在 Django admin 中将外键列显示为详细对象的链接

问题描述

如 link-in-django-admin-to- 中所述foreign-key-object，可以将 ForeignKey 字段显示为管理详细信息页面的链接.

As explained in link-in-django-admin-to-foreign-key-object, one can display a ForeignKey field as a link to the admin detail page.

总结一下，

class Foo(Model):
    bar = models.ForeignKey(Bar)

class FooAdmin(ModelAdmin):
    list_display = ('link_to_bar',)
    def link_to_bar(self, obj):
        link = urlresolvers.reverse('admin:app_bar_change', args=[obj.bar_id])
        return u'<a href="%s">%s</a>' % (link, obj.bar) if obj.bar else None
    link_to_bar.allow_tags = True

问题是:我们可以更自动地做到这一点吗?例如，向 FooAdmin 定义提供一个外键列表，以显示为详细页面的链接:

The question is: can we do it more automatically? For instance, provide to the FooAdmin definition a list of foreign key to display as links to detail page:

class FooAdmin(ModelAdmin):
    ...
    list_foreign_key_links = ('bar',)
    ...

我知道这些 ModelAdmin 类是通过元类编程生成的.那么，应该是可以的.这样做是一个好的开始?

I know that these ModelAdmin classes are generated with metaclass programming. Then, it should be possible. What would be a good start to do so?

推荐答案

下面的解决方案使用这个答案但使其可被所有模型重用，避免了向每个管理类添加方法的需要.

The solution below uses this answer but makes it reusable by all models, avoiding the need to add methods to each admin class.

# models.py
from django.db import models

class Country(models.Model):
    name = models.CharField(max_length=200)
    population = models.IntegerField()

class Career(models.Model):
    name = models.CharField(max_length=200)
    average_salary = models.IntegerField()

class Person(models.Model):
    name = models.CharField(max_length=200)
    age = models.IntegerField()
    country = models.ForeignKey(Country, on_delete=models.CASCADE)
    career = models.ForeignKey(Career, on_delete=models.CASCADE)

示例管理员

# admin.py
from django.utils.html import format_html
from django.urls import reverse

from .models import Person


def linkify(field_name):
    """
    Converts a foreign key value into clickable links.
    
    If field_name is 'parent', link text will be str(obj.parent)
    Link will be admin url for the admin url for obj.parent.id:change
    """
    def _linkify(obj):
        linked_obj = getattr(obj, field_name)
        if linked_obj is None:
            return '-'
        app_label = linked_obj._meta.app_label
        model_name = linked_obj._meta.model_name
        view_name = f'admin:{app_label}_{model_name}_change'
        link_url = reverse(view_name, args=[linked_obj.pk])
        return format_html('<a href="{}">{}</a>', link_url, linked_obj)

    _linkify.short_description = field_name  # Sets column name
    return _linkify



@admin.register(Person)
class PersonAdmin(admin.ModelAdmin):
    list_display = [
        "name",
        "age",
        linkify(field_name="country"),
        linkify(field_name="career"),
    ]

结果

给定一个名为 app 的 App，以及一个 Person 实例 Person(name='Adam' age=20)，其国家和职业外键值以及 ids 123 和 456,列表结果将是:

Results

Given an App named app, and a Person instance Person(name='Adam' age=20) with country and carreer foreign key values with ids 123 and 456, the list result will be:

| Name | Age |                          Country                          |...|
|------|-----|-----------------------------------------------------------|...|
| Adam |  20 | <a href="/admin/app/country/123">Country object(123)</a>  |...|

(继续)

|...|                          Career                         |
|---|---------------------------------------------------------|
|...| <a href="/admin/app/career/456">Career object(456)</a>  |

这篇关于在 Django admin 中将外键列显示为详细对象的链接的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！