Python Django admin+；大数据集多对多字段的自动完成

Django版本：1.7
自动完成照明版本：2.0

我有一个多对多关系：订阅应用程序，其中每个应用程序可以属于多个订阅，每个订阅可以包含多个应用程序。 订阅数量预计将达到数百个，而应用程序数量约为150万

使用Django Admin来管理订阅及其应用程序将是一件非常棒的事情，并且按照建议，使用autocomplete_light来管理大型数据集。 下面是我编写的代码，但它无法尝试一次加载AppMetadata的所有150万条记录，并挂起浏览器。 请告诉我

# model.py
class AppMetadata(models.Model):
    package_key = models.IntegerField(primary_key=True)
    title = models.CharField(max_length=500)
    package = models.CharField(max_length=400)

class Subscription(models.Model):
    name = models.CharField(max_length=256, unique=True)
    is_active = models.BooleanField('active', default=True)

class SubscriptionApp(models.Model):
    subscription_key = models.ForeignKey(to=Subscription, on_delete=models.DO_NOTHING)
    package_key = models.ManyToManyField(to=AppMetadata, db_column='package_key')

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我已经想出了一个折衷的解决办法。从上面的代码中，我删除了SubscriptionAutocomplete，并重写了admin.py类，如下所示：

@admin.register(SubscriptionApp)
class SubscriptionAppAdmin(admin.ModelAdmin):
    form = autocomplete_light.modelform_factory(model=SubscriptionApp)

class SubscriptionAppInline(admin.TabularInline):
    model = SubscriptionApp

@admin.register(Subscription)
class SubscriptionAdmin(admin.ModelAdmin):
    model = Subscription
    ordering = ('name',)

# admin.py
class SubscriptionAppInline(admin.TabularInline):
    model = SubscriptionApp

class SubscriptionAdmin(admin.ModelAdmin):
    form = autocomplete_light.modelform_factory(model=Subscription)
    inlines = [SubscriptionAppInline]
admin.site.register(Subscription, SubscriptionAdmin)

# settings.py
INSTALLED_APPS = (
    'autocomplete_light',
    'django.contrib.admin',
    ...
...

# urls.py
urlpatterns = patterns(
    ...
    # last record
    url(r'^autocomplete/', include('autocomplete_light.urls')),
)

@admin.register(SubscriptionApp)
class SubscriptionAppAdmin(admin.ModelAdmin):
    form = autocomplete_light.modelform_factory(model=SubscriptionApp)

class SubscriptionAppInline(admin.TabularInline):
    model = SubscriptionApp

@admin.register(Subscription)
class SubscriptionAdmin(admin.ModelAdmin):
    model = Subscription
    ordering = ('name',)