Question

我有两张与此无关的表：

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),
)

以下每个查询都可以正常运行。我如何将它们结合起来，这样才能那些价值超过50,000,000,000的公司？

# -*- 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)

如果我尝试使用联接，我会收到错误：

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

我甚至不能做简易版

# This will return more results than you are probably expecting.
s = select([hurst, funda])
run(s)

Answer 1

目前还没有一个环境可以对此进行测试，但有些内容应该有效：

j = join(husrt, funda, (hurst.c.symbol_id == funda.c.symbol_id) & (funda.c.MarketCap > 50000000000))
stmt = select([hurst]).select_from(j)
run(stmt)

在this SQL Alchemy docs page.

上查看sqlalchemy.sql.expression.join

Answer 2

我不相信你可以加入两个数据库。

您的一个表位于一个数据库中，另一个位于另一个数据库中：

dbh = create_engine('sqlite:///hurst.db')
dbf = create_engine('sqlite:///fundamental.db')

您应该将所有表格放在同一个database.db文件中，并拥有一个db = create_engine('sqlite:///database.db')。

更正：您可以这样做：Cross database join in sqlalchemy

但是你真的想把每个表放在一个单独的数据库中吗？

python sqlalchemy结合两个表的查询

2 个答案: