blocks|key|327962|text|在OneToOne关系中使用内联是完全可能的。然而，定义关系的实际字段必须在内联模型上，而不是在父级模型上-就像ForeignKey一样。把它换过来，它就能工作了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|327963|注释后编辑：您说父模型已经注册到管理员:然后取消注册并重新注册。|offset|length|style|BOLD|327964|from+original.satchmo.admin+import+ProductAdmin

class+MyProductInline(admin.StackedInline):
++++model+=+MyProduct

class+ExtendedProductAdmin(ProductAdmin):
++++inlines+=+ProductAdmin.inlines+%2B+(MyProductInline,)

admin.site.unregister(Product)
admin.site.register(Product,+ExtendedProductAdmin)|code-block|syntax|javascript|327965|327966|更新2020+(Django+3.1.1)|327967|这种方法仍然有效，但在新的Django版本中，一些类型发生了变化，因为ExtendedProductAdmin中的inlines现在应该作为列表而不是元组添加，如下所示：|CODE|327968|class+ExtendedProductAdmin(ProductAdmin):
++++inlines+=+ProductAdmin.inlines+%2B+[MyProductInline]|327969|否则，您将得到以下错误：|327970|++++inlines+=+ProductAdmin.inlines+%2B+(MyProductInline,)
TypeError:+can+only+concatenate+list+(not+"tuple")+to+list|327971|entityMap^0|0|0|5|0|0|0|0|L|0|D|6|Z|K|1L|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|16|8|@$D|17|E|18|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|19|8|@$D|1A|E|1B|F|G]|$D|1C|E|1D|F|R]|$D|1E|E|1F|F|R]]|9|@]|A|$]]|$1|S|3|T|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|X|5|J|7|1I|8|@]|9|@]|A|$K|L]]|$1|Y|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|Z|$]]

It's perfectly possible to use an inline for a OneToOne relationship. However, the actual field defining the relationship has to be on the inline model, not the parent one - in just the same way as for a ForeignKey. Switch it over and it will work.
Edit after comment: you say the parent model is already registered with the admin: then unregister it and re-register.
<pre><code>from original.satchmo.admin import ProductAdmin

class MyProductInline(admin.StackedInline):
 model = MyProduct

class ExtendedProductAdmin(ProductAdmin):
 inlines = ProductAdmin.inlines + (MyProductInline,)

admin.site.unregister(Product)
admin.site.register(Product, ExtendedProductAdmin)
</code></pre>
<hr />
Update 2020 (Django 3.1.1)
This method is still working but some types has changed in new Django version since <code>inlines</code> in <code>ExtendedProductAdmin</code> should now be added as list and not tuple, like this:
<pre class="lang-py prettyprint-override"><code>class ExtendedProductAdmin(ProductAdmin):
 inlines = ProductAdmin.inlines + [MyProductInline]
</code></pre>
Or you will get this error:
<pre><code> inlines = ProductAdmin.inlines + (MyProductInline,)
TypeError: can only concatenate list (not &quot;tuple&quot;) to list
</code></pre>

blocks|key|3291988|text|关于最后一个问题，对于多个子类型，什么是最好的解决方案。例如，class+Product包含子类型的class和子类型的class+CD。按照这里所示的方法，您必须编辑一个产品，一般项加上图书的子类型项和CD的子类型项。因此，即使您只想添加一本书，也会获得CD字段。如果您添加子类型，例如DVD，您将获得三个子类型字段组，而您实际上只需要一个子类型组，在前面提到的示例中:book。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3291989|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Referring to the last question, what would be the best solution for multiple sub-types. E.g class Product with sub-type class Book and sub-type class CD. The way shown here you would have to edit a product the general items plus the sub-type items for book AND the sub-type items for CD. So even if you only want to add a book you also get the fields for CD. If you add a sub-type e.g. DVD, you get three sub-type field groups, while you actually only want one sub-type group, in the mentioned example: books.

blocks|key|323238|text|可以使用继承而不是OneToOne关系|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323239|class+Product(models.Model):
++++name+=+models.CharField(max_length=100)
++++...

class+MyProduct(Product):
++++.....|code-block|syntax|javascript|323240|或使用代理类|323241|class+ProductProxy(Product)
++++class+Meta:
++++++++proxy+=+True|323242|在admin.py中|323243|class+MyProductInlines(admin.StackedInline):
++++model+=+MyProduct

class+MyProductAdmin(admin.ModelAdmin):
++++inlines+=+[MyProductInlines]

++++def+queryset(self,+request):
++++++++qs+=+super(MyProductAdmin,+self).queryset(request)
++++++++qs+=+qs.exclude(relatedNameForYourProduct__isnone=True)
++++++++return+qs

admin.site.register(ProductProxy,+MyProductAdmin)|323244|在这个变种中，你的产品将是内联的。|323245|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Maybe use inheritance instead OneToOne relationship

<pre><code>class Product(models.Model):
 name = models.CharField(max_length=100)
 ...

class MyProduct(Product):
 .....
</code></pre>

Or use proxy classes

<pre><code>class ProductProxy(Product)
 class Meta:
 proxy = True
</code></pre>

in admin.py

<pre><code>class MyProductInlines(admin.StackedInline):
 model = MyProduct

class MyProductAdmin(admin.ModelAdmin):
 inlines = [MyProductInlines]

 def queryset(self, request):
 qs = super(MyProductAdmin, self).queryset(request)
 qs = qs.exclude(relatedNameForYourProduct__isnone=True)
 return qs

admin.site.register(ProductProxy, MyProductAdmin)
</code></pre>

In this variant your product will be in inline.

blocks|key|325735|text|你也可以尝试在你的OneToOneField上设置'parent_link=True‘？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|325736|https://docs.djangoproject.com/en/dev/topics/db/models/#specifying-the-parent-link-field|offset|length|325737|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|2G|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

You can also try setting 'parent_link=True' on your OneToOneField?

<a href="https://docs.djangoproject.com/en/dev/topics/db/models/#specifying-the-parent-link-field" rel="nofollow">https://docs.djangoproject.com/en/dev/topics/db/models/#specifying-the-parent-link-field</a>

I am putting together the admin for a satchmo application. Satchmo uses OneToOne relations to extend the base <code>Product</code> model, and I'd like to edit it all on one page.

It is possible to have a OneToOne relation as an Inline? If not, what is the best way to add a few fields to a given page of my admin that will eventually be saved into the OneToOne relation?

for example:

<pre><code>class Product(models.Model):
 name = models.CharField(max_length=100)
 ...

class MyProduct(models.Model):
 product = models.OneToOne(Product)
 ...
</code></pre>

I tried this for my admin but it does not work, and seems to expect a Foreign Key:

<pre><code>class ProductInline(admin.StackedInline):
 model = Product
 fields = ('name',)

class MyProductAdmin(admin.ModelAdmin):
 inlines = (AlbumProductInline,)

admin.site.register(MyProduct, MyProductAdmin)
</code></pre>

Which throws this error: <code>&lt;class 'satchmo.product.models.Product'&gt; has no ForeignKey to &lt;class 'my_app.models.MyProduct'&gt;</code>

Is the only way to do this a <a href="http://docs.djangoproject.com/en/dev/ref/contrib/admin/#form" rel="noreferrer">Custom Form</a>?

edit: Just tried the following code to add the fields directly... also does not work:

<pre><code>class AlbumAdmin(admin.ModelAdmin):
 fields = ('product__name',)
</code></pre>

Django Admin: OneToOne Relation as an Inline?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在把一个satchmo应用程序的管理。Product使用OneToOne关系来扩展基本的Satchmo模型，我想在一个页面上对其进行编辑。可以将OneToOne关系作为内联？如果不是，那么添加几个字段到我的管理页面的最佳方式是什么，这些字段最终将保存到OneToOne关系中？例如：class Product(mod...

问Django Admin:作为内联的OneToOne关系？
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问Django Admin:作为内联的OneToOne关系？
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