I have built out a Django voting application. The models are pretty simple.
I have a category , an entry , and participant model
class Category(models.Model):
name = models.CharField(max_length=30)
slug = models.SlugField()
class Participant(models.Model):
name = models.CharField(max_length=30)
def __unicode__(self):
return self.name
class Entry(models.Model):
votes = models.IntegerField()
category = models.ForeignKey(Category)
participant = models.ForeignKey(Participant)
def __unicode__(self):
output = 'Entry For {0} in Category {1}'.format(self.participant, self.category)
return output
Pretty straightforward. You can add a category an entry and a participant using the default django admin models. This works really well.
Now the question:
In the admin, I want user to click a button and is presented with the listing of all winners for all the categories in the db. I have an idea on how to implement this, where I basically want a user to submit a form in the admin interface. I know the admin interface is implemented via the way all djanog apps are MVC style. But I don't know where I can extend because the adminBaseModel / adminModel acts like a models and view controllers, and url-confs at the same time. It's seems difficult to rewire alot of the internets there.
Can someone point me in the right direction? Just want to simply implement my own view that merely extends the admin view with my own context and method calls.
I hope that was clear. Thanks for all your help guys.
Related
I want add a course to the users purchased_course. I am unable to figure out how to add it. The Buy Button Link I added on the Course Listview and Detailview I want that particular course to be added in in the users purchased_course which is many to many field.
My models
class Course(models.Model):
...
title = models.CharField(max_length=1000)
class User(AbstractUser):
...
purchased = models.ManyToManyField(Course, blank=True, related_name="buyers")
My views
I want the function here
#login_required
def puchases_chaeckout(request):
user=request.user
return redirect('student-gome-page')
Please Help
its simple!
list_of_courses = user.purchased.all()
please read Django documents
I want the admin interface to show disctrict field only if I choose 'B' as the category. If I choose 'W' I want all fields of Offer model to be displayed. Is it possible to show selected (filtered) fields in admin page depending on the choice in other field in the same model? Thanks in advance for your help.
My models:
class Category(models.Model):
NAME_CHOICES = (
('B', 'BLACK'),
('W', 'WHITE'),
)
name = models.CharField(max_length=200, choices=NAME_CHOICES)
class Meta:
verbose_name_plural = 'Categories'
def __unicode__(self):
return self.get_name_display()
class Offer(models.Model):
category = models.ForeignKey(Category, verbose_name='Kategoria')
city = models.CharField(max_length=128, verbose_name='Miasto')
province = models.CharField(max_length=3)
district = models.CharField(max_length=128, verbose_name='Dzielnica')
def __unicode__(self):
return "Offer number %s" % (self.id)
First of all I must to tell, that django works only in sync way. So if you want to choose which input to use, you must send a request and wait a feedback. In my opinion there're no straight way to do this task correctly.
And I see a few solutions:
1) You can use jQuery for that. But the main problem is that django has a own admin system with a built-in widgets. You can try to customize it in two ways:
Take an app with this option (for example, django-admin-tools) and create custom behavior on your form;
manage.py collectstatic and after that going to admin folder and create custom jQuery script.
2) Build a custom admin form for your model with ModelChoiceField. I don't quit sure about this field behavior really help you, but you can investigate that.
If I need to do this task, I choose first way with admin static and custom jQuery.
I'm working on building a django app that extends the Mezzanine project. Mezzanine has a Gallery app (photos). I'd like to create "portfolio" page that acts as a landing page which has a single image and link to each gallery page.
Each gallery (Gallery) can have multiple images (GalleryImage). I'd like to via the admin select a gallery, then select an image to be displayed. However, I can't seem to figure out what to do.
Here's my model:
class GalleriesThumb(models.Model):
relatedlandingpage = models.ForeignKey(LandingPage)
relatedgallery = models.ForeignKey(Galleries)
thumb = models.ManyToManyField(GalleryImage)
def __unicode__(self):
return self.name
class Galleries(models.Model):
landingpage = models.ForeignKey(LandingPage)
tagline = models.CharField(max_length=100)
galleries = models.ManyToManyField(Gallery)
def __unicode__(self):
return self.name
class LandingPage(models.Model):
gallerytitle = models.CharField(max_length=200)
def __unicode__(self):
return self.name
My admin is something like:
class GalleryInline(admin.InlineModelAdmin)
model = Galleries
model = GalleriesThumb
list_display = galleries
list_display = thumb
class LangingPageAdmin(admin.ModelAdmin):
fieldsets = (
(None, {
'fields': ('gallerytitle')
})
inlines = [GalleryInline,]
I realized that this won't do what i want, but how do I get the list_display on the the images that are related to Galleries. I'm pretty sure it needs to be a method, or am I taking a completing wrong approach if the selections that are made will be defining the content on the page. (I realize that I'm also missing my fields to store the selection in.)
I'm sorry if this a dumb question, but this my first real world attempt an app.
I think this link will resolve your problem
Django 1.2.1 Inline Admin for Many To Many Fields
class GalleryInline(admin.InlineModelAdmin)
model = Galleries.galleries.through
models:
class Detail(models.Model):
def __unicode__(self):
return self.title
title = models.CharField(max_length=32)
class Cars(models.Model):
def __unicode__(self):
return self.name
name = models.CharField(max_length=32, unique=True)
details = models.ManyToManyField(Detail)
So, every car has a many details - wheels, engine, etc. How to do this: in Django Admin situated Cars menu, in that menu we have a many lines of details (like in tutorial).
In admin I use:
class DetailInline(admin.TabularInline):
model = Detail
extra = 6
class CarsAdmin(admin.ModelAdmin):
inlines = [DetailInline]
But it has error: Detail has no ForeignKey to Cars. How to fix it?
Django does not natively let you add a reverse inline.
i.e. You can have the Detail page contain an inline admin of all the Cars that contain a ForeignKey to that particular Detail. However, the reverse is not natively possible.
There is a workaround though wherein you have to override the admin template a bit. There is a previous SO question about this here: Inline-like solution for Django Admin where Admin contains ForeignKey to other model
When trying to navigate to 'Add a presentation' in Django admin, I have to wait ~1 minute for the response to render. The problem is that I have ~500 slides in the database and the admin is selecting all the slides three different times to fill in the menues. I am obviously doing something wrong with my model definitions, as I wouldn't expect this amount of data to bring my server to its knees. Any visibility into why I am experiencing this issue with the way I have defined relationships or am using the django admin?
class PresentationTitle(models.Model):
title = models.CharField(max_length=255)
order_number = models.IntegerField(default=0)
def __unicode__(self):
return self.title
class PresentationUser(models.Model):
user = models.OneToOneField(User)
authorized_modules = models.ManyToManyField(PresentationTitle)
class Presentation(models.Model):
title = models.ForeignKey(PresentationTitle)
user = models.ForeignKey(PresentationUser)
presentation_date = models.DateTimeField()
def __unicode__(self):
return self.title.title
class Slide(models.Model):
....
submodule = models.ForeignKey(Submodule)
presentation = models.ManyToManyField(Presentation, through='PresentationSlide')
...
class Meta:
order_with_respect_to = 'submodule'
ordering = ['order']
class PresentationSlide(models.Model):
presentation = models.ForeignKey(Presentation)
slide = models.ForeignKey(Slide)
slide_order = models.IntegerField()
Additionally, my admin contains:
class PresentationSlideInline(admin.StackedInline):
model = PresentationSlide
class PresentationAdmin(admin.ModelAdmin):
inlines = [PresentationSlideInline]
admin.site.register(Presentation, PresentationAdmin)
Understandably, removing just having PresentationAdmin from the admin.site.register makes it load very responsively.
Resolved my issue -- there was another culprit at play: django-debug-toolbar
After removing this from my setup, the admin panel for the Presentation effectively loads.
Thanks to everyone on the interest and the support, be wary of profiling add-ons when you are experience performance issues.