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SQLAlchemy 几种查询方式总结

 

几种常见sqlalchemy查询:
#简单查询
print(session.query(User).all())
print(session.query(User.name, User.fullname).all())
print(session.query(User, User.name).all())

#带条件查询
print(session.query(User).filter_by(name=’user1′).all())
print(session.query(User).filter(User.name == “user”).all())
print(session.query(User).filter(User.name.like(“user%”)).all())

#多条件查询
print(session.query(User).filter(and_(User.name.like(“user%”), User.fullname.like(“first%”))).all())
print(session.query(User).filter(or_(User.name.like(“user%”), User.password != None)).all())

#sql过滤
print(session.query(User).filter(“id>:id”).params(id=1).all())

#关联查询
print(session.query(User, Address).filter(User.id == Address.user_id).all())
print(session.query(User).join(User.addresses).all())
print(session.query(User).outerjoin(User.addresses).all())

#聚合查询
print(session.query(User.name, func.count(‘*’).label(“user_count”)).group_by(User.name).all())
print(session.query(User.name, func.sum(User.id).label(“user_id_sum”)).group_by(User.name).all())

#子查询
stmt = session.query(Address.user_id, func.count(‘*’).label(“address_count”)).group_by(Address.user_id).subquery()
print(session.query(User, stmt.c.address_count).outerjoin((stmt, User.id == stmt.c.user_id)).order_by(User.id).all())

#exists
print(session.query(User).filter(exists().where(Address.user_id == User.id)))
print(session.query(User).filter(User.addresses.any()))

限制返回字段查询
person = DB.session.query(Person.name,Person.created_at,Person.updated_at).filter_by(name=”zhongwei”).order_by(Person.created_at).first()

记录总数查询:
from sqlalchemy import func

# count User records, without
# using a subquery.
session.query(func.count(User.id))

# return count of user “id” grouped
# by “name”
session.query(func.count(User.id)).\
group_by(User.name)

from sqlalchemy import distinct
# count distinct “name” values
session.query(func.count(distinct(User.name)))

 

select two distinct columns with flask and sqlalchemy

 

SELECT DISTINCT gradelevel, beginningorendtest 
FROM wordlist 
ORDER BY gradelevel,beginningorendoftest

 

the equivalent query:

data = wordlist.query.with_entities(wordlist.gradelevel, wordlist.beginningorendtest)
        .distinct().order_by(wordlist.gradelevel).order_by(wordlist.beginningorendtest)