Django管理嵌套內聯

Question

我需要一個嵌套的Django管理內聯， ，我可以在其他內聯如下所示內嵌日期字段內聯。

我有以下型號：

class Person(models.Model): 
    name = models.CharField(max_length=200) 
    id_no = models.IntegerField() 

class Certificate(models.Model): 
    cerfificate_no = models.CharField(max_length=200) 
    certificate_date = models.DateField(max_length=100) 
    person = models.ForeignKey(Person) 
    training = models.CharField(max_length=200) 

class Training_Date(models.Model): 
     date = models.DateField() 
     certificate = models.ForeignKey(Certificate) 

及以下管理：

class CertificateInline(admin.StackedInline): 
    model = Certificate 

class PersonAdmin(admin.ModelAdmin): 
    inlines = [CertificateInline,] 
admin.site.register(Person,PersonAdmin) 

，但我需要包括Training_Date模型直列這是證書管理內嵌的一部分。

有什麼想法？

Answer 1

AFAIK，在默認的Django管理員中不能有第二級內聯。

Django管理員只是一個普通的Django應用程序，所以沒有什麼能阻止你實現第二層嵌套表單，但恕我直言，這將是一種複雜的設計來實現。也許這就是爲什麼沒有規定。

Answer 2

最近在https://code.djangoproject.com/ticket/9025已經有一些動作，但我不會屏住呼吸。

圍繞這一點的一種常見方式是通過同時具有的ModelAdmin和內聯該同樣的模型鏈接到第一和第二（或第二和第三）電平之間的管理員：

給證明書的ModelAdmin與TrainingDate如內聯。爲CertificateInline添加一個額外的字段「Details」，該字段是指向其ModelAdmin更改表單的鏈接。

models.py：

from django.core import urlresolvers 

class Certificate(models.Model): 

    # ... 

    def changeform_link(self): 
     if self.id: 
      # Replace "myapp" with the name of the app containing 
      # your Certificate model: 
      changeform_url = urlresolvers.reverse(
       'admin:myapp_certificate_change', args=(self.id,) 
      ) 
      return u'<a href="%s" target="_blank">Details</a>' % changeform_url 
     return u'' 
    changeform_link.allow_tags = True 
    changeform_link.short_description = '' # omit column header 

admin.py：

# Certificate change form has training dates as inline 

class TrainingDateInline(admin.StackedInline): 
    model = TrainingDate 

class CertificateAdmin(admin.ModelAdmin): 
    inlines = [TrainingDateInline,] 
admin.site.register(Certificate ,CertificateAdmin) 

# Person has Certificates inline but rather 
# than nesting inlines (not possible), shows a link to 
# its own ModelAdmin's change form, for accessing TrainingDates: 

class CertificateLinkInline(admin.TabularInline): 
    model = Certificate 
    # Whichever fields you want: (I usually use only a couple 
    # needed to identify the entry) 
    fields = ('cerfificate_no', 'certificate_date', 'changeform_link') 
    readonly_fields = ('changeform_link',) 

class PersonAdmin(admin.ModelAdmin): 
    inlines = [CertificateLinkInline,] 
admin.site.register(Person, PersonAdmin) 

Answer 3

更通用的解決方案

from django.utils.safestring import mark_safe 
from django.core.urlresolvers import reverse 
class EditLinkToInlineObject(object): 
    def edit_link(self, instance): 
     url = reverse('admin:%s_%s_change' % (
      instance._meta.app_label, instance._meta.model_name), args=[instance.pk]) 
     if instance.pk: 
      return mark_safe(u'<a href="{u}">edit</a>'.format(u=url)) 
     else: 
      return '' 

class MyModelInline(EditLinkToInlineObject, admin.TabularInline): 
    model = MyModel 
    readonly_fields = ('edit_link',) 

class MySecondModelAdmin(admin.ModelAdmin): 
    inlines = (MyModelInline,) 

admin.site.register(MyModel) 
admin.site.register(MySecondModel, MySecondModelAdmin) 

Answer 4

pip install django-nested-inline 

這個包應該做你所需要的。在提供

Answer 5

嵌套內聯： https://github.com/BertrandBordage/django-super-inlines/

pip install django-super-inlines 

Answer 6

我使用@bigzbig提供的解決方案（謝謝）。

我也想回去的第一列表頁面，一旦改變已經保存，以便補充說：

class MyModelInline(EditLinkToInlineObject, admin.TabularInline): 
    model = MyModel 
    readonly_fields = ('edit_link',) 

    def response_post_save_change(self, request, obj): 
     my_second_model_id = MyModel.objects.get(pk=obj.pk).my_second_model_id 
     return redirect("/admin/mysite/mysecondmodel/%s/change/" % (my_second_model_id))