0

我似乎在 Django 中遇到了 ManyToMany 关系的问题,我似乎没有得到正确的管理屏幕。我想为地址字段显示一个stackedinline,但是我只得到一个Dropbox。有人可以帮忙吗,下面是我的代码。

模型.py

class AddressType(models.Model):
    TypeCode = models.CharField('Address Type', max_length=50, blank=False, null = False)
    Description = models.TextField('Description')
    def __unicode__(self):
        return self.TypeCode

class Address(models.Model):
    AddressType = models.ForeignKey(AddressType)
    Address = models.TextField('Address', blank=False, null=False)

    def __unicode__(self):
        return self.AddressType.TypeCode


class BusinessPartner(models.Model):
    FullName = models.CharField('Full Name', max_length=50, blank=False, null=False)
    PartnerType = models.ForeignKey(PartnerType)
    Company = models.CharField('Company', max_length=100, blank=True, null=True)
    Website = models.URLField('Website', blank=True, null=True)
    Address = models.ManyToManyField(Address,through='AddressPartner_Assn')

    def __unicode__(self):
        if self.Company:
            return self.FullName + '-' + self.Company
        else:
            return self.FullName

class AddressPartner_Assn(models.Model):
    Address = models.ForeignKey(Address)
    BusinessPartner = models.ForeignKey(BusinessPartner)

Admins.py
=========
class AddressTypeAdmin(admin.ModelAdmin):
    list_display = ('TypeCode','Description')
    search_fields = ['TypeCode','Description']

class AddressAdmin(admin.ModelAdmin):
    list_display = ['Address']
    search_fields = ['Address']
    list_filter = ['AddressType']

class AddressInlineAdmin(admin.StackedInline):
    model = BusinessPartner.Address.through
    extra = 1

class BusinessPartnerAdmin(admin.ModelAdmin):
    list_display = ('__unicode__','FullName','PartnerType','Company')
    inlines = [AddressInlineAdmin,]
    search_fields = ['FullName','Company']
    list_filter = ('PartnerType',)
    raw_id_fields = ('PartnerType',)
4

1 回答 1

1

由于您使用 through 参数来明确提到中间表,因此您必须将 AddressInlineAdmin 替换为

class AddressInlineAdmin(admin.StackedInline):
    model = AddressPartner_Assn
    extra = 1

有关任何其他信息,请参阅https://docs.djangoproject.com/en/1.4/ref/contrib/admin/#working-with-many-to-many-intermediary-models

于 2013-01-14T08:22:05.150 回答