django - 管理面板中多对多字段的 List_display

Question

我有 2 个应用程序:Visitors和 Meetings ，通过 ManytoMany 链接 field :

访客/models.py:

from django.db import models
from meetings.models import Meeting

class Visitor(models.Model):

    visitor_name = models.CharField(default='name', max_length=128, blank=False, null=False)
    visitor_meetings = models.ManyToManyField(Meeting)

    def __str__(self):
        return self.visitor_name

session /models.py:

from django.db import models
from team.models import Team

class Meeting(models.Model):
    team_member = models.ForeignKey(Team)
    meeting_name = models.CharField(default='name', max_length=128, blank=True, null=True)

    def __str__(self):
        return self.meeting_name

我知道获得 list_display 的正确方法用于 session Visitors管理面板在这里: https://stackoverflow.com/a/18108586/2429989 .

但是，如何显示 list_display在 中为每次 session 的访客Meetings 管理面板？我试过了:

session /admin.py:

from django.contrib import admin
from .models import Meeting
from visitors.models import Visitor

class MeetingAdmin(admin.ModelAdmin):
    list_display = ['id', 'team_member', 'show_visitors' ]

    def show_visitors(self, obj):
        return "\n".join([a.visitor_name for a in obj.visitor.all()])

admin.site.register(Meeting, MeetingAdmin)

这导致 'Meeting' object has no attribute 'visitor' ，我哪里错了？

Answer 1

默认的反向查找关系总是 obj.tablename_set .在您的情况下，您需要使用 obj.visitor_set.all()

class MeetingAdmin(admin.ModelAdmin):
    list_display = ['id', 'team_member', 'show_visitors' ]

    def show_visitors(self, obj):
        return "\n".join([a.visitor_name for a in obj.visitor_set.all()])

但是，您可以通过定义 related_name 来自定义反向查找的名称。 :

class Visitor(models.Model):
    visitor_name = models.CharField(default='name', max_length=128, blank=False, null=False)
    visitor_meetings = models.ManyToManyField(Meeting, related_name='visitors')

    def __str__(self):
        return self.visitor_name

全部 Meeting对象现在可以访问 visitors .像这样使用它:

class MeetingAdmin(admin.ModelAdmin):
    list_display = ['id', 'team_member', 'show_visitors' ]

    def show_visitors(self, obj):
        return "\n".join([a.visitor_name for a in obj.visitors.all()])