Mixing
Using choices used in Models.py to route the page in Django
I started a project in Django where I am making a new restaurant site. What I thought initially is to make a Model like this:
class Menu(models.Model):
user = models.ForeignKey(User,default=1, on_delete=models.CASCADE)
name = models.CharField("Name of the dish",max_length=30)
category = models.CharField(max_length=20,choices=CATEGORY_CHOICES,default='SP')
description = models.TextField(max_length=300)
dish_image = models.ImageField(upload_to='uploads/')
And I have CATEGORY_CHOICES as CATEGORY_CHOICES = [('L','Lunch'),.....]
Now I have this button on the index page that routes to menu page. In Menu page, I am thinking of displaying the [n] number of tiles with the [name] ( n = number of elements in category choices and name = the full name of it ex:Lunch)
After getting lots of error and figuring out each and every error, I have achieved this task. Now, after that, I want to open the menu of particular category when user clicks in it. Eg: If user clicks 'Lunch' tile, it should route to other page where all the list of the foods with that category are listed. My url pattern is like this:
urlpatterns = [
path('', views.index, name="index"),
path('menu/', views.menu, name="menu"),
path('menu/<str:menu_category>/', views.menu_detail, name="menu_detail"),
]
And I have this view :
def menu_detail(request, menu_category):
menu_list = get_list_or_404(Menu ,category=menu_category)
menu_title = Menu.objects.get(category=menu_category)
return render(request,'restaurant/menu_detail.html',{'menu_list':menu_list,'menu_title':menu_title})
In the Menu template, I am calling it as
<a href="{% url 'menu_detail' menu_category = menus.menu_category %}">
I don't know where I have done wrong. But I am getting ReverseMatchException while browsing ..../menu/ . Maybe it's too complicated for me as I am new to this field. Can anybody please help?
django django-models django-templates django-views django-urls
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
add a comment |
I started a project in Django where I am making a new restaurant site. What I thought initially is to make a Model like this:
class Menu(models.Model):
user = models.ForeignKey(User,default=1, on_delete=models.CASCADE)
name = models.CharField("Name of the dish",max_length=30)
category = models.CharField(max_length=20,choices=CATEGORY_CHOICES,default='SP')
description = models.TextField(max_length=300)
dish_image = models.ImageField(upload_to='uploads/')
And I have CATEGORY_CHOICES as CATEGORY_CHOICES = [('L','Lunch'),.....]
Now I have this button on the index page that routes to menu page. In Menu page, I am thinking of displaying the [n] number of tiles with the [name] ( n = number of elements in category choices and name = the full name of it ex:Lunch)
After getting lots of error and figuring out each and every error, I have achieved this task. Now, after that, I want to open the menu of particular category when user clicks in it. Eg: If user clicks 'Lunch' tile, it should route to other page where all the list of the foods with that category are listed. My url pattern is like this:
urlpatterns = [
path('', views.index, name="index"),
path('menu/', views.menu, name="menu"),
path('menu/<str:menu_category>/', views.menu_detail, name="menu_detail"),
]
And I have this view :
def menu_detail(request, menu_category):
menu_list = get_list_or_404(Menu ,category=menu_category)
menu_title = Menu.objects.get(category=menu_category)
return render(request,'restaurant/menu_detail.html',{'menu_list':menu_list,'menu_title':menu_title})
In the Menu template, I am calling it as
<a href="{% url 'menu_detail' menu_category = menus.menu_category %}">
I don't know where I have done wrong. But I am getting ReverseMatchException while browsing ..../menu/ . Maybe it's too complicated for me as I am new to this field. Can anybody please help?
django django-models django-templates django-views django-urls
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
add a comment |
I started a project in Django where I am making a new restaurant site. What I thought initially is to make a Model like this:
class Menu(models.Model):
user = models.ForeignKey(User,default=1, on_delete=models.CASCADE)
name = models.CharField("Name of the dish",max_length=30)
category = models.CharField(max_length=20,choices=CATEGORY_CHOICES,default='SP')
description = models.TextField(max_length=300)
dish_image = models.ImageField(upload_to='uploads/')
And I have CATEGORY_CHOICES as CATEGORY_CHOICES = [('L','Lunch'),.....]
Now I have this button on the index page that routes to menu page. In Menu page, I am thinking of displaying the [n] number of tiles with the [name] ( n = number of elements in category choices and name = the full name of it ex:Lunch)
After getting lots of error and figuring out each and every error, I have achieved this task. Now, after that, I want to open the menu of particular category when user clicks in it. Eg: If user clicks 'Lunch' tile, it should route to other page where all the list of the foods with that category are listed. My url pattern is like this:
urlpatterns = [
path('', views.index, name="index"),
path('menu/', views.menu, name="menu"),
path('menu/<str:menu_category>/', views.menu_detail, name="menu_detail"),
]
And I have this view :
def menu_detail(request, menu_category):
menu_list = get_list_or_404(Menu ,category=menu_category)
menu_title = Menu.objects.get(category=menu_category)
return render(request,'restaurant/menu_detail.html',{'menu_list':menu_list,'menu_title':menu_title})
In the Menu template, I am calling it as
<a href="{% url 'menu_detail' menu_category = menus.menu_category %}">
I don't know where I have done wrong. But I am getting ReverseMatchException while browsing ..../menu/ . Maybe it's too complicated for me as I am new to this field. Can anybody please help?
django django-models django-templates django-views django-urls
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
I started a project in Django where I am making a new restaurant site. What I thought initially is to make a Model like this:
class Menu(models.Model):
user = models.ForeignKey(User,default=1, on_delete=models.CASCADE)
name = models.CharField("Name of the dish",max_length=30)
category = models.CharField(max_length=20,choices=CATEGORY_CHOICES,default='SP')
description = models.TextField(max_length=300)
dish_image = models.ImageField(upload_to='uploads/')
And I have CATEGORY_CHOICES as CATEGORY_CHOICES = [('L','Lunch'),.....]
Now I have this button on the index page that routes to menu page. In Menu page, I am thinking of displaying the [n] number of tiles with the [name] ( n = number of elements in category choices and name = the full name of it ex:Lunch)
After getting lots of error and figuring out each and every error, I have achieved this task. Now, after that, I want to open the menu of particular category when user clicks in it. Eg: If user clicks 'Lunch' tile, it should route to other page where all the list of the foods with that category are listed. My url pattern is like this:
urlpatterns = [
path('', views.index, name="index"),
path('menu/', views.menu, name="menu"),
path('menu/<str:menu_category>/', views.menu_detail, name="menu_detail"),
]
And I have this view :
def menu_detail(request, menu_category):
menu_list = get_list_or_404(Menu ,category=menu_category)
menu_title = Menu.objects.get(category=menu_category)
return render(request,'restaurant/menu_detail.html',{'menu_list':menu_list,'menu_title':menu_title})
In the Menu template, I am calling it as
<a href="{% url 'menu_detail' menu_category = menus.menu_category %}">
I don't know where I have done wrong. But I am getting ReverseMatchException while browsing ..../menu/ . Maybe it's too complicated for me as I am new to this field. Can anybody please help?
django django-models django-templates django-views django-urls
django django-models django-templates django-views django-urls
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
asked Jan 2 at 17:10
Sulove BistaSulove Bista
436
436
add a comment |
add a comment |
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