[list_display有条件地在表中显示行

Question

我有两个桌子

Employee
-----
username
organisation_id
email

Organisation
------------
organisation_id
company_name

User_A->organisation_1
User_B->organisation_1
User_C->organisation_2
User_D->organisation_2

[user_A登录时，他应该只看到User_A和User_B此代码应如下所示，逻辑是首先，我们需要找到当前的用户组织并有条件地在管理页面中显示结果Django管理员代码

admin.site.register(Employee, CustomUserAdmin)
admin.site.register(Organization)
class CustomUserAdmin(UserAdmin):
def filter_view(request):
current_user = request.user
organization_id =employees.objects.get(id=current_user.id).organization_id
List_display = ('username', 'email','Organizations.objects.filter(id=organization_id)')

Answer 1

您可以写list_display是...

list_display = ('username', 'email', 'filtered_organisation')

并如下编写filtered_organisation的方法...
def filtered_organisation(self, obj):
    return Organizations.objects.filter(id=organization_id)

更多参考click here

Answer 2

重写ModelAdmin类的get_list_display--(doc)方法

get_list_display

Answer 3

您可以覆盖

class MyAdmin(admin.ModelAdmin):
    def get_list_display(self, request):
        if request.user.id == 'some_id':
            return ['field_1', 'field_2']
        else:
            return ['field_1', 'field_2', 'field_3', 'field_4', 'field_5']

方法：

get_queryset(request)