姜戈 |如何让应用中的对象在查看列表时取name的值

DJANGO | How to make the object in the application take the value of the name when viewing the list

https://i.stack.imgur.com/30M2y.jpg - 图像管理面板

管理员代码:

from django.contrib import admin
from .models import Destination

admin.site.register(Destination)

夏天: 我做了动态迁移到主页。我希望目的地的名称在管理面板的列表中被接受。 怎么做?

我解决了问题:

from django.contrib import admin
from .models import Destination

class DestinationAdmin(admin.ModelAdmin):
    list_display = ('name',) #I taking value from list of object.

admin.site.register(Destination, DestinationAdmin)

Nekain helped me