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使用alembic升级表时设置列的值

alembic auto-generate upgrade postgresql database

17

Max · 技术社区 · 10 年前

我正在使用PostgreSQL和Alembic进行迁移。当我向用户表中添加新列时,Alembic使用以下脚本生成了迁移:

revision = '4824acf75bf3'
down_revision = '2f0fbdd56de1'

from alembic import op
import sqlalchemy as sa

def upgrade():
    op.add_column(
        'user', 
        sa.Column(
            'username', 
            sa.Unicode(length=255), 
            nullable=False
        )
    )

def downgrade():
    op.drop_column('user', 'username')

我实际上想做的是在升级生产版本时自动生成用户名的值。换句话说,我的生产版本中有很多用户,如果我对其运行上述升级,将出现一个错误,说明用户名不能为NULL,因此我必须删除所有用户,升级User表,然后再次添加用户,这很痛苦。因此,我想通过以下方式更改上述脚本:

revision = '4824acf75bf3'
down_revision = '2f0fbdd56de1'

from alembic import op
import sqlalchemy as sa

def upgrade():
    op.add_column(
        'user', 
        sa.Column(
            'username', 
            sa.Unicode(length=255)
        )
    )
    op.execute(
        'UPDATE "user" set username = <email address with no '@' 
         and everything comes after '@' sign should be removed> 
         WHERE email is not null'
    )
    <only after the above code is executed 'nullable=False' must be set up>

def downgrade():
    op.drop_column('user', 'username')

正如上面在代码中所述,我希望执行一个SQL代码,检查电子邮件地址,如test@example.com并在“@”sign(在本例中为“@example.com”)之后抛出所有内容,然后设置username的值(在本示例中为“test”),使nullable=false。

我该怎么做?脚本必须是什么而不是 username = <email address with no '@' and everything comes after '@' sign should be removed> 和设置 nullable=false

或者如果有其他设置方式 username 默认值为电子邮件地址,不包含@ssing及其后的所有内容?

2 回复 | 直到 10 年前

1

7

Max 10 年前

这是如何解决这个问题的。

def upgrade():
    op.add_column(
        'user',
        sa.Column(
            'username',
            sa.Unicode(length=255)
        )
    )
    op.create_index('ix_user_username', 'user', ['username'], unique=True)
    op.execute(
        '''
        DO
        $do$
        DECLARE uid INTEGER;
        DECLARE username_candidate TEXT;
        BEGIN
        FOR uid, username_candidate IN (
            SELECT
                id,
                lower(
                    substring(email for position('@' in email) - 1)
                )
            FROM "user" WHERE username is null
        ) LOOP
            UPDATE "user"
            SET username = username_candidate
            WHERE
                id = uid AND
                NOT EXISTS (
                SELECT id FROM "user" WHERE username = username_candidate
            );
        END LOOP;
        END
        $do$
        '''
    )
    # Fix name colissions
    op.execute(
        '''
        DO
        $do$
        DECLARE uniqufier INTEGER := 0;
        DECLARE uid INTEGER;
        DECLARE username_candidate TEXT;
        BEGIN
        WHILE EXISTS (SELECT id FROM "user" WHERE username is null) LOOP
            uniqufier := uniqufier + 1;
            FOR uid, username_candidate IN (
                SELECT
                    id,
                    lower(
                        substring(email for position('@' in email) - 1)
                        || uniqufier
                    )
                FROM "user" WHERE username is null
            ) LOOP
                UPDATE "user"
                SET username = username_candidate
                WHERE
                    id = uid AND
                    NOT EXISTS (
                        SELECT id FROM "user" WHERE username = username_candidate
                    );
            END LOOP;
        END LOOP;
        END;
        $do$
        '''
    )
    op.alter_column(
        'user',
        'username',
        nullable=False,
    )


def downgrade():
    op.drop_index('ix_user_username', table_name='user')
    op.drop_column('user', 'username')

2

0

Tim 6 年前

可以使用子查询编写基于同一表的另一列的值更新新列的脚本。唯一的诀窍是,因为您从同一个表中查询,所以需要为表名赋予别名,以确保您从相应的行中进行选择:

update 'user' as target set username = (
  select substring(email from '.+?(?=@)')
  from 'user' as source where source.id = target.id
);