List_display 用于管理面板中的 ManytoMany 字段

List_display for ManytoMany fields in Admin panel

我有 2 个应用程序:VisitorsMeetings,通过 ManytoMany 字段链接:

visitors/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

meetings/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

我知道在 Visitors 管理面板中获得会议 list_display 的正确方法是:.

但是,如何在 Meetings 管理面板中为每次会议的访问者显示 list_display?我试过:

meetings/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',我哪里出错了?

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

所有 Meeting 个对象现在都可以访问 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()])