Question

管理员操作似乎适用于django管理界面的列表视图中选择的多个项目：

在我的情况下，我想在更改（一个项目）视图上有一个简单的操作按钮。

有没有办法在那里提供django管理员操作？

我知道我可以通过转到列表视图来解决此问题，并在那里选择一个项目。但直接提供它会更好。

Answer 1

在您的应用中为您的模型创建模板。

templates/admin/<yourapp>/<yourmodel>/change_form.html

使用此示例内容在更改现有对象时添加按钮。

{% extends "admin/change_form.html" %}
{% block submit_buttons_bottom %}
    {{ block.super }}
    {% if original %} {# Only show if changing #}
        <div class="submit-row">
            <a href="{% url 'custom-model-action' original.pk %}">
                 Another action
            </a>
        </div>
    {% endif %}
{% endblock %}

将该操作链接到任何网址并重定向回模型更改对象视图。 More information about extending admin templates

更新：为现有对象添加了自定义操作的完整常见用例

urls.py

urlpatterns = [
    url(r'^custom_model_action/(?P<object_pk>\d+)/$',
        core_views.custom_model_action, name='custom-model-action')
]

views.py

from django.contrib import messages
from django.http import HttpResponse, HttpResponseRedirect

def custom_model_action(request, object_pk):
    messages.info(request, 'Performed custom action!')
    return HttpResponseRedirect(
       reverse('admin:<yourapp>_<yourmodel>_change', args=[object_pk])
    )

Answer 2

如果你真的需要单个对象，我建议你使用这个解决方案，例如：

class Gallery(TimeStampedModel):
    title = models.CharField(max_length=200)
    attachment = models.FileField(upload_to='gallery/attachment/%Y/%m/%d')

    def __str__(self):
        return self.title

    def process_button(self):
        return ('<button id="%(id)s class="btn btn-default process_btn" '
                'data-value="%(value)s>Process</button>' % {'id': self.pk, 'value': self.attachment.url})
    process_button.short_description = 'Action'
    process_button.allow_tags = True

在admin.py中，将process_button插入list_display;

class GalleryAdmin(admin.ModelAdmin):
    list_display = ['title', 'process_button', 'created']
    search_fields = ['title', 'pk']
    ....

    class Media:
        js = ('path/to/yourfile.js', )

然后，在yourfile.js内，你也可以处理它..

$('.process_btn').click(function(){
    var id = $(this).attr('id');       // single object id
    var value = $(this).data('value'); // single object value
    ...
});

希望它有用......

Answer 3

内置管理操作对查询集进行操作。

您可以使用calable进行您所做的操作或显示其他内容：

class ProductAdmin(admin.ModelAdmin):
    list_display ('name' )
    readonly_fields('detail_url)


def detail_url(self, instance):
    url = reverse('product_detail', kwargs={'pk': instance.slug})
    response = format_html("""<a href="{0}">{0}</a>""", product_detail)
    return response

或使用表格

class ProductForm(forms.Form):
    name = forms.Charfield()

    def form_action(self, product, user):
        return Product.value(
            id=product.pk,
            user= user,
        .....

        )


@admin.register(Product)
class ProductAdmin(admin.ModelAdmin):

# render buttons and links to
def product_actions(self, obj):
        return format_html(
            '<a class="button" href="{}">Action1</a>&nbsp;'
            '<a class="button" href="{}">Action 2</a>',
            reverse('admin:product-action-1', args=[obj.pk]),
            reverse('admin:aproduct-action-3', args=[obj.pk]),
        )

for more details about using forms

Answer 4

与主题入门要求的不同，但是此摘录允许使用最少的代码量

在列表页面上执行“单个对象”操作

BaseAction代码

class AdminActionError(Exception):
    pass


class AdminObjectAction:
    """Base class for Django Admin actions for single object"""

    short_description = None
    exp_obj_state = {}

    def __init__(self, modeladmin, request, queryset):
        self.admin = modeladmin
        self.request = request
        self.queryset = queryset
        self.__call__()

    def validate_qs(self):
        count = self.queryset.count()
        if count != 1:
            self.error("You must select one object for this action.")

        if self.exp_obj_state:
            if self.queryset.filter(**self.exp_obj_state).count() != 1:
                self.error(f'Selected object does not meet the requirements: {self.exp_obj_state}')

    def error(self, msg):
        raise AdminActionError(msg)

    def get_object(self):
        return self.queryset.get()

    def process_object_action(self, obj):
        pass

    def validate_obj(self, obj):
        pass

    def __call__(self, *args, **kwargs):
        try:
            self.validate_qs()
            obj = self.get_object()
            self.validate_obj(obj)
        except AdminActionError as e:
            self.admin.message_user(self.request, f"Failed: {e}", level=messages.ERROR)
        else:
            with transaction.atomic():
                result = self.process_object_action(obj)

            self.admin.message_user(self.request, f"Success: {self.short_description}, {result}")

自定义操作[最少代码量]

class RenewSubscriptionAction(AdminObjectAction):
    short_description = 'Renew subscription'

    exp_obj_state = {
        'child': None,
        'active_status': True,        
    }

    def process_object_action(self, obj):
        manager = RenewManager(user=obj.user, subscription=obj)
        return manager.process()

AdminClass

class SomeAdmin(admin.ModelAdmin):
    actions = [RenewSubscriptionAction]

Django Admin对单个对象的操作

4 个答案: