合并查询我有两个不相关类似这样的表格:蟒蛇SQLAlchemy的两个表中的
hurst = Table('Hurst', metadata,
Column('symbol_id' , Integer, primary_key=True),
Column('Date' , String(20), nullable=False),
Column('symbol' , String(40), nullable=False),
Column('HurstExp' , Float, nullable=False),
)
fundamental = Table('Fundamental', metadata,
Column('symbol_id' , Integer, primary_key=True),
Column('Date' , String(20), nullable=False),
Column('symbol' , String(40), nullable=False),
Column('MarketCap' , Float, nullable=False),
)
以下查询中的每个工作正常。我如何结合他们,所以我可以得到只有价值超过50,000,000,000公司的hurst?
# -*- coding: utf-8 -*-
"""
Created on Sun Dec 13 19:22:35 2015
@author: idf
"""
from sqlalchemy import *
def run(stmt):
rs = stmt.execute()
return rs
# Let's re-use the same database as before
dbh = create_engine('sqlite:///hurst.db')
dbf = create_engine('sqlite:///fundamental.db')
dbh.echo = True # We want to see the SQL we're creating
dbf.echo = True # We want to see the SQL we're creating
metadatah = MetaData(dbh)
metadataf = MetaData(dbf)
# The users table already exists, so no need to redefine it. Just
# load it from the database using the "autoload" feature.
hurst = Table('Hurst', metadatah, autoload=True)
funda = Table('Fundamental', metadataf, autoload=True)
hurstQ = hurst.select(hurst.c.HurstExp < .5)
run(hurstQ)
fundaQ = funda.select(funda.c.MarketCap > 50000000000)
run(fundaQ)
如果我尝试使用一个连接,我得到一个错误:
j = join(hurst, funda, hurst.c.symbol == funda.c.symbol)
stmt = select([hurst]).select_from(j)
theJoin = run(stmt)
Traceback (most recent call last):
File "/home/idf/anaconda3/lib/python3.5/site-packages/sqlalchemy/engine/base.py", line 1139, in _execute_context
context)
File "/home/idf/anaconda3/lib/python3.5/site-packages/sqlalchemy/engine/default.py", line 450, in do_execute
cursor.execute(statement, parameters)
cursor.execute(statement, parameters)
sqlite3.OperationalError: no such table: Fundamental
我甚至不能做简单的版本
# This will return more results than you are probably expecting.
s = select([hurst, funda])
run(s)
请参阅编辑的文章。桌子是分开的。也许我需要以某种方式将它们联系起来,但两个表格是分开创建的。 – Ivan
您需要使用'sqlalchemy'的'backref'功能并使用外键或关系创建另一个表的引用。 – Abhinav