Django给admin添加Action的步骤详解

$ django-admin startproject DjangoActions
$ cd DjangoActions
$ python3 manage.py startapp mysite

from django.db import models


class Article(models.Model):
 # 文章状态，类似枚举
 STATUS_CHOICES = (
  ('d', '草稿'),
  ('p', '发布'),
 )
 # 标题
 title = models.CharField(max_length=30)
 # 内容
 body = models.TextField()
 # 状态
 status = models.CharField(max_length=1, choices=STATUS_CHOICES, default='d')

 def __str__(self):
  return self.title

from django.contrib import admin
from .models import Article


# 指令处理函数，参数固定写法
def make_article_published(modeladmin, request, queryset):
 # queryset 是从数据库查询到的model set
 # 更新status状态为p，在此处就是Article
 queryset.update(status='p')


# 此指令的名称
make_article_published.short_description = '更改状态为发布'


#将Article注册到后台
@admin.register(Article)
class ArticleAdmin(admin.ModelAdmin):
 # 后台列表展示的字段
 list_display = ('title', 'status')
 # 动作集合
 actions = [make_article_published]

$ python3 manage.py makemigrations
$ python3 manage.py migrate

$ python3 manage.py createsuperuser

$ python3 manage.py runserver