【发布时间】:2021-01-30 23:24:46
【问题描述】:
我正在开发一个 Django 项目,特别是 Wagtail 项目。我想切换下面的代码,以包含一个预填充的管理字段,该字段将由 API 响应填充。这是我目前使用的代码:
"""Flexible page."""
from django.db import models
from wagtail.admin.edit_handlers import FieldPanel
from wagtail.core.models import Page
from wagtail.core.fields import RichTextField
from wagtail.images.edit_handlers import ImageChooserPanel
class FlexPage(Page):
"""Flexibile page class."""
template = "flex/flex_page.html"
# @todo add streamfields
# content = StreamField()
subtitle = models.CharField(max_length=100, null=True, blank=True)
Flexbody = RichTextField(blank=True)
bannerImage = models.ForeignKey(
"wagtailimages.Image",
null=True,
blank=True,
on_delete=models.SET_NULL,
related_name="+"
)
audioUrl = models.URLField()
content_panels = Page.content_panels + [
FieldPanel("subtitle"),
FieldPanel('Flexbody', classname="full"),
ImageChooserPanel('bannerImage'),
FieldPanel('audioUrl'),
]
class Meta: # noqa
verbose_name = "Flex Page"
verbose_name_plural = "Flex Pages"
这将使我能够创建一个标准的 Wagtail URL 字段,并且我可以设置一个指向 MP3 文件的 URL。我想做的是从 API 响应中预填充一个下拉菜单,如下所示:
{
"id":"83ee98f6-3207-4130-9508-8f4d15ed7d5c",
"title":"some random description",
"description":"some random description.",
"audio":"https://somerandomurl.mp3",
"slug":"some random description",
"draft":false
},
{
"id":"83ee98f6-3207-4130-9508-8f4d15ed7d5c2",
"title":"some random description2",
"description":"some random description2.",
"audio":"https://somerandomurl2.mp3",
"slug":"some random description2",
"draft":false2
},
我想知道如何使用 JSON 响应中的 URL 填充管理字段?我猜 A)这是可能的,B)我将如何编写一个类模型来完成这样的事情?
【问题讨论】: