How to show all fields of model in admin page?

Question

here is the models page

In this picture, only the title shows up on here, I used:

 def __unicode__(self):
        return self.title;  

Answer 1

Every solution found here raises an error like this

The value of 'list_display[n]' must not be a ManyToManyField.

If the model contains a Many to Many field.

A possible solution that worked for me is:

list_display = [field.name for field in MyModel._meta.get_fields() if not x.many_to_many]

Answer 2

I like this answer and thought I'd post the complete admin.py code (in this case, I wanted all the User model fields to appear in admin)

from django.contrib import admin
from django.contrib.auth.models import User
from django.db.models import ManyToOneRel, ForeignKey, OneToOneField


MySpecialAdmin = lambda model: type('SubClass'+model.__name__, (admin.ModelAdmin,), {
    'list_display': [x.name for x in model._meta.fields],
    'list_select_related': [x.name for x in model._meta.fields if isinstance(x, (ManyToOneRel, ForeignKey, OneToOneField,))]
})

admin.site.unregister(User)
admin.site.register(User, MySpecialAdmin(User))

Answer 3

I'm using Django 3.1.4 and here is my solution.

I have a model Qualification

model.py

from django.db import models

TRUE_FALSE_CHOICES = (
    (1, 'Yes'),
    (0, 'No')
)


class Qualification(models.Model):
    qual_key = models.CharField(unique=True, max_length=20)
    qual_desc = models.CharField(max_length=255)
    is_active = models.IntegerField(choices=TRUE_FALSE_CHOICES)
    created_at = models.DateTimeField()
    created_by = models.CharField(max_length=255)
    updated_at = models.DateTimeField()
    updated_by = models.CharField(max_length=255)

    class Meta:
        managed = False
        db_table = 'qualification'

admin.py

from django.contrib import admin
from models import Qualification


@admin.register(Qualification)
class QualificationAdmin(admin.ModelAdmin):
    list_display = [field.name for field in Qualification._meta.fields if field.name not in ('id', 'qual_key', 'qual_desc')]
    list_display.insert(0, '__str__')

here i am showing all fields in list_display excluding 'id', 'qual_key', 'qual_desc' and inserting '__str__' at the beginning.

This answer is helpful when you have large number of modal fields, though i suggest write all fields one by one for better functionality.

Answer 4

By default, the admin layout only shows what is returned from the object's unicode function. To display something else you need to create a custom admin form in app_dir/admin.py.

See here: https://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_display

You need to add an admin form, and setting the list_display field.

In your specific example (admin.py):

class BookAdmin(admin.ModelAdmin):
    list_display = ('title', 'author', 'price')
admin.site.register(Book, BookAdmin)

Answer 5

Here is my approach, will work with any model class:

MySpecialAdmin = lambda model: type('SubClass'+model.__name__, (admin.ModelAdmin,), {
    'list_display': [x.name for x in model._meta.fields],
    'list_select_related': [x.name for x in model._meta.fields if isinstance(x, (ManyToOneRel, ForeignKey, OneToOneField,))]
})

This will do two things:

1. Add all fields to model admin
2. Makes sure that there is only a single database call for each related object (instead of one per instance)

Then to register you model:

admin.site.register(MyModel, MySpecialAdmin(MyModel))

Note: if you are using a different default model admin, replace 'admin.ModelAdmin' with your admin base class

Answer 6

Show all fields:

list_display = [field.attname for field in BookModel._meta.fields]