你能为Django自定义管理操作添加参数吗？

For example (not the case, but just to illustrate) if I wanted to add an action to set a specific field in the selected items to X. Is it possible to add an action to allow X to be entered as opposed to hard coded?

例如（不是这种情况，只是为了说明），如果我想添加一个动作来将所选项目中的特定字段设置为X.是否可以添加一个动作以允许输入X而不是硬编码？

2 个解决方案

#1

See: Actions that Provide Intermediate Pages

请参阅：提供中间页面的操作

You can't do it right from the changelist page, but on your intermediate page, you could have a form that would allow the user to enter a value, and then use that value for the action.

您无法从更改列表页面直接执行此操作，但在中间页面上，您可以使用允许用户输入值的表单，然后将该值用于操作。

#2

It is possible with Django 1.5, though its a little hackish. I am not sure with what other older Django versions it's possible.

有可能使用Django 1.5，虽然它有点hackish。我不确定其他更旧的Django版本是否可行。

You write your own ModelAdmin subclass. ModelAdmin has an attribute called action_form which determines the form shown before Go button on changelist page. You can write your own form, and set it as action_form on your ModelAdmin subclass.

您编写自己的ModelAdmin子类。 ModelAdmin有一个名为action_form的属性，它确定在更改列表页面上的Go按钮之前显示的表单。您可以编写自己的表单，并在ModelAdmin子类上将其设置为action_form。

from django.contrib.admin.helpers import ActionForm
# ActionForm is the default form used by Django
# You can extend this class

class XForm(ActionForm):
    x_field = forms.CharField()


class YourModelAdmin(admin.ModelAdmin):
    action_form = XForm

With these changes, you will have a CharField, where you can put value for X.

通过这些更改，您将拥有一个CharField，您可以在其中为X赋值。

And use x_field in your action function.

并在动作函数中使用x_field。

def set_x_on_objects(modeladmin, request, queryset):
    for obj in queryset:
        obj.something = request.POST['x_field']
    # do whatever else you want

class YourModelAdmin(admin.ModelAdmin):
    action_form = XForm
    actions = [set_x_on_objects]

#1