如何在 Django 管理页面（在 formfield_for_foreignkey 内）中获取实际的对象 ID？

Question

我已经解决了使用此代码获取正在编辑的对象 id 的问题：

class CompanyUserInline(admin.StackedInline):
    """
    Defines tabular rules for editing company users direct in company admin
    """
    model = CompanyUser

    def formfield_for_foreignkey(self, db_field, request, **kwargs):

        if db_field.name == "user":
            users = User.objects.filter( Q(is_superuser=False) )
            query = Q()
            for u in users:
                aux = CompanyUser.objects.filter(user=u)
                if aux.count() == 0:
                    query |= Q(pk=u.id)

            try:
                cpu = CompanyUser.objects.filter(company__id=int(request.path.split('/')[4]))
                for p in cpu:
                    query |= Q(pk=p.user.id)
            except:
                pass

            kwargs["queryset"] = User.objects.filter(query).order_by('username')

        return super(CompanyUserInline, self).formfield_for_foreignkey(db_field, request, **kwargs)

但是，int(request.path.split('/')[4]) 真的很丑。我想知道如何从 Django AdminModel 获取 id。我确定它在里面的某个地方，有人知道吗？

提前感谢您！ ;D

Answer 1

据我所知，无法通过formfield_for_...-方法访问当前实例，因为它们只会被单个字段实例调用！

更好的方法是使用get_form 来访问可以访问整个实例/表单的逻辑。您还可以在那里覆盖表单字段的查询集！

Answer 2

我正在处理类似的情况，并意识到我需要从请求中获得的 id，因为它是该模型的外键，所以我可以来自模型本身。所以它会是这样的：

cpu = CompanyUser.objects.filter(company__id=self.company_id)

或者你的模型结构所要求的。

Answer 3

经过一番挖掘，我们能够获取传递给管理视图的参数（在被 django 管理员的 urls.py 解析之后）并使用该 (self_pub_id) 来获取对象：

class PublicationAdmin(admin.ModelAdmin):

    def formfield_for_manytomany(self, db_field, request, **kwargs):
        if db_field.name == "authors":
            #this line below got the proper primary key for our object of interest
            self_pub_id = request.resolver_match.args[0]

            #then we did some stuff you don't care about
            pub = Publication.objects.get(id=self_pub_id)
            kwargs["queryset"] = pub.authors.all()
        return super(PublicationAdmin, self).formfield_for_manytomany(db_field, request, **kwargs)

更优雅的解决方案是使用接受的答案推荐并利用 get_form ModelAdmin 成员函数。像这样：

class ProfileAdmin(admin.ModelAdmin):
    my_id_for_formfield = None
    def get_form(self, request, obj=None, **kwargs):
        if obj:
            self.my_id_for_formfield = obj.id
        return super(ProfileAdmin, self).get_form(request, obj, **kwargs)

    def formfield_for_foreignkey(self, db_field, request, **kwargs):
        if db_field.name == "person":
            kwargs["queryset"] = Person.objects.filter(profile=self.my_id_for_formfield)
        return super(ProfileAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs)

Answer 4

我通过在 model.py 中创建一个返回 ID 的 property() 使其工作

models.py：

class MyModel(models.Model):
    myfield = models.CharField(max_length=75)
    ...
    def get_id(self):
        return str(self.id)
    getid = property(get_id)

admin.py:

from myapp.models import MyModel

class MyModelAdmin(admin.ModelAdmin):
    list_display = ['mylink',]
    def mylink(self, object):
        return '<a href="http://mywebsite/'+object.getid+'/">Edit</a>'
    mylink.allow_tags = True

admin.site.register(MyModel, MyModelAdmin)

Answer 5

我通过重写change_view()使其工作

class CartAdmin(admin.ModelAdmin):

def change_view(self, request, object_id, form_url='', extra_context=None):
    self.object_id = object_id
    return self.changeform_view(request, object_id, form_url, extra_context)


def formfield_for_foreignkey(self, db_field, request, **kwargs):
    print self.object_id
    return super(CartAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs)

那么你可以在formfield_for_foreignkey()里面调用self.object_id

Answer 6

以下代码 sn -p 将为您提供对象 id：

request.resolver_match.kwargs['object_id']

示例用法：（我正在过滤显示的电话号码，只显示客户的电话号码）

def formfield_for_foreignkey(self, db_field, request, **kwargs):
    if db_field.name == 'preferred_contact_number':
        kwargs['queryset'] = CustomerPhone.objects.filter(customer__pk=request.resolver_match.kwargs['object_id'])
    return super().formfield_for_foreignkey(db_field, request, **kwargs)

P.S：通过调试和遍历可访问变量找到它。

Answer 7

更通用的方法是编写一个辅助方法来获取模型实例（如果有），就像您通常使用（有界）ModelForm 所做的那样，并从中检索 id 或任何其他属性：

from django.contrib import admin

class MyModelAdmin(admin.ModelAdmin):

    def get_instance(self, request):
        try:
            object_id = request.resolver_match.kwargs['object_id']
            obj = self.get_object(request, object_id)
        except:
            obj = None
        return obj