【发布时间】:2015-01-13 04:56:59
【问题描述】:
我想和你分享这个案例:
我有 Albums、Artists 和 Tracks 的模型
-
一个
Artist可能有许多Albums -
一个
Album可能有许多Tracks -
很多
Tracks在里面一个Album(也可能是ManyToMany ..)
在Albums 模型中,我想添加一个SlugField 类型的字段。这是以下内容:
from django.db import models
from artists.models import Artists
class Album(models.Model):
title = models.CharField(max_length=255)
cover = models.ImageField(upload_to='albums')
slug = models.SlugField(max_length=100)
artist = models.ForeignKey(Artists)
def __unicode__(self):
return self.title
我向南执行迁移:
(myvenv)➜ myvenv ./manage.py syncdb
Syncing...
Creating tables ...
Installing custom SQL ...
Installing indexes ...
Installed 0 object(s) from 0 fixture(s)
Synced:
> django.contrib.admin
> django.contrib.auth
> django.contrib.contenttypes
> django.contrib.sessions
> django.contrib.messages
> django.contrib.staticfiles
> south
> albums
Not synced (use migrations):
- django_extensions
- djcelery
- tracks
- artists
- userprofiles
(use ./manage.py migrate to migrate these)
(myenv)➜ myenv ./manage.py convert_to_south albums
Creating migrations directory at '/home/bgarcial/workspace/myenv/sfotipy/albums/migrations'...
Creating __init__.py in '/home/bgarcial/workspace/myenv/sfotipy/albums/migrations'...
+ Added model albums.Album
Created 0001_initial.py. You can now apply this migration with: ./manage.py migrate albums
- Soft matched migration 0001 to 0001_initial.
Running migrations for albums:
- Nothing to migrate.
- Loading initial data for albums.
Installed 0 object(s) from 0 fixture(s)
App 'albums' converted. Note that South assumed the application's models matched the database
(i.e. you haven't changed it since last syncdb); if you have, you should delete the albums/migrations directory, revert models.py so it matches the database, and try again.
(myenv)➜ myenv ./manage.py migrate albums
Running migrations for albums:
- Nothing to migrate.
- Loading initial data for albums.
Installed 0 object(s) from 0 fixture(s)
如果我输入命令 ./manage.py sqlall,专辑模型已经出现在数据库中,并带有 slug 字段
(Sfoti.py)➜ sfotipy ./manage.py sqlall albums
BEGIN;
CREATE TABLE "albums_album" (
"id" integer NOT NULL PRIMARY KEY,
"title" varchar(255) NOT NULL,
"cover" varchar(100) NOT NULL,
"slug" varchar(100) NOT NULL,
"artist_id" integer NOT NULL REFERENCES "artists_artists" ("id")
);
CREATE INDEX "albums_album_f52cfca0" ON "albums_album" ("slug");
CREATE INDEX "albums_album_7904f807" ON "albums_album" ("artist_id");
COMMIT;
但是,当我去数据库时,直接到Django给我带来的数据库结构,我看到slug字段无效......这个可以在这个url中详细说明https://cldup.com/-F9SQ2D3W8.jpeg
为了测试这个 slug 是否有效,我创建了指向 AlbumListView 基于类的视图的 url “专辑”
from django.conf.urls import patterns, url
from artists.views import AlbumListView
urlpatterns = patterns('',
url(r'^albums/$', AlbumListView.as_view(), name='album_list'),
url(r'^albums/(?P<artist>[\w\-]+)/$', AlbumListView.as_view(), name='album_list'),
)
基于类的视图AlbumListView 如下:这里我定义了一个查询集,用于恢复艺术家的专辑,并使用kwargs 变量是如何采取的方式
class AlbumListView(ListView):
model = Album
template_name = 'album_list.html'
def get_queryset(self):
if self.kwargs.get('artist'):
queryset = self.model.objects.filter(artist__slug=self.kwargs['artist'])
else:
queryset = super(AlbumListView, self).get_queryset()
return queryset
当我在浏览器中转到视图 /albums 时,我看到以下消息:
no such column: albums_album.slug
这是我浏览器中的错误图片,请查看此网址:
我的问题可能是什么?为什么迁移不起作用? 感谢您的指导:)
【问题讨论】: