将参数传递给 django admin 中的内联表单

Question

我有一个从 ModelAdmin 继承的管理类：

class TemplateAdmin (admin.ModelAdmin):
    inlines = (TemplateAttributeInline, CompanyAttributeInline)
    list_display = ("name", "created", "updated","departments")
    list_filter = ['companies__company']
    list_editable = ("departments",)
    search_fields = ("name", "companies__company",)
    exclude = ("companies",)
    save_as = True

我想传递一个参数，TemplateAttributeInline然后将参数传递给TemplateAttributeForm. 做这个的最好方式是什么？

模板属性内联：

class TemplateAttributeInline (admin.TabularInline):
    model = TemplateAttribute
    extra = 0
    sortable_field_name = "display"
    form = TemplateAttributeForm

模板属性表单

class TemplateAttributeForm(forms.ModelForm):
    class Meta:
        model = Template
    def __init__(self,*args, **kwargs):
        super(TemplateAttributeForm, self).__init__(*args, **kwargs) 
        self.fields['attribute'].queryset = Attribute.objects.filter(#WANT TO FILTER BY THE ID OF THE COMPANY THAT OWNS THE TEMPLATE WE ARE EDITING ON THE ADMIN PAGE)

Answer 1

您可以创建一个返回表单类的函数：

def TemplateAttributeForm(param):

    class MyTemplateAttributeForm(forms.ModelForm):
        class Meta:
            model = Template
        def __init__(self,*args, **kwargs):
            super(TemplateAttributeForm, self).__init__(*args, **kwargs) 
            #do what ever you want with param

    return MyTemplateAttributeForm

在另一个函数中使用它来定义 TemplateAttributeInline

def TemplateAttributeInline(param):

        class MyTemplateAttributeInline (admin.TabularInline):
            model = TemplateAttribute
            extra = 0
            sortable_field_name = "display"
            form = TemplateAttributeForm(param)

        return MyTemplateAttributeInline 

最后，在你的TemplateAdmin定义中使用这个函数：

class TemplateAdmin (admin.ModelAdmin):
    inlines = (TemplateAttributeInline(param), CompanyAttributeInline)
    ....