No sir please! Gourav is just fine.
Also, you would need to go edit/add code in the admin.py file for this to work.
Try this part of the docs:
https://docs.djangoproject.com/en/2.2/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_filter
Hope, it helps.
On Mon, 22 Apr, 2019, 1:15 PM Balaji Shetty, <balajishetty@gmail.com> wrote:
Dear Gaurav SirI am very much thankful to you for your reply to my query.My questions is How can i make the customization in Django Admin Panel itself to implement necessary logic after user of particular department do the login.( How and where to embed proposed Logic T1.objects.filter(user=current_user, department=current_user.department)If i design separate pages for login using templates and showing necessary content user-wise (Login wise),Your suggested logic will definitely work (after making necessary changes in schema)On Sun, Apr 21, 2019 at 11:00 AM Gourav Chawla <gauravchawla.chawla281@gmail.com> wrote:First of all, use appropriate model fields for linking objects. For example users should be a foreign key to your(or Django's) user table.
Then whenever you want to show data on any interface to any user, just filter content/rows by user department.
Something like:
T1.objects.filter(user=current_user, department=current_user.department)
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