How can I customize the display of a model using contenttypes in the admin?

孤人 提交于 2020-01-13 09:52:11

问题


I have these models:

class App(models.Model):
  name = models.CharField(max_length=100)

class ProjectA(models.Model):
  name = models.CharField(max_length=100)
  app  = models.ForeignKey(App)

class ProjectB(ProjectA):
  pass

class Attachment(models.Model):
  content_type    = models.ForeignKey(ContentType)
  object_id       = models.PositiveIntegerField()
  project         = generic.GenericForeignKey("content_type","object_id")
  file            = models.FileField(upload_to=".")

I'm registering all the models for the admin, and I'm unregistering Group, User and Site. The thing is, when I access the Attachment in the admin, I see it rendered like this:

In the Content type select, I see this list:

The reason Attachment has a GenericForeignKey is because both ProjectA and ProjectB need to access it. I know that ProjectA and ProjectB are identical, but it's a requirement that they are stored in 2 separate tables. How could I made the Attachment class useable from the admin? I know how to use contenttypes from normal views, but from the admin not.

In the Attachment class I would only like to have a select for Project A or Project B, and then a list of all Project A's or all Project B's, followed by the file that I want to attach.

Is such a thing possible from the Admin? Will I need to show the user the Object Id column?


回答1:


if I'm not wrong, you want this. http://code.google.com/p/django-genericadmin/

my advice will work differently. you will add a little more form in ProjectA, ProjectB as inline. in your admin.py

from django.contrib import admin
from django.contrib.contenttypes import generic

from myproject.myapp.models import Attachment, ProjectA, ProjectB

class Attachmentline(generic.GenericTabularInline): #or generic.GenericStackedInline, this has different visual layout.
    model = Attachment

class ProjectAdmin(admin.ModelAdmin):
    inlines = [
        Attachmentline,
    ]

admin.site.register(ProjectA, ProjectAdmin)
admin.site.register(ProjectB, ProjectAdmin)

go your ProjectA or ProjectB admin and see new admin.

this isn't what you want but it can help you. otherwise you need check first link.




回答2:


You should notice that

" know that ProjectA and ProjectB are identical, but it's a requirement that they are stored in 2 separate tables"

is not really correct. All the data is stored in your app_projecta table, and (only) some pointers are kept in table app_projectb. If you are already going in this path, I would suggest starting with this instead:

class App(models.Model):
  name = models.CharField(max_length=100)

class Project(models.Model):
    name = models.CharField(max_length=100)
    app = models.ForeignKey(App)

class ProjectA(Project):
  pass

class ProjectB(Project):
  pass

class Attachment(models.Model):
  project = models.ForeignKey(Project)
  file = models.FileField(upload_to=".")  

This already gets you a bit closer to where you want to get...



来源:https://stackoverflow.com/questions/6335565/how-can-i-customize-the-display-of-a-model-using-contenttypes-in-the-admin

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