在 Django admin 中重新排序应用程序和模型

问题描述 投票：0回答：6

我想在 Django 管理面板中重新排序我的应用程序，我在 SO 中看到了另一个类似问题的一些回复，所以我选择安装此方法：django-modeladmin-reorder

我按照所有步骤操作，但不起作用。这是我实际的 Django 面板

#settings.py

INSTALLED_APPS = [
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.messages',
    # Disable Django's own staticfiles handling in favour of WhiteNoise, for
    # greater consistency between gunicorn and `./manage.py runserver`. See:
    # http://whitenoise.evans.io/en/stable/django.html#using-whitenoise-in-development
    'whitenoise.runserver_nostatic',
    'django.contrib.staticfiles',
    'admin_reorder',
    'organization_owners',
    'panel',
    'clients',
]

MIDDLEWARE_CLASSES = (
    'admin_reorder.middleware.ModelAdminReorder',
)


ADMIN_REORDER = (
    # Keep original label and models
    'organization_owners',
    'panel',
    'clients',
)

并且也在我的requirements.txt中

Django==2.0.1
django-extensions==1.9.8
django-modeladmin-reorder==0.2
djangorestframework==3.7.7
flake8==3.5.0
python django django-models django-admin
6个回答
2
投票

我检查了他们的 github 存储库，它在两个月前更新以支持 Django 2.0，但它的 Python Package Index 版本 0.2 上次上传于 2016-09-08。

所以有可能pip安装仍然会安装不支持Django 2.0的版本

请注意，urlresolvers 模块在 Django 1.10 中已被弃用，并在 2.0 中被删除（django-modeladmin-reorder 仍然依赖于 urlresolvers）

你可以做什么：

  1. 如果您已经了解 Django==1.8，请先尝试 django-modeladmin-reorder。

  2. 在您的项目中使用 GitHub 上的代码。这是最新提交的链接。 https://github.com/mishbahr/django-modeladmin-reorder/commit/f21929480c398c2628291d74af2f319421f651f3

2
投票

如果您使用的是 Django 2.0，则必须使用 this commit 编辑 middleware.py。即使存储库也支持 Django 2.0，pip 正在停滞且版本较旧（正如 @Dhaval Savalia 所说）。

之后，这一步：

将 ModelAdminReorder 添加到 MIDDLEWARE_CLASSES：

MIDDLEWARE_CLASSES = (
    ...
    'admin_reorder.middleware.ModelAdminReorder',
    ...
)

应替换为： 

MIDDLEWARE = [ 
        ...
        'admin_reorder.middleware.ModelAdminReorder',
        ...
]

仅此而已。

2
投票

我遵循了anjaneyulubatta505答案，但它只更改了索引页面中的顺序来更改所有管理页面中的顺序 覆盖

app_list.html
而不是
index.html

在任何应用程序中应用anjaneyulubatta505答案中的模板标签

from django import template
from django.conf import settings


register = template.Library()


def pop_and_get_app(apps, key, app_label):
    for index, app in enumerate(apps):
        if app[key] == app_label:
            obj = apps.pop(index)
            return obj
    return None

@register.filter
def sort_apps(apps):
    new_apps = []
    order = settings.APP_ORDER
    for app_label in order.keys():
        obj = pop_and_get_app(apps, "app_label", app_label)
        new_apps.append(obj) if obj else None
    apps = new_apps + apps
    for app in apps:
        models = app.get("models")
        app_label = app.get("app_label")
        new_models = []
        order_models = settings.APP_ORDER.get(app_label, [])
        for model in order_models:
            obj = pop_and_get_app(models, "object_name", model)
            new_models.append(obj) if obj else None
        models = new_models + models
        app["models"] = models
    return apps

并将订单添加到您的 

settings.py

from collections import OrderedDict

APP_ORDER = OrderedDict([
  ("app1", ["Model2", "Model1", "Model3"]),
  ("app2", ["Model2", "Model5", "Model3"]),
  ("app3", ["Model1", "Model6", "Model3"]),
])

这是我的

app_list.html

{% load i18n admin_tags %}

{% if app_list %}
  {% for app in app_list|sort_apps %}
    <div class="app-{{ app.app_label }} module{% if app.app_url in request.path|urlencode %} current-app{% endif %}">
      <table>
        <caption>
          <a href="{{ app.app_url }}" class="section" title="{% blocktranslate with name=app.name %}Models in the {{ name }} application{% endblocktranslate %}">{{ app.name }}</a>
        </caption>
        {% for model in app.models %}
          <tr class="model-{{ model.object_name|lower }}{% if model.admin_url in request.path|urlencode %} current-model{% endif %}">
            {% if model.admin_url %}
              <th scope="row"><a href="{{ model.admin_url }}"{% if model.admin_url in request.path|urlencode %} aria-current="page"{% endif %}>{{ model.name }}</a></th>
            {% else %}
              <th scope="row">{{ model.name }}</th>
            {% endif %}

            {% if model.add_url %}
              <td><a href="{{ model.add_url }}" class="addlink">{% translate 'Add' %}</a></td>
            {% else %}
              <td></td>
            {% endif %}

            {% if model.admin_url and show_changelinks %}
              {% if model.view_only %}
                <td><a href="{{ model.admin_url }}" class="viewlink">{% translate 'View' %}</a></td>
              {% else %}
                <td><a href="{{ model.admin_url }}" class="changelink">{% translate 'Change' %}</a></td>
              {% endif %}
            {% elif show_changelinks %}
              <td></td>
            {% endif %}
          </tr>
        {% endfor %}
      </table>
    </div>
  {% endfor %}
{% else %}
  <p>{% translate 'You don’t have permission to view or edit anything.' %}</p>
{% endif %}
1
投票

我最近写了一篇关于此的文章。这可能有帮助。 我们需要使用模板标签来重新排序应用程序和模型，如下所示。

在设置中添加以下代码

from collections import OrderedDict

APP_ORDER = OrderedDict([
  ("app1", ["Model2", "Model1", "Model3"]),
  ("app2", ["Model2", "Model5", "Model3"]),
  ("app3", ["Model1", "Model6", "Model3"]),
])

添加以下模板标签

from django import template
from django.conf import settings


register = template.Library()


def pop_and_get_app(apps, key, app_label):
    for index, app in enumerate(apps):
        if app[key] == app_label:
            obj = apps.pop(index)
            return obj
    return None

@register.filter
def sort_apps(apps):
    new_apps = []
    order = settings.APP_ORDER
    for app_label in order.keys():
        obj = pop_and_get_app(apps, "app_label", app_label)
        new_apps.append(obj) if obj else None
    apps = new_apps + apps
    for app in apps:
        models = app.get("models")
        app_label = app.get("app_label")
        new_models = []
        order_models = settings.APP_ORDER.get(app_label, [])
        for model in order_models:
            obj = pop_and_get_app(models, "object_name", model)
            new_models.append(obj) if obj else None
        models = new_models + models
        app["models"] = models
    return apps

像下面一样覆盖 admin/index.html 

{% for app in app_list|sort_apps %}

参考：https://learnbatta.com/blog/how-to-re-order-apps-models-django/

1
投票

项目/admin.py

from django.contrib.admin import AdminSite

class MyAdminSite(AdminSite):
    site_header = 'My Site'
    index_title = ''

    def get_app_list(self, request):
        app_order = [
            'app_1',
            'app_2',
            'auth',
        ]
        app_order_dict = dict(zip(app_order, range(len(app_order))))
        app_list = list(self._build_app_dict(request).values())
        app_list.sort(key=lambda x: app_order_dict.get(x['app_label'], 0))
        for app in app_list:
            if app['app_label'] == 'app_1':
                model_order = [
                    'Model1 verbose name',
                    'Model2 verbose name',
                    'Model3 verbose name',
                ]
                model_order_dict = dict(zip(model_order, range(len(model_order))))
                app['models'].sort(key=lambda x: model_order_dict.get(x['name'], 0))
        return app_list

项目/apps.py

from django.contrib.admin.apps import AdminConfig

class MyAdminConfig(AdminConfig):
    default_site = 'project.admin.MyAdminSite'

项目/设置.py

INSTALLED_APPS = [
    'project.apps.MyAdminConfig',
    <instead of django.contrib.admin>
    ...
0
投票

@tulsluper 给出的回复对我来说几乎是完美的。我发现当我点击管理面板中的类别（应用程序名称）而不是模型时，它死了，因为它正在寻找带有 3 个参数而不是两个参数的 get_app_list 。通过将该重载（从 contrib.admin.sites.py 复制）添加到 project/admin.py 文件中，它能够找到重载，并且一切正常。就我而言，这是我添加到 project/admin.py 程序中的内容：

def get_app_list(self, request, app_label=None):
    """
    Return a sorted list of all the installed apps that have been
    registered in this site.
    """
    app_dict = self._build_app_dict(request, app_label)

    # Sort the apps alphabetically.
    app_list = sorted(app_dict.values(), key=lambda x: x["name"].lower())

    # Sort the models alphabetically within each app.
    for app in app_list:
        app["models"].sort(key=lambda x: x["name"])

    return app_list
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