运行Flask-Migrate时保留数据库中的现有表 [英] Preserve existing tables in database when running Flask-Migrate

问题描述

我有一个包含现有表的数据库.我的代码具有 User 模型.我使用Flask-Migrate生成了一个修订版本并运行它，并且它在创建用户表时删除了我现有的表.如何在不删除现有表的情况下运行迁移?

I have a database with existing tables. My code has a User model. I generated a revision using Flask-Migrate and ran it, and it deleted my existing tables while creating the user table. How can I run migrations without removing the existing tables?

from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_script import Manager
from flask_migrate import Migrate, MigrateCommand

app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'my_data'

db = SQLAlchemy(app)
migrate = Migrate(app, db)

manager = Manager(app)
manager.add_command('db', MigrateCommand)

class User(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(128))

if __name__ == '__main__':
    manager.run()

推荐答案

如果数据库中已有表，并且代码中没有相应的模型，则Alembic(Flask-Migrate)仅知道存在数据库和代码之间的差异.(默认情况下)它不知道您要保留这些表不变.

If you have existing tables in your database, and they don't have a corresponding model in your code, Alembic (Flask-Migrate) only knows that there is a difference between your database and your code. It can't know (by default) that you meant to leave those tables untouched.

传递 include_object 函数对环境产生影响，以影响Alembic将为其生成命令的数据库对象.下面的示例跳过列出的表名，但允许其他所有内容.

Pass an include_object function to the environment to effect what database objects Alembic will generate commands for. The following example skips the listed table names, but allows everything else.

def include_object(object, name, type_, reflected, compare_to):
    if type_ == 'table' and name in ('table', 'names', 'to', 'skip'):
        return False

    return True

# in env.py
context.configure(
    # ...
    include_object=include_object
)

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