如何使用south将数据从字段迁移到表中的另一个字段

Question

我正在使用南方的Django项目进行模式迁移.我有以下场景:

模式

class Book(model.Models):
    name = models.CharField(max_length=255, blank=True)
    bid = models.IntegerField(blank=True, null=True)

class Author(model.Models):
   name = models.CharField(max_length=255, blank=True)
   book_id = models.ForeignKey(Book, null=True, to_field="bid", db_column="bookID")

我想将作者模型更改为以下内容:

class Author(model.Models):
   name = models.CharField(max_length=255, blank=True)
   book = models.ForeignKey(Book, null=True, db_column="book_id")

但没有松散的数据.我想按出价搜索每本书,并将作者发现的那一个分配给作者模型中的新字段.

谢谢

Answer 1

您必须进行 3 次迁移。添加新书 FK 的 schemamigration，然后是数据迁移，然后是删除和重命名字段的 schemamigration。

因此，您需要将 models.py 文件更改为：

class Book(model.Models):
    name = models.CharField(max_length=255, blank=True)
    bid = models.IntegerField(blank=True, null=True)

class Author(model.Models):
   name = models.CharField(max_length=255, blank=True)
   book_id = models.ForeignKey(Book, null=True, to_field="bid", db_column="bookID")
   # You don't need the db_column="book_id" since that's what it does at the DB level by default.
   # I'm not sure if things will break if you name it book with another field as book_id.
   book_new = models.ForeignKey(Book, null=True) 

然后运行python manage.py schemamigration APP_NAME auto 然后运行```python manage.py datamigration APP_NAME populate_book_id

然后编辑新创建的数据迁移并循环遍历 Author 实例，使用 book_id 字段设置新的 book 字段。不要忘记在向后方法中删除 book 字段值。

然后将 models.py 文件更改为以下内容：

class Book(model.Models):
    name = models.CharField(max_length=255, blank=True)
    bid = models.IntegerField(blank=True, null=True)

class Author(model.Models):
   name = models.CharField(max_length=255, blank=True)
   # You don't need the db_column="book_id" since that's what it does at the DB level by default.
   book = models.ForeignKey(Book, null=True) 

然后运行python manage.py schemamigration APP_NAME auto 您将需要检查最后一个架构迁移，以确保它重命名book_new为book而不是删除和创建列。这是解释如何更改它的答案。