blocks|key|1611715|text|不能，但您可以复制管理模板并在那里定义应用程序名称。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1611716|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|$]]

No, but you can copy admin template and define app name there.

blocks|key|1396935|text|给它们一个verbose_name属性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1396936|别抱太大希望。您还需要将索引视图从django.contrib.admin.sites复制到您自己的ProjectAdminSite视图中，并将其包含在您自己的自定义管理实例中：|1396937|class+ProjectAdminSite(AdminSite):
++++def+index(self,+request,+extra_context=None):
++++++++copied+stuff+here...

admin.site+=+ProjectAdminSite()|code-block|syntax|javascript|1396938|然后调整复制的视图，使其使用verbose_name属性作为应用程序的标签。|1396939|我是通过在复制的视图中添加一些类似如下的内容来实现的：|1396940|++++++++try:
++++++++++++app_name+=+model_admin.verbose_name
++++++++except+AttributeError:
++++++++++++app_name+=+app_label|1396941|当你调整索引视图时，为什么不添加一个'order‘属性呢？|1396942|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Give them a verbose_name property.

Don't get your hopes up. You will also need to copy the index view from django.contrib.admin.sites into your own ProjectAdminSite view and include it in your own custom admin instance:

<pre><code>class ProjectAdminSite(AdminSite):
 def index(self, request, extra_context=None):
 copied stuff here...

admin.site = ProjectAdminSite()
</code></pre>

then tweak the copied view so that it uses your verbose_name property as the label for the app.

I did it by adding something a bit like this to the copied view:

<pre><code> try:
 app_name = model_admin.verbose_name
 except AttributeError:
 app_name = app_label
</code></pre>

While you are tweaking the index view why not add an 'order' property too.

blocks|key|1396969|text|如果应用程序中有多个模型，只需使用元信息创建一个模型，并为所有模型创建该类的子类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1396970|class+MyAppModel(models.Model):
++++class+Meta:
++++++++app_label+=+'My+App+Label'
++++++++abstract+=+True

class+Category(MyAppModel):
+++++name+=+models.CharField(max_length=50)|code-block|syntax|javascript|1396971|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you have more than one model in the app just create a model with the Meta information and create subclasses of that class for all your models.

<pre><code>class MyAppModel(models.Model):
 class Meta:
 app_label = 'My App Label'
 abstract = True

class Category(MyAppModel):
 name = models.CharField(max_length=50)
</code></pre>

blocks|key|1397004|text|对于Django+1.4+(尚未发布，但trunk相当稳定)，您可以使用以下方法。它依赖于这样一个事实:+AdminSite现在返回一个TemplateResponse，您可以在呈现它之前对其进行更改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1397005|在这里，我们做了一点小补丁来插入我们的行为，如果你使用一个自定义的AdminSite子类，这是可以避免的。|1397006|from+functools+import+wraps
def+rename_app_list(func):
++++m+=+{'Sites':+'Web+sites',
+++++++++'Your_app_label':+'Nicer+app+label',
++++}

++++@wraps(func)
++++def+_wrapper(*args,+**kwargs):
++++++++response+=+func(*args,+**kwargs)
++++++++app_list+=+response.context_data.get('app_list')

++++++++if+app_list+is+not+None:
++++++++++++for+a+in+app_list:
++++++++++++++++name+=+a['name']
++++++++++++++++a['name']+=+m.get(name,+name)
++++++++title+=+response.context_data.get('title')
++++++++if+title+is+not+None:
++++++++++++app_label+=+title.split('+')[0]
++++++++++++if+app_label+in+m:
++++++++++++++++response.context_data['title']+=+"%25s+administration"+%25+m[app_label]
++++++++return+response
++++return+_wrapper

admin.site.__class__.index+=+rename_app_list(admin.site.__class__.index)
admin.site.__class__.app_index+=+rename_app_list(admin.site.__class__.app_index)|code-block|syntax|javascript|1397007|这将修复索引和app_index视图。它不会修复所有其他管理视图中的面包屑。|1397008|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

For Django 1.4 (not yet released, but trunk is pretty stable), you can use the following method. It relies on the fact that AdminSite now returns a TemplateResponse, which you can alter before it is rendered.

Here, we do a small bit of monkey patching to insert our behaviour, which can be avoided if you use a custom AdminSite subclass.

<pre><code>from functools import wraps
def rename_app_list(func):
 m = {'Sites': 'Web sites',
 'Your_app_label': 'Nicer app label',
 }

 @wraps(func)
 def _wrapper(*args, **kwargs):
 response = func(*args, **kwargs)
 app_list = response.context_data.get('app_list')

 if app_list is not None:
 for a in app_list:
 name = a['name']
 a['name'] = m.get(name, name)
 title = response.context_data.get('title')
 if title is not None:
 app_label = title.split(' ')[0]
 if app_label in m:
 response.context_data['title'] = "%s administration" % m[app_label]
 return response
 return _wrapper

admin.site.__class__.index = rename_app_list(admin.site.__class__.index)
admin.site.__class__.app_index = rename_app_list(admin.site.__class__.app_index)
</code></pre>

This fixes the index and the app_index views. It doesn't fix the bread crumbs in all other admin views.

blocks|key|1397053|text|有一种不需要任何迁移就可以完成的黑客攻击。摘自Ionel的博客，功劳归于他：http://blog.ionelmc.ro/2011/06/24/custom-app-names-in-the-django-admin/|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1397054|还有一个应该在Django1.7+https://code.djangoproject.com/ticket/3591中解决的问题|1397055|""“|1397056|假设你有一个这样的模型：|1397057|class+Stuff(models.Model):
++++class+Meta:
++++++++verbose_name+=+u'The+stuff'
++++++++verbose_name_plural+=+u'The+bunch+of+stuff'|code-block|syntax|javascript|1397058|你有verbose_name，但你想要定制的管理中的不同显示app_label也。不幸的是，有一些任意的字符串(带空格)不起作用，而且也不是用来显示的。|1397059|原来管理员使用的是app_label。title+()用于显示，所以我们可以做一些修改:使用重写的title方法的Str子类：|1397060|class+string_with_title(str):
++++def+__new__(cls,+value,+title):
++++++++instance+=+str.__new__(cls,+value)
++++++++instance._title+=+title
++++++++return+instance

++++def+title(self):
++++++++return+self._title

++++__copy__+=+lambda+self:+self
++++__deepcopy__+=+lambda+self,+memodict:+self|1397061|现在我们可以有这样的模型：|1397062|class+Stuff(models.Model):
++++class+Meta:
++++++++app_label+=+string_with_title("stuffapp",+"The+stuff+box")
++++++++#+'stuffapp'+is+the+name+of+the+django+app
++++++++verbose_name+=+'The+stuff'
++++++++verbose_name_plural+=+'The+bunch+of+stuff'|1397063|管理员会在应用程序名中显示"The+stuff+box“。|1397064|1397065|entityMap|0|LINK|mutability|MUTABLE|url|http://blog.ionelmc.ro/2011/06/24/custom-app-names-in-the-django-admin/|1|https://code.djangoproject.com/ticket/3591^0|12|1Z|0|0|H|16|1|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]]|C|$]]|$1|D|3|E|5|6|7|1F|8|@]|9|@$A|1G|B|1H|1|1I]]|C|$]]|$1|F|3|G|5|6|7|1J|8|@]|9|@]|C|$]]|$1|H|3|I|5|6|7|1K|8|@]|9|@]|C|$]]|$1|J|3|K|5|L|7|1L|8|@]|9|@]|C|$M|N]]|$1|O|3|P|5|6|7|1M|8|@]|9|@]|C|$]]|$1|Q|3|R|5|6|7|1N|8|@]|9|@]|C|$]]|$1|S|3|T|5|L|7|1O|8|@]|9|@]|C|$M|N]]|$1|U|3|V|5|6|7|1P|8|@]|9|@]|C|$]]|$1|W|3|X|5|L|7|1Q|8|@]|9|@]|C|$M|N]]|$1|Y|3|Z|5|6|7|1R|8|@]|9|@]|C|$]]|$1|10|3|G|5|6|7|1S|8|@]|9|@]|C|$]]|$1|11|3|-4|5|6|7|1T|8|@]|9|@]|C|$]]]|12|$13|$5|14|15|16|C|$17|18]]|19|$5|14|15|16|C|$17|1A]]]]

There is a hack that can be done that does not require any migrations.
Taken from Ionel's blog and credit goes to him: <a href="http://blog.ionelmc.ro/2011/06/24/custom-app-names-in-the-django-admin/" rel="nofollow">http://blog.ionelmc.ro/2011/06/24/custom-app-names-in-the-django-admin/</a>

There is also a ticket for this that should be fixed in Django 1.7 <a href="https://code.djangoproject.com/ticket/3591" rel="nofollow">https://code.djangoproject.com/ticket/3591</a>

"""

Suppose you have a model like this:

<pre><code>class Stuff(models.Model):
 class Meta:
 verbose_name = u'The stuff'
 verbose_name_plural = u'The bunch of stuff'
</code></pre>

You have verbose_name, however you want to customise app_label too for different display in admin. Unfortunatelly having some arbitrary string (with spaces) doesn't work and it's not for display anyway.

Turns out that the admin uses app_label. title () for display so we can make a little hack: str subclass with overriden title method:

<pre><code>class string_with_title(str):
 def __new__(cls, value, title):
 instance = str.__new__(cls, value)
 instance._title = title
 return instance

 def title(self):
 return self._title

 __copy__ = lambda self: self
 __deepcopy__ = lambda self, memodict: self
</code></pre>

Now we can have the model like this:

<pre><code>class Stuff(models.Model):
 class Meta:
 app_label = string_with_title("stuffapp", "The stuff box")
 # 'stuffapp' is the name of the django app
 verbose_name = 'The stuff'
 verbose_name_plural = 'The bunch of stuff'
</code></pre>

and the admin will show "The stuff box" as the app name.

"""

blocks|key|1611883|text|正如rhunwicks对OP的评论所述，从Django+1.7开始，这现在是可能的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1611884|摘自docs|offset|length|1611885|#+in+yourapp/apps.py
from+django.apps+import+AppConfig

class+YourAppConfig(AppConfig):
++++name+=+'yourapp'
++++verbose_name+=+'Fancy+Title'|code-block|syntax|javascript|1611886|然后将default_app_config变量设置为YourAppConfig|style|CODE|1611887|#+in+yourapp/__init__.py
default_app_config+=+'yourapp.apps.YourAppConfig'|1611888|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/1.7/ref/applications/#for-application-authors^0|0|2|4|0|0|0|3|I|Q|D|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@$D|15|E|16|M|N]|$D|17|E|18|M|N]]|9|@]|A|$]]|$1|O|3|P|5|H|7|19|8|@]|9|@]|A|$I|J]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

As stated by rhunwicks' comment to OP, this is now possible out of the box since Django 1.7

Taken from the <a href="https://docs.djangoproject.com/en/1.7/ref/applications/#for-application-authors" rel="noreferrer">docs</a>:



<pre class="lang-py prettyprint-override"><code># in yourapp/apps.py
from django.apps import AppConfig

class YourAppConfig(AppConfig):
 name = 'yourapp'
 verbose_name = 'Fancy Title'
</code></pre>

then set the <code>default_app_config</code> variable to <code>YourAppConfig</code>

<pre class="lang-py prettyprint-override"><code># in yourapp/__init__.py
default_app_config = 'yourapp.apps.YourAppConfig'
</code></pre>

blocks|key|1397151|text|下面这段即插即用的代码在Django+1.7之后工作得很好。您所要做的就是复制特定应用程序的__init__.py文件中的以下代码，并更改VERBOSE_APP_NAME参数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1397152|from+os+import+path
from+django.apps+import+AppConfig

VERBOSE_APP_NAME+=+"YOUR+VERBOSE+APP+NAME+HERE"


def+get_current_app_name(file):
++++return+path.dirname(file).replace('\\',+'/').split('/')[-1]


class+AppVerboseNameConfig(AppConfig):
++++name+=+get_current_app_name(__file__)
++++verbose_name+=+VERBOSE_APP_NAME


default_app_config+=+get_current_app_name(__file__)+%2B+'.__init__.AppVerboseNameConfig'|code-block|syntax|javascript|1397153|如果您将此函数用于多个应用程序，则应将get_current_app_name函数分解为一个帮助文件。|1397154|entityMap^0|C|A|1A|B|1X|G|0|0|J|K|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

The following plug-and-play piece of code works perfectly since <code>Django 1.7</code>. All you have to do is copy the below code in the <code>__init__.py</code> file of the specific app and change the <code>VERBOSE_APP_NAME</code> parameter.

<pre><code>from os import path
from django.apps import AppConfig

VERBOSE_APP_NAME = "YOUR VERBOSE APP NAME HERE"


def get_current_app_name(file):
 return path.dirname(file).replace('\\', '/').split('/')[-1]


class AppVerboseNameConfig(AppConfig):
 name = get_current_app_name(__file__)
 verbose_name = VERBOSE_APP_NAME


default_app_config = get_current_app_name(__file__) + '.__init__.AppVerboseNameConfig'
</code></pre>

If you use this for multiple apps, you should factor out the <code>get_current_app_name</code> function to a helper file.

blocks|key|1397229|text|好吧，这对我很有效。在app.py中使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1397230|class+MainConfig(AppConfig):
++++name+=+'main'
++++verbose_name="Fancy+Title"|code-block|syntax|javascript|1397231|在setting.py中，添加应用程序文件夹中的app.py文件中存在的应用程序名称和类名称|1397232|INSTALLED_APPS+=+[
++++'main.apps.MainConfig',
++++'django.contrib.admin',
++++'django.contrib.auth',
++++'django.contrib.contenttypes',
++++'django.contrib.sessions',
++++'django.contrib.messages',
++++'django.contrib.staticfiles',|1397233|]|1397234|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Well, this works for me. In the app.py use this:

<pre><code>class MainConfig(AppConfig):
 name = 'main'
 verbose_name="Fancy Title"
</code></pre>

In setting.py add the name of App and the class name present in app.py file in App folder

<pre><code>INSTALLED_APPS = [
 'main.apps.MainConfig',
 'django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
</code></pre>

]

blocks|key|1612010|text|这很简单，只需在appName/apps上添加如下内容即可|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1612011|class+AppNameConfig(AppConfig):
++++default_auto_field+=+'default+Django'
++++name+=+'AppName'
++++verbose_name+=+'Name+you+Want'|code-block|syntax|javascript|1612012|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

That's simple just add as follow on your appName/apps
<pre><code>class AppNameConfig(AppConfig):
 default_auto_field = 'default Django'
 name = 'AppName'
 verbose_name = 'Name you Want'
</code></pre>

In the same way that you can give fields and models verbose names that appear in the Django admin, can you give an app a custom name?

Can you give a Django app a verbose name for use throughout the admin?

Django

以同样的方式，您可以为字段和模型提供显示在Django管理中的详细名称，您可以为应用程序提供自定义名称吗？

问你能给一个Django应用一个详细的名字，以便在整个管理过程中使用吗？
EN

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问你能给一个Django应用一个详细的名字，以便在整个管理过程中使用吗？
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