在相同的Flask

本文介绍了在相同的Flask-SQLAlchemy模型中使用多个POSTGRES数据库和架构的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我将在这里非常具体，因为已经提出了类似的问题，但是没有一个解决方案可以解决这个问题。

I'm going to be very specific here, because similar questions have been asked, but none of the solutions work for this problem.

我正在研究一个有四个postgres数据库的项目，但为简单起见，我们说有2个。即A& B

I'm working on a project that has four postgres databases, but let's say for the sake of simplicity there are 2. Namely, A & B

A，B代表两个地理位置，但是数据库中的表和模式是相同的。

A,B represent two geographical locations, but the tables and schemas in the database are identical.

示例模型：

from flask_sqlalchemy import SQLAlchemy from sqlalchemy import * from sqlalchemy.ext.declarative import declarative_base db = SQLAlchemy() Base = declarative_base() class FRARecord(Base): __tablename__ = 'tb_fra_credentials' recnr = Column(db.Integer, primary_key = True) fra_code = Column(db.Integer) fra_first_name = Column(db.String)

此模型已在两个数据库中复制，但是具有不同的模式，因此要使其在A中工作，我需要这样做：

This model is replicated in both databases, but with different schemas, so to make it work in A, I need to do:

__table_args__ = {'schema' : 'A_schema'}

我想使用一个可以访问数据库但具有相同方法的单一内容提供程序：

I'd like to use a single content provider that is given the database to access, but has identical methods:

class ContentProvider(): def __init__(self, database): self.database = database def get_fra_list(): logging.debug("Fetching fra list") fra_list = db.session.query(FRARecord.fra_code)

两个问题是，我该怎么办确定要指向的数据库，以及如何不为不同的模式复制模型代码（这是Postgres特定的问题）

Two problems are, how do I decide what db to point to and how do I not replicate the model code for different schemas (this is a postgres specific problem)

这是到目前为止我尝试过的方法：

Here's what I've tried so far:

1）我为每个模型制作了单独的文件并继承了它们，因此：

1) I've made separate files for each of the models and inherited them, so:

class FRARecordA(FRARecord): __table_args__ = {'schema' : 'A_schema'}

这似乎不起作用，因为出现错误：

This doesn't seem to work, because I get the error:

"Can't place __table_args__ on an inherited class with no table."

意思是我无法在db.Model（在其父级中）之后设置该参数声明的

Meaning that I can't set that argument after the db.Model (in its parent) was already declared

2）所以我尝试对多重继承进行相同的操作，

2) So I tried to do the same with multiple inheritance,

class FRARecord(): recnr = Column(db.Integer, primary_key = True) fra_code = Column(db.Integer) fra_first_name = Column(db.String) class FRARecordA(Base, FRARecord): __tablename__ = 'tb_fra_credentials' __table_args__ = {'schema' : 'A_schema'}

，但得到了可预测的错误：

but got the predictable error:

"CompileError: Cannot compile Column object until its 'name' is assigned."

很明显，我不能将Column对象移动到FRARecordA模型，而不必为B重复它们好（实际上有4个数据库和更多模型）。

Obviously I can't move the Column objects to the FRARecordA model without having to repeat them for B as well (and there are actually 4 databases and a lot more models).

3）最后，我正在考虑进行某种分片（这似乎是正确的）方法），但我找不到如何进行此操作的示例。我的感觉是，我只会使用这样的单个对象：

3) Finally, I'm considering doing some sort of sharding (which seems to be the correct approach), but I can't find an example of how I'd go about this. My feeling is that I'd just use a single object like this:

class FRARecord(Base): __tablename__ = 'tb_fra_credentials' @declared_attr def __table_args__(cls): #something where I go through the values in bind keys like for key, value in self.db.app.config['SQLALCHEMY_BINDS'].iteritems(): # Return based on current session maybe? And then have different sessions in the content provider? recnr = Column(db.Integer, primary_key = True) fra_code = Column(db.Integer) fra_first_name = Column(db.String)

要清楚一点，我访问不同数据库的意图如下：

Just to be clear, my intention for accessing the different databases was as follows:

app.config['SQLALCHEMY_DATABASE_URI']='postgresql://%(user)s:\ %(pw)s@%(host)s:%(port)s/%(db)s' % POSTGRES_A app.config['SQLALCHEMY_BINDS']={'B':'postgresql://%(user)s:%(pw)s@%(host)s:%(port)s/%(db)s' % POSTGRES_B, 'C':'postgresql://%(user)s:%(pw)s@%(host)s:%(port)s/%(db)s' % POSTGRES_C, 'D':'postgresql://%(user)s:%(pw)s@%(host)s:%(port)s/%(db)s' % POSTGRES_D }

假设POSTGRES词典包含所有用于连接数据的键

Where the POSTGRES dictionaries contained all the keys to connect to the data

我假设使用继承的对象，那么我将连接到正确的键，例如thi s（因此sqlalchemy查询将自动知道）：

I assumed with the inherited objects, I'd just connect to the correct one like this (so the sqlalchemy query would automatically know):

class FRARecordB(FRARecord): __bind_key__ = 'B' __table_args__ = {'schema' : 'B_schema'}

推荐答案

最终找到了解决方案。

基本上，我没有为每个数据库创建新类，而是为每个数据库使用了不同的数据库连接。

Essentially, I didn't create new classes for each database, I just used different database connections for each.

这种方法本身很常见，棘手的部分（我找不到示例）处理模式差异。我最终这样做：

This method on its own is pretty common, the tricky part (which I couldn't find examples of) was handling schema differences. I ended up doing this:

from sqlalchemy import create_engine from sqlalchemy.orm import sessionmaker Session = sessionmaker() class ContentProvider(): db = None connection = None session = None def __init__(self, center): if center == A: self.db = create_engine('postgresql://%(user)s:%(pw)s@%(host)s:%(port)s/%(db)s' % POSTGRES_A, echo=echo, pool_threadlocal=True) self.connection = self.db.connect() # It's not very clean, but this was the extra step. You could also set specific connection params if you have multiple schemas self.connection.execute('set search_path=A_schema') elif center == B: self.db = create_engine('postgresql://%(user)s:%(pw)s@%(host)s:%(port)s/%(db)s' % POSTGRES_B, echo=echo, pool_threadlocal=True) self.connection = self.db.connect() self.connection.execute('set search_path=B_schema') def get_fra_list(self): logging.debug("Fetching fra list") fra_list = self.session.query(FRARecord.fra_code) return fra_list