django-admin

问题 It gives you filter by staff status and superuser status, but what about groups? 回答1: Since version 1.3 it can be done using this: list_filter = ('groups__name') Of course as @S.Lott explains you must register your customized class in the admin.py file: from django.contrib import admin from django.contrib.auth.admin import UserAdmin from django.contrib.auth.models import User class MyUserAdmin(UserAdmin): list_filter = UserAdmin.list_filter + ('groups__name',) admin.site.unregister(User)

问题 I've got a ModelAdmin class that includes a foreign key field in its list_display . But the admin list page for that model is doing hundreds of queries, one query per row to get the data from the other table instead of a join ( select_related() ). The Django docs indicate you can add list_select_related = True as an attribute to your ModelAdmin to make this go away, but it doesn't seem to work at all for me. This SO question seems to give a similar problem, but his resolution is unclear, and

来源： https://stackoverflow.com/questions/64520274/show-related-data-inside-foreign-key-object-in-django-admin

来源： https://stackoverflow.com/questions/64520274/show-related-data-inside-foreign-key-object-in-django-admin

来源： https://stackoverflow.com/questions/59937531/django-call-id-expected-a-number-but-got-string

来源： https://stackoverflow.com/questions/39964672/django-admin-template-override-not-working

来源： https://stackoverflow.com/questions/25748396/django-extend-admin-index

来源： https://stackoverflow.com/questions/25748396/django-extend-admin-index

来源： https://stackoverflow.com/questions/62303040/user-change-password-got-an-unexpected-keyword-argument-extra-context

来源： https://stackoverflow.com/questions/36284422/django-admin-is-staff-based-on-group