flask-sqlacodegen 生成基于db.Model而非Base的Model

在flask中，我们想要生成的model如下：

# coding: utf-8
from flask_sqlalchemy import SQLAlchemy


db = SQLAlchemy()



class Clientset(db.Model):
    __tablename__ = 'clientset'

    id = db.Column(db.Integer, primary_key=True)
    compid = db.Column(db.Integer, info=' 企业ID')
    n = db.Column(db.String(20, 'utf8mb4_0900_ai_ci'), info='名称')
    v = db.Column(db.Text(collation='utf8mb4_0900_ai_ci'), info=' 内容')
    i = db.Column(db.Integer, info='整数内容')

可有时生成的确是：

# coding: utf-8
from sqlalchemy import Column, Date, DateTime, Float, Index, String, Text
from sqlalchemy.dialects.mysql import BIGINT, BIT, INTEGER, TEXT, TINYINT, VARCHAR
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()
metadata = Base.metadata


class Clientset(Base):
    __tablename__ = 'clientset'
    __table_args__ = {'comment': '服务器企业设置'}

    id = Column(INTEGER(11), primary_key=True)
    compid = Column(INTEGER(11), comment=' 企业ID')
    n = Column(VARCHAR(20), comment='名称')
    v = Column(TEXT, comment=' 内容')
    i = Column(INTEGER(11), comment='整数内容')

经过一个下午的折腾，终于弄好了。记录一下：上面第二种基于Base的样式是sqlacodegen生成的，第一种才是flask应有的样式。安装包是注意不要安装 sqlacodegen。 只安装flask-sqlacodegen.

同时生成命令如下：

python -m sqlacodegen.main --flask --outfile models.py mysql+pymysql://<username>:<password>@<database-ip>:<port>/<database-name> [--tables <tablenames>] [--notables]