如何加上「+&引用;使用一对多关系时Django Admin中的按钮,如果该模型具有ForeignKey?
这个问题是问题的延伸 modles.py如何加上「+&引用;使用一对多关系时Django Admin中的按钮,如果该模型具有ForeignKey?,django,django-models,django-admin,Django,Django Models,Django Admin,这个问题是问题的延伸 modles.py from django.db import models # Create your models here. class Dude(models.Model): name = models.CharField(blank=False, null=False, max_length=100, unique=True) def phones(self): # use reverse relation to get a
from django.db import models
# Create your models here.
class Dude(models.Model):
name = models.CharField(blank=False, null=False, max_length=100, unique=True)
def phones(self):
# use reverse relation to get a list ofall recorded numbers
phone_numbers = self.phonenumber_set.values_list('number', flat=True)
phone_count = self.phonenumber_set.count()
return "This user have %s numbers recorded: %s" % (phone_count, ', '.join(phone_numbers))
def __unicode__(self):
return u"%s" % self.name
class PhoneType(models.Model):
name = models.CharField(blank=False, null=False, max_length=100, unique=True)
class PhoneNumber(models.Model):
dude = models.ForeignKey(Dude)
number = models.CharField(blank=False, null=False, max_length=100, unique=True)
type = models.ForeignKey(PhoneType)
from django.contrib import admin
# Register your models here.
from TestingDjango.apps.one_to_many.models \
import Dude, PhoneNumber
class PhoneNumberInline(admin.TabularInline):
model = PhoneNumber
extra = 1
class DudeAdmin(admin.ModelAdmin):
pass
list_display = ('name', 'phones')
inlines = [ PhoneNumberInline, ]
class PhoneNumberAdmin(admin.ModelAdmin):
pass
list_display = ('dude', 'number')
admin.site.register(Dude, DudeAdmin)
admin.py
from django.db import models
# Create your models here.
class Dude(models.Model):
name = models.CharField(blank=False, null=False, max_length=100, unique=True)
def phones(self):
# use reverse relation to get a list ofall recorded numbers
phone_numbers = self.phonenumber_set.values_list('number', flat=True)
phone_count = self.phonenumber_set.count()
return "This user have %s numbers recorded: %s" % (phone_count, ', '.join(phone_numbers))
def __unicode__(self):
return u"%s" % self.name
class PhoneType(models.Model):
name = models.CharField(blank=False, null=False, max_length=100, unique=True)
class PhoneNumber(models.Model):
dude = models.ForeignKey(Dude)
number = models.CharField(blank=False, null=False, max_length=100, unique=True)
type = models.ForeignKey(PhoneType)
from django.contrib import admin
# Register your models here.
from TestingDjango.apps.one_to_many.models \
import Dude, PhoneNumber
class PhoneNumberInline(admin.TabularInline):
model = PhoneNumber
extra = 1
class DudeAdmin(admin.ModelAdmin):
pass
list_display = ('name', 'phones')
inlines = [ PhoneNumberInline, ]
class PhoneNumberAdmin(admin.ModelAdmin):
pass
list_display = ('dude', 'number')
admin.site.register(Dude, DudeAdmin)
当我添加新Dude时,我只能选择现有的电话类型。没有像其他外键字段那样的“+”按钮。我怀疑这是一对多关系的结果。那么,如何为PhoneType添加“+”按钮呢?您需要注册PhoneNumberAdmin以使Django的管理员可以访问它,然后它将为您提供“+”选项以添加更多内容 插入
admin.site.register(PhoneNumber, PhoneNumberAdmin)
在文件底部您需要注册PhoneNumberAdmin以使Django的管理员可以访问它,然后它将为您提供“+”选项以添加更多内容 插入
admin.site.register(PhoneNumber, PhoneNumberAdmin)
在文件的底部