如何迁移具有新日期时间列(不可为空)的表中数据的表? [英] How can I migrate a table with data in a table which got a new datetime column which is non-nullable?

问题描述

我已添加一列

registered_on = db.Column(db.DateTime, nullable=False)

到我的users表.自动创建的迁移是

to my users table. The migration which was automatically created is

def upgrade():
    # ### commands auto generated by Alembic - please adjust! ###
    ...
    op.add_column('users', sa.Column('registered_on', sa.DateTime(), nullable=False))
    ...

当我执行它时，我得到了一个关于无效值0000-00-00 00:00:00的异常(等等).

When I execute it, I got an exception about an invalid value 0000-00-00 00:00:00 (or so).

我应该如何调整迁移脚本以解决此问题?

How should I adjust the migration script to not have this problem?

(在这种情况下，最好填写一个虚拟值)

(In this situation, filling in a dummy value would probably have been the best)

推荐答案

我可以想到两种方法:

1. 在add_column操作中添加server_default:

1. Add a server_default to the add_column operation:

op.add_column('users', sa.Column('registered_on', sa.DateTime(), nullable=False, server_default=func.now()))

1. 首先将列插入为可空值，然后将所有行编辑为具有值.给所有行赋予非空值后，再进行一次迁移以将列更改为非空.

这篇关于如何迁移具有新日期时间列(不可为空)的表中数据的表?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！