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更多推荐
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