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Create custom ModelAdmin class

use_django_admin
Alexandre Iooss 7 years ago
parent
commit
040b15dcd1
  1. 10
      useradmin/admin.py
  2. 9
      users/admin.py

10
useradmin/admin.py

@ -1,10 +1,16 @@
from django.contrib.admin.sites import AdminSite
from django.contrib import admin
class UserAdmin(AdminSite):
class UserAdmin(admin.sites.AdminSite):
def has_permission(self, request):
"""This admin site doesn't require being staff"""
return request.user.is_active
class ModelUserAdmin(admin.ModelAdmin):
# Display actions on changelist bottom bu default
actions_on_bottom = True
# Defines new management site
user_admin_site = UserAdmin(name='user_admin')

9
users/admin.py

@ -32,8 +32,7 @@ from django.contrib.auth.models import Group
from django.contrib.auth.admin import UserAdmin as BaseUserAdmin
from reversion.admin import VersionAdmin
from useradmin.admin import user_admin_site
from useradmin.admin import ModelUserAdmin, user_admin_site
from .models import (
User,
EMailAddress,
@ -85,10 +84,9 @@ class LdapServiceUserGroupAdmin(admin.ModelAdmin):
search_fields = ('name',)
class SchoolAdmin(VersionAdmin):
class SchoolAdmin(VersionAdmin, ModelUserAdmin):
"""Administration, gestion des écoles"""
list_display = ('name',)
actions_on_bottom = True
class ListRightAdmin(VersionAdmin):
@ -97,10 +95,9 @@ class ListRightAdmin(VersionAdmin):
list_display = ('unix_name',)
class ListShellAdmin(VersionAdmin):
class ListShellAdmin(VersionAdmin, ModelUserAdmin):
"""Gestion de la liste des shells coté admin"""
list_display = ('shell',)
actions_on_bottom = True
class RequestAdmin(admin.ModelAdmin):

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