我有一个简单的表格,其中包含 int[] 类型的列,我想能够选择其中任意一个数组元素与我拥有的值相匹配的行,但我无法在不使用原始查询的情况下使用 SQLAlchemy 实现这一点。
以下是该表格("testuser")的架构:
Column | Type |
---------+------------------------+
id | integer |
name | character varying(250) |
numbers | integer[] |
以下是使用示例数据的样子:
id | name | numbers
----+---------+---------------
1 | David | {25,33,42,55}
2 | Salazar | {11,33,7,19}
3 | Belinda | {32,6,20,23}
4 | Casey | {19,20,27,8}
5 | Kathie | {25,31,10,40}
6 | Dianne | {25,20,40,39}
7 | Cortez | {1,20,18,38}
这里有一条SQL语句可以生成我想要的结果;我希望能够在Python中完成这个操作,而不是简单地编写原始查询(25只是一个示例)。
scrape=# select * from testuser where 25 = ANY(numbers);
id | name | numbers
----+--------+---------------
5 | Kathie | {25,31,10,40}
6 | Dianne | {25,20,40,39}
1 | David | {25,33,42,55}
(3 rows)
我发现另一种编写方式:
scrape=# select * from testuser where numbers @> ARRAY[25];
id | name | numbers
----+--------+---------------
5 | Kathie | {25,31,10,40}
6 | Dianne | {25,20,40,39}
1 | David | {25,33,42,55}
(3 rows)
以下是我用来生成表格的Python代码:
from sqlalchemy import Column, Integer, String
from sqlalchemy.dialects import postgresql
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker
Base = declarative_base()
class TestUser(Base):
__tablename__ = 'testuser'
id = Column(Integer, primary_key=True)
name = Column(String(250))
numbers = Column(postgresql.ARRAY(Integer))
engine = create_engine('postgresql://postgres:{pw}@localhost:5432/scrape'.format(pw=POSTGRES_PASSWORD))
Base.metadata.create_all(engine)
DBSession = sessionmaker(bind=engine)
session = DBSession()
testcases = [{"numbers": [25, 33, 42, 55], "name": "David"}, {"numbers": [11, 33, 7, 19 ], "name": "Salazar"}, {"numbers": [32, 6, 20, 23 ], "name": "Belinda"}, {"numbers": [19, 20, 27, 8 ], "name": "Casey"}, {"numbers": [25, 31, 10, 40 ], "name": "Kathie"}, {"numbers": [25, 20, 40, 39 ], "name": "Dianne"}, {"numbers": [1, 20, 18, 38 ], "name": "Cortez"} ]
for t in testcases:
session.add(TestUser(name=t['name'], numbers=t['numbers']))
session.commit()