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Python 显示django管理站点中许多对象的字段名而不是整个对象_Python_Django_Django Admin - Fatal编程技术网

Python 显示django管理站点中许多对象的字段名而不是整个对象

Python 显示django管理站点中许多对象的字段名而不是整个对象,python,django,django-admin,Python,Django,Django Admin,我的模型如下: class Retailer(BaseModel): brand = models.ManyToManyField('brands.Brand',blank=True) class Brand(BaseModel): name = models.CharField(max_length=100, unique=True) website = models.URLField(max_length=500, blank=True, default='')

我的模型如下:

class Retailer(BaseModel):
    brand = models.ManyToManyField('brands.Brand',blank=True)

class Brand(BaseModel):
    name = models.CharField(max_length=100, unique=True)
    website = models.URLField(max_length=500, blank=True, default='')
class RetailerAdmin(admin.ModelAdmin):
    filter_horizontal = ('brand',)
我的管理类如下:

class Retailer(BaseModel):
    brand = models.ManyToManyField('brands.Brand',blank=True)

class Brand(BaseModel):
    name = models.CharField(max_length=100, unique=True)
    website = models.URLField(max_length=500, blank=True, default='')
class RetailerAdmin(admin.ModelAdmin):
    filter_horizontal = ('brand',)

管理员网站确实为我显示了多选字段,但品牌列表中的每个条目都显示为
品牌对象
,这对我来说毫无意义。我希望每个条目都显示为该品牌的名称字段。我该怎么办?

您可以将
\uuuuUnicode\uuuu
(python 2)或
\uuuuu str\uuuu
(python 3)方法添加到您的模型中,这样看起来就像这样

class Brand(BaseModel):
    name = models.CharField(max_length=100, unique=True)
    website = models.URLField(max_length=500, blank=True, default='')

    def __unicode__(self):
        return self.name