显示在Django管理引用表，而不是对象的列

Question

我怎样才能显示列ID，标题和书，而不是“丛书对象”的一年？

该截图显示当前状态：

我model.py看起来是这样的：

from __future__ import unicode_literals
from django.db import models


class Authors(models.Model):
    name = models.CharField(max_length=45, blank=True, null=True)
    birthday = models.DateField(blank=True, null=True)

    class Meta:
        managed = True
        db_table = 'authors'


class AuthorsBooks(models.Model):
    author_id = models.OneToOneField('Authors', models.DO_NOTHING, db_column='author_id', primary_key=True)
    book_id = models.OneToOneField('Books', models.DO_NOTHING, db_column='book_id', primary_key=True)

    class Meta:
        managed = True
        db_table = 'authors_books'
        unique_together = (('author_id', 'book_id'),)


class Awards(models.Model):
    author = models.OneToOneField('Authors', models.DO_NOTHING, db_column='author', primary_key=True)
    award_name = models.CharField(max_length=45)
    year = models.IntegerField(blank=True, null=True)

    class Meta:
        managed = True
        db_table = 'awards'
        unique_together = (('author', 'award_name'),)


class Books(models.Model):
    titel = models.CharField(max_length=45, blank=True, null=True)
    year = models.IntegerField(blank=True, null=True)

    class Meta:
        managed = True
        db_table = 'books'

在类AuthorsBooks我已经改变了两个外键OneToOneFields。

我的admin.py如下所示：

from django.contrib import admin
from myapp.models import Authors
...

class AwardsInline(admin.TabularInline):
    model = Awards

class AuthorsBooksInline(admin.TabularInline):
    model = AuthorsBooks

class AuthorsAdmin(admin.ModelAdmin):
    list_display = ("name", "birthday" )
    inlines = (AwardsInline, AuthorsBooksInline)


admin.site.register(Authors, AuthorsAdmin)

Answer 1

在每个模型的models.py添加一个Unicode功能。

class Authors(models.Model):
    name = models.CharField(max_length=45, blank=True, null=True)
    birthday = models.DateField(blank=True, null=True)

    class Meta:
        managed = True
        db_table = 'authors'

    def __unicode__(self):
        return self.name

Answer 2

在你的models.py文件，你可以用你的关注类特殊__str__方法，使你的对象更多的描述。 （这通常是像大多数程序员做到这一点）

def __str__(self):
    return self.title #if your using 'title' as the attribute to identify your objects

你可以选择任何其他属性，使你的对象描述。

祝好运！