientInline(admin.TabularInline):
  model = Ingredient
  extra = 2

class RecipeAdmin(admin.ModelAdmin):
  fieldsets = [
    ('Information', {'fields': ['name', 'description', 'servings']}),
    ('Tools', {'fields': ['tools']})
  ]
  inlines = [IngredientInline]
  list_display = ('name', 'description', 'servings')
  search_fields = ['name']
  list_filter = ['servings']

admin.site.register(Recipe, RecipeAdmin)
admin.site.register(Ingredient)
admin.site.register(Instruction)
admin.site.register(Tool)

Answer 1

如果你遇到这样的错误。

确保在添加到模型后迁移数据库。

从多对多字段中选择项目到列表中的 link 个特定项目到管理面板中的模型

Selecting items from a many to many field to link specific items in the list to a model in the admin panel

django

django-models

django-admin