如何在 Django 管理中显示 ManyToMany 关系的 raw_id 值？

Question

我有一个应用程序在ForeignKeyField 和ManyToManyField 上使用raw_id。管理员在编辑框右侧显示外键的值。

不幸的是，它不适用于ManyToMany。我检查了代码，我认为这是正常行为。但是我想知道是否有人有一个简单的技巧来改变这种行为？

提前致谢

更新：我尝试对 ManyToManyRawIdWidget 进行子类化，但我不知道如何说 raw_id_fields 应该使用我的自定义小部件。 formfield_overrides 似乎不适用于 raw_id 字段

Answer 1

最后我成功地让它发挥作用。这是更新的 Django2.0 版本

from django.contrib.admin.widgets import ManyToManyRawIdWidget
from django.utils.encoding import smart_str
from django.urls import reverse
from django.utils.html import escape, mark_safe


class VerboseManyToManyRawIdWidget(ManyToManyRawIdWidget):
    """
    A Widget for displaying ManyToMany ids in the "raw_id" interface rather 
    than in a <select multiple> box. Display user-friendly value like the ForeignKeyRawId widget
    """

    def __init__(self, remote_field, attrs=None, *args, **kwargs):
        super().__init__(remote_field, attrs, *args, **kwargs)

    def label_and_url_for_value(self, value):
        values = value
        str_values = []
        field = self.rel.get_related_field()
        key = field.name
        fk_model = self.rel.model
        app_label = fk_model._meta.app_label
        class_name = fk_model._meta.object_name.lower()
        for the_value in values:
            try:
                obj = fk_model._default_manager.using(self.db).get(**{key: the_value})
                url = reverse('admin:{0}_{1}_change'.format(app_label, class_name), args=[obj.id])
                label = escape(smart_str(obj))
                elt = '<a href="{0}" {1}>{2}</a>'.format(
                    url,
                    'onclick="return showAddAnotherPopup(this);" target="_blank"',
                    label
                )
                str_values += [elt]
            except fk_model.DoesNotExist:
                str_values += [u'???']
        return mark_safe(', '.join(str_values)), ''


class MyAdmin(admin.ModelAdmin):
     ...
     def formfield_for_dbfield(self, db_field, **kwargs):
         if db_field.name in ('groups', ):
             kwargs['widget'] = VerboseManyToManyRawIdWidget(db_field.remote_field, self.admin_site)
         else:
             return super().formfield_for_dbfield(db_field, **kwargs)
         kwargs.pop('request')
         return db_field.formfield(**kwargs)

不幸的是，我白白花了一笔赏金 ;-)

更新：这个片段现在与 Django 2.0 兼容。另请参阅 http://djangosnippets.org/snippets/2108/

Answer 2

这适用于 Django 1.11 及更高版本

from django.contrib.admin.sites import site
from django.contrib.admin.widgets import ManyToManyRawIdWidget
from django.core.urlresolvers import reverse, NoReverseMatch
from django.utils.safestring import mark_safe


class VerboseManyToManyRawIdWidget(ManyToManyRawIdWidget):

    def label_and_url_for_value(self, value):
        result = []
        for v in value:
            key = self.rel.get_related_field().name
            try:
                obj = self.rel.model._default_manager.using(self.db).get(**{key: v})
            except (ValueError, self.rel.model.DoesNotExist):
                return '', ''

            try:
                url = reverse(
                    '{}:{}_{}_change'.format(self.admin_site.name, obj._meta.app_label,
                                             obj._meta.object_name.lower()),
                    args=(obj.pk,))
            except NoReverseMatch:
                url = ''  # Admin not registered for target model.

            result.append('<strong><a href="{}">{}</a></strong>'.format(url,  str(obj)))

        return mark_safe('; '.join(result)), ''


class VerboseRawIdManyToManyAdminMixin:
    def formfield_for_dbfield(self, db_field, **kwargs):
        if db_field.name in self.raw_id_fields:
            kwargs.pop('request', None)
            if db_field.rel.__class__.__name__ == 'ManyToManyRel':
                kwargs['widget'] = VerboseManyToManyRawIdWidget(db_field.rel, site)
            return db_field.formfield(**kwargs)
        return super().formfield_for_dbfield(db_field, **kwargs)

还有用于此目的的 Django 应用 django-沙门氏菌