python - 如何在 SQLAlchemy 中为 MSSQL 设置架构？

Question

我目前这样做:

#!/usr/bin/env python

# 3rd party modules
from sqlalchemy import create_engine  # requires pymssql

# local modules
from config import cfg

connection_string = 'mssql+pymssql://{user}:{password}@{host}:{port}/{db}'

engine = create_engine(connection_string
                       .format(host=cfg['db']['host'],
                               db=cfg['db']['database'],
                               user=cfg['db']['user'],
                               password=cfg['db']['password'],
                               port=cfg['db']['port'],
                               schema=cfg['db']['schema']))

with engine.begin() as conn:
    sql = ('SELECT foo FROM bar;')
    rows = conn.execute(sql)
    print(rows)

但是我明白了

  File "/usr/local/lib/python3.5/dist-packages/sqlalchemy/engine/default.py", line 470, in do_execute
    cursor.execute(statement, parameters)
  File "pymssql.pyx", line 464, in pymssql.Cursor.execute (pymssql.c:7491)
sqlalchemy.exc.ProgrammingError: (pymssql.ProgrammingError) (208, b"Invalid object name 'bar'.DB-Lib error message 20018, severity 16:\nGeneral SQL Server error: Check messages from the SQL Server\n") [SQL: 'SELECT foo FROM bar;']

我认为问题在于我必须使用架构 exampleschema。我可以通过 DBeaver 使用相同的凭据在架构 exampleschema 中使用列 foo 访问表 bar。

但是当我将 /{schema} 添加到连接字符串时，我得到了

sqlalchemy.exc.OperationalError: (pymssql.OperationalError)
    (18456,
     b"Login failed for user 'exampleuser'.DB-Lib error message 20018,
     severity 14:\nGeneral SQL Server error: Check messages from the SQL Server\nDB-Lib error message 20002,
     severity 9:\nAdaptive Server connection failed (192.168.123.456:1433)\n")

如何设置模式？

Answer 1

您还可以在类定义中指定架构名称(这不是您的具体情况，但我认为这是常见情况)。

例如，如果您将表“dog”放入“animal”模式中:

from sqlalchemy import Column, Integer, String
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

class Notification(Base):
    __tablename__ = "dog"
    __table_args__ = {"schema": "animal"}
    id = Column(Integer, primary_key=True)
    name = Column(String)