Django 编辑模型实例找不到实例
Django editing a model instance fails to find the instance
我正在制作食谱书。出于某种原因,每当我尝试从数据库中提取食谱进行编辑时,我总是收到错误消息,找不到我指定的食谱。我正在使用 slugs,我的逻辑是我要从已经提取数据库信息的 detailView 转到 updateView。我试图将我已经从 detailView 中提取的食谱对象传递给 updateView,但是当我这样做时,它一直告诉我找不到指定的食谱。
views.py:
我在这里调用的基本视图只提供默认的 post 方法来处理搜索,这样我就不必为我创建的每个视图都放入它,所以我有一些代码可重用性
class RecipeDetailView(BaseDetailView):
model = Recipe
template_name = 'RecipeBook/recipe_detail.html'
context_object_name = 'recipe_view'
queryset = None
slug_field = 'slug'
slug_url_kwarg = 'slug'
def get_context_data(self, *args, **kwargs):
context = super(RecipeDetailView, self).get_context_data()
recipe = self.object
recipe.ingredients = recipe.ingredients_list.split('\n')
context['recipe'] = recipe
return context
class RecipeEditView(BaseUpdateView):
model = Recipe
template_name = 'RecipeBook/edit_recipe.html'
context_object_name = 'recipe_edit'
queryset = None
slug_field = 'slug'
slug_url_kwarg = 'slug'
form_class = RecipeForm
def get_context_data(self, *args, **kwargs):
context = super(RecipeEditView, self).get_context_data()
recipe = self.object
print(recipe.name)
recipe.ingredients = recipe.ingredients_list.split('\n')
recipe.categories_list = ""
categories = Category.objects.filter(recipe=recipe)
for category in categories:
if category != categories[-1]:
recipe.categories_list += (category + ", ")
else:
recipe.categories_list += category
recipe_edit_form = RecipeForm(initial={'name': recipe.name, 'ingredients_list': recipe.ingredients,
'directions': recipe.directions, 'prep_time': recipe.prep_time,
'cook_time': recipe.cook_time, 'servings': recipe.servings,
'source': recipe.source, 'category_input': recipe.categories_list})
context['recipe'] = recipe
context['recipe_edit_form'] = recipe_edit_form
return context
models.py:
class Recipe(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=100, default="")
ingredients_list = models.TextField(default="")
servings = models.IntegerField(default=0, null=True, blank=True)
prep_time = models.IntegerField(default=0, null=True, blank=True)
cook_time = models.IntegerField(default=0, null=True, blank=True)
directions = models.TextField(default="")
source = models.CharField(max_length=100, default="", null=True, blank=True)
categories = models.ManyToManyField(Category, blank=True)
slug = models.CharField(max_length=200, default="")
def __str__(self):
return self.name
urls.py
# ex: /Recipes/Grilled_Chicken/
path('Recipes/<slug>/', views.RecipeDetailView.as_view(), name='view_recipe'),
path('Recipes/<path:slug>/', views.RecipeDetailView.as_view(), name='view_recipe'),
# ex: /Recipes/edit/Steak/
path('Recipes/edit/<slug>/', views.RecipeEditView.as_view(), name='edit_recipe'),
path('Recipes/edit/<path:slug>/', views.RecipeEditView.as_view(), name='edit_recipe'),
link 在 recipe_detail.html:
<a href="{% url 'RecipeBook:edit_recipe' recipe.slug %}" style="float: right">Edit Recipe</a>
我一直在疯狂地想弄明白。根据我在这里的所有内容,我在 detailView 中提取的食谱应该能够传递给 editView,但每次我尝试打开 edit_recipe 页面时,它都会告诉我它找不到指定的食谱。它生成的 URL 显示了正确的 slug 和它应该显示的 link 。我不知道我现在错过了什么...
试试这个方法:
<a href="{% url 'RecipeBook:edit_recipe' recipe_view.slug %}" style="float: right">Edit Recipe</a>
我最终不得不返回并将视图更改为 DetailView。这是我可以让食谱实例被推送的唯一方法。对于不太清楚的模型使用更新视图有一些非常具体的事情......
一旦我更改为 DetailView,页面将填充初始化为配方值的表单。然后我可以进行调整以确保一切正常。
感谢那些回复的人,它至少让我的大脑转向了不同的方向来解决这个问题。
我正在制作食谱书。出于某种原因,每当我尝试从数据库中提取食谱进行编辑时,我总是收到错误消息,找不到我指定的食谱。我正在使用 slugs,我的逻辑是我要从已经提取数据库信息的 detailView 转到 updateView。我试图将我已经从 detailView 中提取的食谱对象传递给 updateView,但是当我这样做时,它一直告诉我找不到指定的食谱。
views.py: 我在这里调用的基本视图只提供默认的 post 方法来处理搜索,这样我就不必为我创建的每个视图都放入它,所以我有一些代码可重用性
class RecipeDetailView(BaseDetailView):
model = Recipe
template_name = 'RecipeBook/recipe_detail.html'
context_object_name = 'recipe_view'
queryset = None
slug_field = 'slug'
slug_url_kwarg = 'slug'
def get_context_data(self, *args, **kwargs):
context = super(RecipeDetailView, self).get_context_data()
recipe = self.object
recipe.ingredients = recipe.ingredients_list.split('\n')
context['recipe'] = recipe
return context
class RecipeEditView(BaseUpdateView):
model = Recipe
template_name = 'RecipeBook/edit_recipe.html'
context_object_name = 'recipe_edit'
queryset = None
slug_field = 'slug'
slug_url_kwarg = 'slug'
form_class = RecipeForm
def get_context_data(self, *args, **kwargs):
context = super(RecipeEditView, self).get_context_data()
recipe = self.object
print(recipe.name)
recipe.ingredients = recipe.ingredients_list.split('\n')
recipe.categories_list = ""
categories = Category.objects.filter(recipe=recipe)
for category in categories:
if category != categories[-1]:
recipe.categories_list += (category + ", ")
else:
recipe.categories_list += category
recipe_edit_form = RecipeForm(initial={'name': recipe.name, 'ingredients_list': recipe.ingredients,
'directions': recipe.directions, 'prep_time': recipe.prep_time,
'cook_time': recipe.cook_time, 'servings': recipe.servings,
'source': recipe.source, 'category_input': recipe.categories_list})
context['recipe'] = recipe
context['recipe_edit_form'] = recipe_edit_form
return context
models.py:
class Recipe(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=100, default="")
ingredients_list = models.TextField(default="")
servings = models.IntegerField(default=0, null=True, blank=True)
prep_time = models.IntegerField(default=0, null=True, blank=True)
cook_time = models.IntegerField(default=0, null=True, blank=True)
directions = models.TextField(default="")
source = models.CharField(max_length=100, default="", null=True, blank=True)
categories = models.ManyToManyField(Category, blank=True)
slug = models.CharField(max_length=200, default="")
def __str__(self):
return self.name
urls.py
# ex: /Recipes/Grilled_Chicken/
path('Recipes/<slug>/', views.RecipeDetailView.as_view(), name='view_recipe'),
path('Recipes/<path:slug>/', views.RecipeDetailView.as_view(), name='view_recipe'),
# ex: /Recipes/edit/Steak/
path('Recipes/edit/<slug>/', views.RecipeEditView.as_view(), name='edit_recipe'),
path('Recipes/edit/<path:slug>/', views.RecipeEditView.as_view(), name='edit_recipe'),
link 在 recipe_detail.html:
<a href="{% url 'RecipeBook:edit_recipe' recipe.slug %}" style="float: right">Edit Recipe</a>
我一直在疯狂地想弄明白。根据我在这里的所有内容,我在 detailView 中提取的食谱应该能够传递给 editView,但每次我尝试打开 edit_recipe 页面时,它都会告诉我它找不到指定的食谱。它生成的 URL 显示了正确的 slug 和它应该显示的 link 。我不知道我现在错过了什么...
试试这个方法:
<a href="{% url 'RecipeBook:edit_recipe' recipe_view.slug %}" style="float: right">Edit Recipe</a>
我最终不得不返回并将视图更改为 DetailView。这是我可以让食谱实例被推送的唯一方法。对于不太清楚的模型使用更新视图有一些非常具体的事情......
一旦我更改为 DetailView,页面将填充初始化为配方值的表单。然后我可以进行调整以确保一切正常。
感谢那些回复的人,它至少让我的大脑转向了不同的方向来解决这个问题。