使用 SQLAlchemy 反映来自 postgres DB 的每个模式

reflecting every schema from postgres DB using SQLAlchemy

我有一个现有数据库,它有两个模式,名为 schoolsstudents,包含在 declarative_base 的实例中,并通过两个不同的 类 继承自那个例子

class DirectorioEstablecimiento(Base):
    __table_args__ = {'schema': 'schools'}
    __tablename__ = 'addresses'
    # some Columns are defined here

class Matricula(Base):
    __table_args__ = {'schema': 'students'}
    __tablename__ = 'enrollments'
    # some Columns are defined here

我可以使用 Base 实例作为 Base.metadata.create_all(bind=engine) 在我在 postgres 中的测试数据库中重新创建它。如果我查询 pg_namespace

,我可以确认这是没有问题的
In [111]: engine.execute("SELECT * FROM pg_namespace").fetchall()
2017-12-13 18:04:01,006 INFO sqlalchemy.engine.base.Engine SELECT * FROM pg_namespace
2017-12-13 18:04:01,006 INFO sqlalchemy.engine.base.Engine {}

Out[111]: 
[('pg_toast', 10, None),
 ('pg_temp_1', 10, None),
 ('pg_toast_temp_1', 10, None),
 ('pg_catalog', 10, '{postgres=UC/postgres,=U/postgres}'),
 ('public', 10, '{postgres=UC/postgres,=UC/postgres}'),
 ('information_schema', 10, '{postgres=UC/postgres,=U/postgres}'),
 ('schools', 16386, None),
 ('students', 16386, None)]

以及来自 psql CLI

user# select * from pg_tables;

     schemaname     |          tablename           | tableowner | tablespace | hasindexes | hasrules | hastriggers | rowsecurity 
--------------------+------------------------------+------------+------------+------------+----------+-------------+-------------
 schools            | addresses                    | diego      |            | t          | f        | f           | f
 students           | enrollments                  | diego      |            | t          | f        | f           | f
 pg_catalog         | pg_statistic                 | postgres   |            | t          | f        | f           | f
 pg_catalog         | pg_type                      | postgres   |            | t          | f        | f           | f
 pg_catalog         | pg_authid                    | postgres   | pg_global  | t          | f        | f           | f
 pg_catalog         | pg_user_mapping              | postgres   |            | t          | f        | f           | f
-- other tables were omitted

但是,如果我想在 declarative_base 的某些其他实例中反映该数据库,则不会反映任何内容。

类似

In [87]: Base.metadata.tables.keys()
Out[87]: dict_keys(['schools.addresses', 'students.enrollments'])

In [88]: new_base = declarative_base()

In [89]: new_base.metadata.reflect(bind=engine)

In [90]: new_base.metadata.tables.keys()
Out[90]: dict_keys([])

我知道 reflect 接受一个 schema 作为参数,但我想在反思期间一次获得所有这些参数。出于某种原因,我一次可以做到这一点。

有办法吗?

当您调用 metadata.reflect() 时,它只会反映默认架构(您有权访问 search_path 中的第一个架构)。因此,如果您的 search_path 是 public,students,school,它只会反映模式 public 中的 table。如果您没有模式 public 的权限,public 模式将被跳过并且默认仅反映 students.

默认架构由 SELECT current_schema();

检索

为了反映其他模式 您需要为每个模式调用 metadata.reflect()

metadata.reflect(schema='public')  # will reflect even if you do not have permissions on the tables in schema `public`, as long as you have access to pg_* system tables
metadata.reflect(schema='students')
metadata.reflect(schema='schools')

Note: When you reflect with an explicit schema

  • Reflected tables in metadata.tables will have the keys with the tables fully qualified schema name as in schema1.mytable, schema2.mytable
  • Any conflicting table names will be replaced with the later one. If you have any tables with the same name, you should implement your the function classname_for_table to prefix the names with the schema name.
  • table 名称前缀示例

    def classname_for_table(base, tablename, table):
        schema_name = table.schema
        fqname = '{}.{}'.format(schema_name, tablename)
        return fqname
    
    
    Base.prepare(classname_for_table=classname_for_table)
    

    **作为奖励,这里有一个小片段,它将在每个模式的动态子模块中公开所有 table,以便您可以访问它 **

    创建一个文件,即。 db.py 并放置以下内容

    from types import ModuleType
    
    
    def register_classes(base, module_dict):
        for name, table in base.classes.items():
            schema_name, table_name = name.split('.')
            class_name = table_name.title().replace('_', '')
            if schema_name not in module_dict:
                module = module_dict[schema_name] = ModuleType(schema_name)
            else:
                module = module_dict[schema_name]
            setattr(module, class_name, table)
    

    使用自动映射库和您要注册模式的模块的 __dict__ 调用此函数。

     register_classes(base, globals())
    

    import db
    db.register_classes(base, db.__dict__)
    

    然后你会得到

    import db
    db.students.MyTable
    db.schools.MyTable