Model name of objects in django templates

Is there any way to get the model name of any objects in django templates. Manually, we can try it by defining methods in models or using template tags... But 开发者_StackOverflow中文版is there any built-in way?

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object.__class__.__name__ or object._meta.object_name should give you the name of the model class. However, this cannot be used in templates because the attribute names start with an underscore.

There isn't a built in way to get at that value from the templates, so you'll have to define a model method that returns that attribute, or for a more generic/reusable solution, use a template filter:

@register.filter
def to_class_name(value):
    return value.__class__.__name__

which you can use in your template as:

{{ obj | to_class_name }}

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You cannot access the class name directly. Doing something like this:

{{ object.__class__ }}

will cause a TemplateSyntaxError: Variables and attributes may not begin with underscores. Django doesn't let you access those sorts of attributes - Python conventions means they are hidden implementation details, not part of the object's API.

Create a template filter instead, and then you can use it as follows:

{{ object|model_name_filter }}

Creating filters is very easy: https://docs.djangoproject.com/en/dev/howto/custom-template-tags/

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Django added a publicly accessible API to model options called _meta, but you still can't access variables with an underscore in the template. Use a template filter:

@register.filter
def verbose_name(instance):
    return instance._meta.verbose_name

In the template:

{{ instance|verbose_name }}

I even like to chain the title filter to capitalize the words in the my template:

{{ instance|verbose_name|title }}

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You can very easily access the content type of the page, this way you don't even need to poke into the specific:

{% if pub.content_type.app_labeled_name == 'home | publication' %}
    {% include "home/publication.html" with pub=pub.specific %}
{% endif %}

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Since 1.2 version and may be early Django has an attribute opts into templates. The atribute is link to model._meta For evidence you should look at source code into Github

It used into template very simple: {{opts}} or {{opts.db_table}}

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This answer demonstrates a practical use case. In my app each model has a generic.CreateView named name=”<model>-create” in urls.py; namespaced/referenced as {% url <app>:<model>-create %} in those templates which need to link to it. Each model has a generic.ListView rendered by <app>/templates/<app>/<model>_list.html. I wanted to include a button at the bottom of each rendered list view to link to the corresponding create view. To ensure consistency across multiple list views I wanted to do this in a base template which each list view template could extend upon. To do this I needed to be able to reference the model name in a template. Here’s how I did it.

$ mkdir <app>/templatetags
$ touch <app>/templatetags/__init__.py
$ vi <app>/templatetags/tags.py

from django import template
register = template.Library()
# this filter allows any template to reference an instance’s model name as <instance>|class_name
@register.filter(name="class_name")
def class_name(instance):
    return instance._meta.model.__name__

$ vi <app>/templates/base_list.html

~
{% with name=object_list|first|class_name|lower %}
        <button onclick="location.href='{% url request.resolver_match.app_name|add:':'|add:name|add:'-create' %}';" title="create"><i class="fa fa-plus"></i></button>
{% endwith  %}

* substitute <app>, <model>, and <instance> for your own names.