如何在Django Admin中将模型重新分配给其他应用程序以仅用于显示 [英] How to reassign a model to a different app for display only purposes in Django Admin

问题描述

我知道我可以使用

  class Meta：
 verbose_name更改Django Admin中模型的显示标题。 ='此处为自定义模型名称'
  

但是，有没有一种方法可以显示哪个应用程序是模型的标题显示在下面？

例如，如果我在也称为 Users >用户，则默认用户模型来自身份验证和授权>用户到用户>用户。

我想将其保留在原始标题身份验证和授权>用户。

我已阅读

解决方案

没有简单的方法可以完成您想要的工作，请查看生成视图的Django代码/模板：

  @never_cache 
 def index（self，request，extra_context = None）：
 
显示主管理员索引页面，其中列出了所有
 
 app_list = self.get_app_list（request）
 
 context = {
 ** self.each_context（request），
'title'：self.index_title，
'app_list'：app_list，
 **（extra_context或{}），
} 
 
 request.current_app = self.name 
 
 return TemplateResponse（request，self.index_template或'admin / in dex.html'，上下文）
  

templates / admin / index.html ：

  {％for app_list％}中的应用程序
< div class = app -{{app.app_label}}模块> 
< table> 
< caption> 
< a href = {{app.app_url}} class = section title = {％blocktrans with name = app.name％} {{name}}应用程序中的模型{％endblocktrans％ }> {{app.name}}< / a> 
< / caption> 
 {app.models中模型的%%} 
< tr class = model-{{model.object_name | lower}}> 
 {％if model.admin_url％} 
< th scope = row>< a href = {{model.admin_url}}> {{model.name}}< ; / a< / th> 
 {％else％} 
< th scope = row> {{model.name}}< / th> 
 {％endif％} 
 
 {％if model.add_url％} 
< td>< a href = {{model.add_url}} class = addlink> {％trans'Add'％}< / a< / td> 
 {％else％} 
< / td> 
 {％endif％} 
 
 {％如果model.admin_url％} 
 {％如果model.view_only％} 
< td>< a href =  {{model.admin_url}} class = viewlink> {％trans'View'％}< / a>< / td> 
 {％else％} 
< td>< a href = {{model.admin_url}} class = changelink> {％trans'Change'％}< / a> ;< / td> 
 {％endif％} 
 {％else％} 
< td>& nbsp;< / td> 
 {％endif％} 
< / tr> 
 {％endfor％} 
< / table> 
< / div> 
 {％endfor％} 
  

如您所见，Django admin模板正在循环您的app_list直接，因此唯一的方法就是覆盖 admin / index.html 模板，以将模型放置在所需的顺序中。

从Django文档中：

对于无法以这种方式覆盖的模板，您仍然可以覆盖它们您的整个项目。只需将新版本放在您的template / admin目录中即可。这对于创建自定义404页和500页特别有用。

https://docs.djangoproject.com/en/2.2/ref/contrib/admin/ ＃overriding-vs-replacing-an-admin-template

I know I can change the display title for a model in Django Admin using

    class Meta:
        verbose_name='Custom Model Name Here'

However, is there a way to display which app heading a model is displayed under?

For example, if I create a custom user model Users in a new app also called users then the default user model goes from Authentication and Authorization > Users to Users > Users.

I would like to retain it under the original heading Authentication and Authorization > Users.

I have read this answer which suggests changes the app verbose_name, however it only changes the verbose name of the app associated with the model. I want to show the model in a different group on the admin panel. You can see the issue that approach takes here:

解决方案

There is no simple way of doing what you're intending, check out the Django code/template that generates the view:

@never_cache
def index(self, request, extra_context=None):
    """
    Display the main admin index page, which lists all of the installed
    apps that have been registered in this site.
    """
    app_list = self.get_app_list(request)

    context = {
        **self.each_context(request),
        'title': self.index_title,
        'app_list': app_list,
        **(extra_context or {}),
    }

    request.current_app = self.name

    return TemplateResponse(request, self.index_template or 'admin/index.html', context)

templates/admin/index.html:

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

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

        {% if model.admin_url %}
            {% if model.view_only %}
            <td><a href="{{ model.admin_url }}" class="viewlink">{% trans 'View' %}</a></td>
            {% else %}
            <td><a href="{{ model.admin_url }}" class="changelink">{% trans 'Change' %}</a></td>
            {% endif %}
        {% else %}
            <td>&nbsp;</td>
        {% endif %}
        </tr>
    {% endfor %}
    </table>
    </div>
{% endfor %}

As you can see, Django admin template is looping your app_list directy, so the only way of doing it would be to override the admin/index.html template to place your models in your desired order.

From Django docs:

For those templates that cannot be overridden in this way, you may still override them for your entire project. Just place the new version in your templates/admin directory. This is particularly useful to create custom 404 and 500 pages.

https://docs.djangoproject.com/en/2.2/ref/contrib/admin/#overriding-vs-replacing-an-admin-template

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