blocks|key|1662174|text|这只是一个暗中的尝试，但是调用admin.site.register(<+Model+class+>，<+ModelAdmin类>)的顺序是否有可能决定显示顺序呢？实际上，我怀疑这是否会起作用，因为我相信Django维护了一个作为标准->字典实现的模型Python+ModelAdmin对象的注册表，它不维护迭代顺序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662175|如果这不是您想要的行为方式，您可以在django/contrib/admin中尝试源代码。如果需要保持迭代顺序，可以将AdminSite类(在admin/sites.py中)中的_registry对象替换为保持键插入顺序的UserDict或DictMixin。(但请对此建议持保留态度，因为我自己从来没有做过这样的更改，而且我只是对Django如何迭代ModelAdmin对象的集合做了一个有根据的猜测。我确实认为django/contrib/admin/sites.py是查找此代码的地方，尤其是AdminSite类以及注册()和索引()方法。)|1662176|显然，这里最好的方法是在您自己的/admin.py模块中指定一个简单的选项。我相信这就是你希望得到的答案。不过，我不确定这些选项是否存在。|1662177|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

This is just a wild stab in the dark, but is there any chance that the order in which you call admin.site.register(&lt; Model class >, &lt; ModelAdmin class >) can determine the display order? Actually, I doubt that would work because I believe Django maintains a registry of the Model -> ModelAdmin objects implemented as a standard Python dictionary, which does not maintain iteration ordering.

If that doesn't behave the way you want, you can always play around with the source in django/contrib/admin. If you need the iteration order maintained, you could replace the _registry object in the AdminSite class (in admin/sites.py) with a UserDict or DictMixin that maintains insertion order for the keys. (But please take this advice with a grain of salt, since I've never made these kinds of changes myself and I'm only making an educated guess at how Django iterates over the collection of ModelAdmin objects. I do think that django/contrib/admin/sites.py is the place to look for this code, though, and the AdminSite class and register() and index() methods in particular are what you want.)

Obviously the nicest thing here would be a simple option for you to specify in your own /admin.py module. I'm sure that's the kind of answer you were hoping to receive. I'm not sure if those options exist, though.

blocks|key|1447428|text|我把它添加到了Django的跟踪中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1447429|http://code.djangoproject.com/ticket/9928|offset|length|1447430|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|15|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]]]]

I added it to Django's trac:

<a href="http://code.djangoproject.com/ticket/9928" rel="nofollow noreferrer">http://code.djangoproject.com/ticket/9928</a>

blocks|key|1447486|text|我的解决方案是创建django.contrib.admin.sites.AdminSite和django.contrib.admin.options.ModelAdmin的子类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1447487|我这样做是为了可以为每个应用程序显示一个更具描述性的标题，并对每个应用程序中的模型进行排序。因此，我在我的settings.py中有一个字典，它将app_labels映射到描述性名称和它们应该出现的顺序，当我在管理站点注册模型时，模型是按我在每个ModelAdmin中提供的序号字段排序的。|1447488|尽管在文档中鼓励创建自己的AdminSite和ModelAdmin子类，但我的解决方案最终看起来像是一个丑陋的技巧。|1447489|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

My solution was to make subclasses of django.contrib.admin.sites.AdminSite and django.contrib.admin.options.ModelAdmin .

I did this so I could display a more descriptive title for each app and order the appearance of models in each app. So I have a dict in my settings.py that maps app_labels to descriptive names and the order in which they should appear, the models are ordered by an ordinal field I provide in each ModelAdmin when I register them with the admin site.

Although making your own subclasses of AdminSite and ModelAdmin is encouraged in the docs, my solution looks like an ugly hack in the end.

blocks|key|1662342|text|您可以尝试的解决方法是调整models.py，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662343|class+Topping(models.Model):
++++.
++++.
++++.
++++class+Meta:
++++++++verbose_name_plural+=+"2.+Toppings"

class+Pizza(models.Model):
++++.
++++.
++++.
++++class+Meta:
++++++++verbose_name_plural+=+"1.+Pizzas"|code-block|syntax|javascript|1662344|不确定它是否违反了django的最佳实践，但它是有效的(使用django主干进行了测试)。|1662345|祝好运!|1662346|PS:如果这个答案发布的太晚了，我很抱歉，但它可以在未来类似的情况下帮助其他人。|1662347|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

A workaround that you can try is tweaking your models.py as follows:

<pre><code>class Topping(models.Model):
 .
 .
 .
 class Meta:
 verbose_name_plural = "2. Toppings"

class Pizza(models.Model):
 .
 .
 .
 class Meta:
 verbose_name_plural = "1. Pizzas"
</code></pre>

Not sure if it is against the django's best practices but it works (tested with django trunk).

Good luck!

PS: sorry if this answer was posted too late but it can help others in future similar situations.

blocks|key|1447587|text|如果你想在10秒内解决这个问题，只需在verbose_name_plural中使用空格，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1447588|class+Topping(models.Model):
++++class+Meta:
++++++++verbose_name_plural+=+"++Toppings"+#+2+spaces

class+Pizza(models.Model):
++++class+Meta:
++++++++verbose_name_plural+=+"+Pizzas"+#+1+space|code-block|syntax|javascript|1447589|当然，它并不高明，但在我们得到更好的解决方案之前，它可能会工作一段时间。|1447590|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you want to solve this in 10 seconds just use spaces in verbose_name_plural, for example:

<pre><code>class Topping(models.Model):
 class Meta:
 verbose_name_plural = " Toppings" # 2 spaces

class Pizza(models.Model):
 class Meta:
 verbose_name_plural = " Pizzas" # 1 space
</code></pre>

Of course it isn't ellegant but may work for a while before we get a better solution.

blocks|key|1662438|text|多亏了这个Django+snippet，我最终做到了这一点，你只需要知道ADMIN_REORDER的设置：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1662439|ADMIN_REORDER+=+(
++++('app1',+('App1Model1',+'App1Model2',+'App1Model3')),
++++('app2',+('App2Model1',+'App2Model2')),
)|code-block|syntax|javascript|1662440|app1不能以项目名称为前缀，即使用app1而不是mysite.app1。|1662441|entityMap|0|LINK|mutability|MUTABLE|url|http://djangosnippets.org/snippets/1939/#comments^0|10|D|5|E|0|0|0|0|4|I|4|P|B|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

I eventually managed to do it thanks to this <a href="http://djangosnippets.org/snippets/1939/#comments">Django snippet</a>, you just need to be aware of the <code>ADMIN_REORDER</code> setting:

<pre><code>ADMIN_REORDER = (
 ('app1', ('App1Model1', 'App1Model2', 'App1Model3')),
 ('app2', ('App2Model1', 'App2Model2')),
)
</code></pre>

<code>app1</code> must not be prefixed with the project name, i.e. use <code>app1</code> instead of <code>mysite.app1</code>.

blocks|key|1447695|text|如果您使用的是AdminSite的Suit，则可以使用menu+tag进行菜单定制。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1447696|entityMap|0|LINK|mutability|MUTABLE|url|http://djangosuit.com/|1|http://django-suit.readthedocs.org/en/develop/configuration.html#id1^0|H|4|0|R|8|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

If you're using <a href="http://djangosuit.com" rel="nofollow">Suit</a> for the AdminSite you can do menu customization using the <a href="http://django-suit.readthedocs.org/en/develop/configuration.html#id1" rel="nofollow">menu tag</a>.

blocks|key|1662551|text|以下是Emmanuel使用的片段，针对Django+1.8进行了更新：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662552|在templatetags/admin_reorder.py中：|1662553|from+django+import+template
from+django.conf+import+settings
from+collections+import+OrderedDict

register+=+template.Library()

#+from+http://www.djangosnippets.org/snippets/1937/
def+register_render_tag(renderer):
++++"""
++++Decorator+that+creates+a+template+tag+using+the+given+renderer+as+the+
++++render+function+for+the+template+tag+node+-+the+render+function+takes+two+
++++arguments+-+the+template+context+and+the+tag+token
++++"""
++++def+tag(parser,+token):
++++++++class+TagNode(template.Node):
++++++++++++def+render(self,+context):
++++++++++++++++return+renderer(context,+token)
++++++++return+TagNode()
++++for+copy_attr+in+("__dict__",+"__doc__",+"__name__"):
++++++++setattr(tag,+copy_attr,+getattr(renderer,+copy_attr))
++++return+register.tag(tag)

@register_render_tag
def+admin_reorder(context,+token):
++++"""
++++Called+in+admin/base_site.html+template+override+and+applies+custom+ordering
++++of+apps/models+defined+by+settings.ADMIN_REORDER
++++"""
++++#+sort+key+function+-+use+index+of+item+in+order+if+exists,+otherwise+item
++++sort+=+lambda+order,+item:+(order.index(item),+"")+if+item+in+order+else+(
++++++++len(order),+item)
++++if+"app_list"+in+context:
++++++++#+sort+the+app+list
++++++++order+=+OrderedDict(settings.ADMIN_REORDER)
++++++++context["app_list"].sort(key=lambda+app:+sort(order.keys(),
++++++++++++app["app_url"].strip("/").split("/")[-1]))
++++++++for+i,+app+in+enumerate(context["app_list"]):
++++++++++++#+sort+the+model+list+for+each+app
++++++++++++app_name+=+app["app_url"].strip("/").split("/")[-1]
++++++++++++if+not+app_name:
++++++++++++++++app_name+=+app["name"].lower()
++++++++++++model_order+=+[m.lower()+for+m+in+order.get(app_name,+[])]
++++++++++++context["app_list"][i]["models"].sort(key=lambda+model:
++++++++++++sort(model_order,+model["admin_url"].strip("/").split("/")[-1]))
++++return+""|code-block|syntax|javascript|1662554|在settings.py中：|1662555|ADMIN_REORDER+=+(
++++('app1',+('App1Model1',+'App1Model2',+'App1Model3')),
++++('app2',+('App2Model1',+'App2Model2')),
)|1662556|(在此处插入您自己的应用程序名称。管理员会将缺少的应用程序或型号放在列表的末尾，只要您在每个应用程序中列出至少两个型号。)|1662557|在您的base_site.html副本中：|1662558|{%25+extends+"admin/base.html"+%25}
{%25+load+i18n+admin_reorder+%25}

{%25+block+title+%25}{{+title+}}+%7C+{%25+trans+'Django+site+admin'+%25}{%25+endblock+%25}

{%25+block+branding+%25}
{%25+admin_reorder+%25}
<h1+id="site-name">{%25+trans+'Django+administration'+%25}</h1>
{%25+endblock+%25}

{%25+block+nav-global+%25}{%25+endblock+%25}|1662559|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Here's the snippet Emmanuel used, updated for Django 1.8:

In templatetags/admin_reorder.py:

<pre><code>from django import template
from django.conf import settings
from collections import OrderedDict

register = template.Library()

# from http://www.djangosnippets.org/snippets/1937/
def register_render_tag(renderer):
 """
 Decorator that creates a template tag using the given renderer as the 
 render function for the template tag node - the render function takes two 
 arguments - the template context and the tag token
 """
 def tag(parser, token):
 class TagNode(template.Node):
 def render(self, context):
 return renderer(context, token)
 return TagNode()
 for copy_attr in ("__dict__", "__doc__", "__name__"):
 setattr(tag, copy_attr, getattr(renderer, copy_attr))
 return register.tag(tag)

@register_render_tag
def admin_reorder(context, token):
 """
 Called in admin/base_site.html template override and applies custom ordering
 of apps/models defined by settings.ADMIN_REORDER
 """
 # sort key function - use index of item in order if exists, otherwise item
 sort = lambda order, item: (order.index(item), "") if item in order else (
 len(order), item)
 if "app_list" in context:
 # sort the app list
 order = OrderedDict(settings.ADMIN_REORDER)
 context["app_list"].sort(key=lambda app: sort(order.keys(),
 app["app_url"].strip("/").split("/")[-1]))
 for i, app in enumerate(context["app_list"]):
 # sort the model list for each app
 app_name = app["app_url"].strip("/").split("/")[-1]
 if not app_name:
 app_name = app["name"].lower()
 model_order = [m.lower() for m in order.get(app_name, [])]
 context["app_list"][i]["models"].sort(key=lambda model:
 sort(model_order, model["admin_url"].strip("/").split("/")[-1]))
 return ""
</code></pre>

In settings.py:

<pre><code>ADMIN_REORDER = (
 ('app1', ('App1Model1', 'App1Model2', 'App1Model3')),
 ('app2', ('App2Model1', 'App2Model2')),
)
</code></pre>

(insert your own app names in here. Admin will place missing apps or models at the end of the list, so long as you list at least two models in each app.)

In your copy of base_site.html:

<pre><code>{% extends "admin/base.html" %}
{% load i18n admin_reorder %}

{% block title %}{{ title }} | {% trans 'Django site admin' %}{% endblock %}

{% block branding %}
{% admin_reorder %}
&lt;h1 id="site-name"&gt;{% trans 'Django administration' %}&lt;/h1&gt;
{% endblock %}

{% block nav-global %}{% endblock %}
</code></pre>

blocks|key|1662593|text|现在有一个很好的Django包：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662594|https://pypi.python.org/pypi/django-modeladmin-reorder|offset|length|1662595|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1I|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]]]]

There's now a nice Django package for that:

<a href="https://pypi.python.org/pypi/django-modeladmin-reorder">https://pypi.python.org/pypi/django-modeladmin-reorder</a>

blocks|key|1662645|text|我正在寻找一个简单的解决方案，在那里我可以在管理面板中按名称排序应用程序。我想出了以下模板标记：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662646|from+django+import+template
from+django.conf+import+settings
register+=+template.Library()

@register.filter
def+sort_apps(apps):
++++apps.sort(
++++++++key+=+lambda+x:
++++++++settings.APP_ORDER.index(x['app_label'])
++++++++if+x['app_label']+in+settings.APP_ORDER
++++++++else+len(apps)
++++)
++++print+[x['app_label']+for+x+in+apps]
++++return+apps|code-block|syntax|javascript|1662647|然后，只需覆盖templates/admin/index.html并添加该模板标记：|offset|length|style|CODE|1662648|{%25+extends+"admin/index.html"+%25}
{%25+block+content+%25}
{%25+load+i18n+static+sort_apps+%25}
<div+id="content-main">

{%25+if+app_list+%25}
++++{%25+for+app+in+app_list%7Csort_apps+%25}
++++++++<div+class="app-{{+app.app_label+}}+module">
++++++++<table>
++++++++<caption>
++++++++++++<a+href="{{+app.app_url+}}"+class="section"+title="{%25+blocktrans+with+name=app.name+%25}Models+in+the+{{+name+}}+application{%25+endblocktrans+%25}">{{+app.name+}}</a>
++++++++</caption>
++++++++{%25+for+model+in+app.models+%25}
++++++++++++<tr+class="model-{{+model.object_name%7Clower+}}">
++++++++++++{%25+if+model.admin_url+%25}
++++++++++++++++<th+scope="row"><a+href="{{+model.admin_url+}}">{{+model.name+}}</a></th>
++++++++++++{%25+else+%25}
++++++++++++++++<th+scope="row">{{+model.name+}}</th>
++++++++++++{%25+endif+%25}

++++++++++++{%25+if+model.add_url+%25}
++++++++++++++++<td><a+href="{{+model.add_url+}}"+class="addlink">{%25+trans+'Add'+%25}</a></td>
++++++++++++{%25+else+%25}
++++++++++++++++<td>&nbsp;</td>
++++++++++++{%25+endif+%25}

++++++++++++{%25+if+model.admin_url+%25}
++++++++++++++++<td><a+href="{{+model.admin_url+}}"+class="changelink">{%25+trans+'Change'+%25}</a></td>
++++++++++++{%25+else+%25}
++++++++++++++++<td>&nbsp;</td>
++++++++++++{%25+endif+%25}
++++++++++++</tr>
++++++++{%25+endfor+%25}
++++++++</table>
++++++++</div>
++++{%25+endfor+%25}
{%25+else+%25}
++++{%25+trans+"You+don't+have+permission+to+edit+anything."+%25}
{%25+endif+%25}
</div>
{%25+endblock+%25}|1662649|然后在settings.py中自定义APP_ORDER：|1662650|APP_ORDER+=+[
++++'app1',
++++'app2',
++++#+and+so+on...
]|1662651|它在Django+1.10上运行得很好|1662652|entityMap^0|0|0|7|Q|0|0|I|9|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$I|13|J|14|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

I was looking for a simple solution where I could order the apps by their name in the admin panel. I came up with the following template tag:

<pre><code>from django import template
from django.conf import settings
register = template.Library()

@register.filter
def sort_apps(apps):
 apps.sort(
 key = lambda x:
 settings.APP_ORDER.index(x['app_label'])
 if x['app_label'] in settings.APP_ORDER
 else len(apps)
 )
 print [x['app_label'] for x in apps]
 return apps
</code></pre>

Then, just override <code>templates/admin/index.html</code> and add that template tag:

<pre><code>{% extends "admin/index.html" %}
{% block content %}
{% load i18n static sort_apps %}
&lt;div id="content-main"&gt;

{% if app_list %}
 {% for app in app_list|sort_apps %}
 &lt;div class="app-{{ app.app_label }} module"&gt;
 &lt;table&gt;
 &lt;caption&gt;
 &lt;a href="{{ app.app_url }}" class="section" title="{% blocktrans with name=app.name %}Models in the {{ name }} application{% endblocktrans %}"&gt;{{ app.name }}&lt;/a&gt;
 &lt;/caption&gt;
 {% for model in app.models %}
 &lt;tr class="model-{{ model.object_name|lower }}"&gt;
 {% if model.admin_url %}
 &lt;th scope="row"&gt;&lt;a href="{{ model.admin_url }}"&gt;{{ model.name }}&lt;/a&gt;&lt;/th&gt;
 {% else %}
 &lt;th scope="row"&gt;{{ model.name }}&lt;/th&gt;
 {% endif %}

 {% if model.add_url %}
 &lt;td&gt;&lt;a href="{{ model.add_url }}" class="addlink"&gt;{% trans 'Add' %}&lt;/a&gt;&lt;/td&gt;
 {% else %}
 &lt;td&gt;&amp;nbsp;&lt;/td&gt;
 {% endif %}

 {% if model.admin_url %}
 &lt;td&gt;&lt;a href="{{ model.admin_url }}" class="changelink"&gt;{% trans 'Change' %}&lt;/a&gt;&lt;/td&gt;
 {% else %}
 &lt;td&gt;&amp;nbsp;&lt;/td&gt;
 {% endif %}
 &lt;/tr&gt;
 {% endfor %}
 &lt;/table&gt;
 &lt;/div&gt;
 {% endfor %}
{% else %}
 &lt;p&gt;{% trans "You don't have permission to edit anything." %}&lt;/p&gt;
{% endif %}
&lt;/div&gt;
{% endblock %}
</code></pre>

Then customized the <code>APP_ORDER</code> in settings.py:

<pre><code>APP_ORDER = [
 'app1',
 'app2',
 # and so on...
]
</code></pre>

It works great on Django 1.10

blocks|key|1662680|text|custom_admin.py|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662681|from+django.contrib.admin+import+AdminSite


class+CustomAdminSite(AdminSite):
++++def+get_urls(self):
++++++++urls+=+super(MyAdminSite,+self).get_urls()
++++++++return+urls

++++def+get_app_list(self,+request):
++++++++app_dict+=+self._build_app_dict(request)
++++++++ordered_app_list+=+[]

++++++++if+app_dict:
++++++++++++#+TODO:+change+this+dict
++++++++++++admin_ordering+=+{
++++++++++++++++'app1':+('Model1',+'Model2'),
++++++++++++++++'app2':+('Model7',+'Model4'),
++++++++++++++++'app3':+('Model3',),
++++++++++++}

++++++++++++ordered_app_list+=+[]
++++++++++++for+app_key+in+admin_ordering:
++++++++++++++++app+=+app_dict[app_key]
++++++++++++++++app_ordered_models+=+[]
++++++++++++++++for+model_name+in+admin_ordering[app_key]:
++++++++++++++++++++for+model+in+app_dict[app_key]['models']:
++++++++++++++++++++++++if+model['object_name']+==+model_name:
++++++++++++++++++++++++++++app_ordered_models.append(model)
++++++++++++++++++++++++++++break

++++++++++++++++app['models']+=+app_ordered_models
++++++++++++++++ordered_app_list.append(app)

++++++++return+ordered_app_list|code-block|syntax|javascript|1662682|admin_site+=+CustomAdminSite()|1662683|urls.py|1662684|from+custom_admin+import+admin_site

urlpatterns+=+[
++++path('admin/',+admin_site.urls),
]|1662685|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

custom_admin.py
<pre><code>from django.contrib.admin import AdminSite


class CustomAdminSite(AdminSite):
 def get_urls(self):
 urls = super(MyAdminSite, self).get_urls()
 return urls

 def get_app_list(self, request):
 app_dict = self._build_app_dict(request)
 ordered_app_list = []

 if app_dict:
 # TODO: change this dict
 admin_ordering = {
 'app1': ('Model1', 'Model2'),
 'app2': ('Model7', 'Model4'),
 'app3': ('Model3',),
 }

 ordered_app_list = []
 for app_key in admin_ordering:
 app = app_dict[app_key]
 app_ordered_models = []
 for model_name in admin_ordering[app_key]:
 for model in app_dict[app_key]['models']:
 if model['object_name'] == model_name:
 app_ordered_models.append(model)
 break

 app['models'] = app_ordered_models
 ordered_app_list.append(app)

 return ordered_app_list
</code></pre>
admin_site = CustomAdminSite()
urls.py
<pre><code>from custom_admin import admin_site

urlpatterns = [
 path('admin/', admin_site.urls),
]
</code></pre>

blocks|key|1447945|text|将以下代码添加到您的settings.py：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1447946|def+get_app_list(self,+request):
++++"""
++++Return+a+sorted+list+of+all+the+installed+apps+that+have+been
++++registered+on+this+site.
++++"""
++++ordering+=+{
++++++++#+for+superuser
++++++++'Group':+1,
++++++++'User':+2,

++++++++#+fist+app
++++++++'TopMessage':+101,
++++++++'Slider':+102,
++++++++'ProductCategory':+103,
++++++++'Product':+104,
++++++++'ProductPicture':+105,
++++++++
++++++++#+2nd+app
++++++++'ProductReview':+201,
++++++++'Promotion':+202,
++++++++'BestSeller':+203,
++++}
++++app_dict+=+self._build_app_dict(request)
++++app_list+=+sorted(app_dict.values(),+key=lambda+x:+x['name'].lower())

++++for+app+in+app_list:
++++++++app['models'].sort(key=lambda+x:+ordering[x['object_name']])

++++return+app_list


admin.AdminSite.get_app_list+=+get_app_list|code-block|syntax|javascript|1447947|并对排序字典进行更改以匹配您的应用程序和型号。就这样。|1447948|我的解决方案的好处是，如果用户是超级用户，它将显示“auth”模型。|1447949|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Add the below code to your settings.py:
<pre><code>def get_app_list(self, request):
 &quot;&quot;&quot;
 Return a sorted list of all the installed apps that have been
 registered on this site.
 &quot;&quot;&quot;
 ordering = {
 # for superuser
 'Group': 1,
 'User': 2,

 # fist app
 'TopMessage': 101,
 'Slider': 102,
 'ProductCategory': 103,
 'Product': 104,
 'ProductPicture': 105,
 
 # 2nd app
 'ProductReview': 201,
 'Promotion': 202,
 'BestSeller': 203,
 }
 app_dict = self._build_app_dict(request)
 app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())

 for app in app_list:
 app['models'].sort(key=lambda x: ordering[x['object_name']])

 return app_list


admin.AdminSite.get_app_list = get_app_list
</code></pre>
And make changes to the ordering dictionary to match your apps and models. That's it.
The benefit of my solution is that it will show the 'auth' models if the user is a superuser.

blocks|key|1662767|text|在Settings.py文件中设置管理中列出的应用程序/模块的显示顺序|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1662768|设置我的排序列表|1662769|ADMIN_ORDERING+=+[
++++('Your_App1',+'*'),
++++('Your_App2',+'*'),
]|code-block|syntax|javascript|1662770|在ADMIN_ORDERING下面创建一个排序函数|1662771|def+get_app_list(self,+request):
++++"""
++++You+Can+Set+Manually+Ordering+For+Your+Apps+And+Models+++

++++++++ADMIN_ORDERING+=+[
++++++++++++('Your_App1',+[
++++++++++++++++'module_1',
++++++++++++++++'module_2'
++++++++++++]),
++++++++++++('Your_App2',+'*'),
++++++++]
++++"""

++++app_dict+=+self._build_app_dict(request)

++++#+Let's+start+by+sorting+the+apps+alphabetically+on+a+list:
++++app_list+=+sorted(app_dict.values(),+key=lambda+x:+x['name'].lower())

++++#+Sorting+the+models+alphabetically+within+each+app.
++++for+app+in+app_list:
++++++++if+app['app_label']+in+[el[0]+for+el+in+ADMIN_ORDERING]:
++++++++++++app_list.remove(app)
++++++++else:
++++++++++++app['models'].sort(key=lambda+x:+x['name'])

++++#+Now+we+order+the+app+list+in+our+defined+way+in+ADMIN_ORDERING+(which+could+be+a+subset+of+all+apps).
++++my_ordered_apps+=+[]
++++if+app_dict:
++++++++for+app_name,+object_list+in+ADMIN_ORDERING:
++++++++++++app+=+app_dict[app_name]
++++++++++++if+object_list+==+'*':
++++++++++++++++app['models'].sort(key=lambda+x:+x['name'])
++++++++++++else:
++++++++++++++++app['models'].sort(key=lambda+x:+object_list.index(x['object_name']))
++++++++++++my_ordered_apps.append(app)

++++++++#+Now+we+combine+and+arrange+the+2+lists+together
++++++++my_ordered_apps.extend(app_list)

++++return+my_ordered_apps

#+Covering+django.contrib.admin.AdminSite.get_app_list
from+django.contrib+import+admin
admin.AdminSite.get_app_list+=+get_app_list|1662772|entityMap^0|1|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

In <code>Settings.py</code> file setting the order in which the apps/modules show up listed on Admin
set my ordering list
<pre><code>ADMIN_ORDERING = [
 ('Your_App1', '*'),
 ('Your_App2', '*'),
]
</code></pre>
Creating a sort function below ADMIN_ORDERING
<pre><code>def get_app_list(self, request):
 &quot;&quot;&quot;
 You Can Set Manually Ordering For Your Apps And Models 

 ADMIN_ORDERING = [
 ('Your_App1', [
 'module_1',
 'module_2'
 ]),
 ('Your_App2', '*'),
 ]
 &quot;&quot;&quot;

 app_dict = self._build_app_dict(request)

 # Let's start by sorting the apps alphabetically on a list:
 app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())

 # Sorting the models alphabetically within each app.
 for app in app_list:
 if app['app_label'] in [el[0] for el in ADMIN_ORDERING]:
 app_list.remove(app)
 else:
 app['models'].sort(key=lambda x: x['name'])

 # Now we order the app list in our defined way in ADMIN_ORDERING (which could be a subset of all apps).
 my_ordered_apps = []
 if app_dict:
 for app_name, object_list in ADMIN_ORDERING:
 app = app_dict[app_name]
 if object_list == '*':
 app['models'].sort(key=lambda x: x['name'])
 else:
 app['models'].sort(key=lambda x: object_list.index(x['object_name']))
 my_ordered_apps.append(app)

 # Now we combine and arrange the 2 lists together
 my_ordered_apps.extend(app_list)

 return my_ordered_apps

# Covering django.contrib.admin.AdminSite.get_app_list
from django.contrib import admin
admin.AdminSite.get_app_list = get_app_list
</code></pre>

blocks|key|1662817|text|ADMIN_ORDERING+=+{
++++"PizzaApp":+[
++++++++"Pizzas",
++++++++"Toppings",
++++],
}


def+get_app_list(self,+request):
++++app_dict+=+self._build_app_dict(request)
++++for+app_name,+object_list+in+app_dict.items():
++++++++if+app_name+in+ADMIN_ORDERING:
++++++++++++app+=+app_dict[app_name]
++++++++++++app["models"].sort(
++++++++++++++++key=lambda+x:+ADMIN_ORDERING[app_name].index(x["object_name"])
++++++++++++)
++++++++++++app_dict[app_name]
++++++++++++yield+app
++++++++else:
++++++++++++yield+app_dict[app_name]


admin.AdminSite.get_app_list+=+get_app_list|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1662818|这个解决方案适用于我，修改了来自林伟雄的解决方案。|unstyled|1662819|您可以保留默认的auth顺序并指定您自己的顺序。|1662820|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>ADMIN_ORDERING = {
 &quot;PizzaApp&quot;: [
 &quot;Pizzas&quot;,
 &quot;Toppings&quot;,
 ],
}


def get_app_list(self, request):
 app_dict = self._build_app_dict(request)
 for app_name, object_list in app_dict.items():
 if app_name in ADMIN_ORDERING:
 app = app_dict[app_name]
 app[&quot;models&quot;].sort(
 key=lambda x: ADMIN_ORDERING[app_name].index(x[&quot;object_name&quot;])
 )
 app_dict[app_name]
 yield app
 else:
 yield app_dict[app_name]


admin.AdminSite.get_app_list = get_app_list
</code></pre>
This solution works for me, modified the one from 林伟雄.
You get to keep the default auth ordering AND specify your own.

blocks|key|1662876|text|一种非常简单且自包含的方法是使用装饰器模式以这种方式重新排序您的应用程序和模块：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662877|#+admin.py

from+django.contrib+import+admin

def+app_resort(func):++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++def+inner(*args,+**kwargs):++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++app_list+=+func(*args,+**kwargs)
++++++++#+Useful+to+discover+your+app+and+module+list:
++++++++#import+pprint++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++#pprint.pprint(app_list)

++++++++app_sort_key+=+'name'
++++++++app_ordering+=+{
++++++++++++"APP_NAME1":+1,
++++++++++++"APP_NAME2":+2,
++++++++++++"APP_NAME3":+3,
++++++++}

++++++++resorted_app_list+=+sorted(app_list,+key=lambda+x:+app_ordering[x[app_sort_key]]+if+x[app_sort_key]+in+app_ordering+else+1000)

++++++++model_sort_key+=+'object_name'
++++++++model_ordering+=+{
++++++++++++"Model1":+1,
++++++++++++"Model2":+2,
++++++++++++"Model3":+3,
++++++++++++"Model14":+4,
++++++++}
++++++++for+app+in+resorted_app_list:
++++++++++++app['models'].sort(key=lambda+x:+model_ordering[x[model_sort_key]]+if+x[model_sort_key]+in+model_ordering+else+1000)
++++++++return+resorted_app_list
++++return+inner++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+++++++++++++++++++
admin.site.get_app_list+=+app_resort(admin.site.get_app_list)|code-block|syntax|javascript|1662878|此代码仅在顶部对排序字典中定义的键进行排序，而将所有其他键留在底部。该方法是干净的，但它需要在...之前执行get_app_list。当性能很重要并且app_list很大时，这可能不是一个好主意。|1662879|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Quite a simple and self contained approach can be using the decorator pattern to resort your app and modules in this way:
<pre><code># admin.py

from django.contrib import admin

def app_resort(func): 
 def inner(*args, **kwargs): 
 app_list = func(*args, **kwargs)
 # Useful to discover your app and module list:
 #import pprint 
 #pprint.pprint(app_list)

 app_sort_key = 'name'
 app_ordering = {
 &quot;APP_NAME1&quot;: 1,
 &quot;APP_NAME2&quot;: 2,
 &quot;APP_NAME3&quot;: 3,
 }

 resorted_app_list = sorted(app_list, key=lambda x: app_ordering[x[app_sort_key]] if x[app_sort_key] in app_ordering else 1000)

 model_sort_key = 'object_name'
 model_ordering = {
 &quot;Model1&quot;: 1,
 &quot;Model2&quot;: 2,
 &quot;Model3&quot;: 3,
 &quot;Model14&quot;: 4,
 }
 for app in resorted_app_list:
 app['models'].sort(key=lambda x: model_ordering[x[model_sort_key]] if x[model_sort_key] in model_ordering else 1000)
 return resorted_app_list
 return inner 
 
admin.site.get_app_list = app_resort(admin.site.get_app_list)
</code></pre>
This code sorts at the top only the defined keys in the ordering dict, leaving at the bottom all the other.
The approach is clean but it requires get_app_list to be executed before... which is probably not a good idea when performance is important and you have a large app_list.

Let's say I have my pizza application with Topping and Pizza classes and they show in Django Admin like this:

<pre><code>PizzaApp
-
Toppings &gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt; Add / Change

Pizzas &gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt; Add / Change
</code></pre>

But I want them like this:

<pre><code>PizzaApp
-
Pizzas &gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt; Add / Change

Toppings &gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt; Add / Change
</code></pre>

How do I configure that in my admin.py?

Ordering admin.ModelAdmin objects

Django

Python

假设我有一个带有Topping和pizza类的披萨应用程序，它们在Django Admin中显示如下：PizzaApp-Toppings      >>>>>>>>>>      Add / ChangePizzas        >>>>>>>>>>      Add / Change但我希望他们是这样的：Pizza...

问对admin.ModelAdmin对象进行排序
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