如何将SQLAlchemy行对象转换为Python字典?

是否有一种简单的方法来遍历列名和值对?

我的SQLAlchemy版本是0.5.6

下面是我尝试使用dict(row)的示例代码:

import sqlalchemy
from sqlalchemy import *
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker

print "sqlalchemy version:",sqlalchemy.__version__ 

engine = create_engine('sqlite:///:memory:', echo=False)
metadata = MetaData()
users_table = Table('users', metadata,
     Column('id', Integer, primary_key=True),
     Column('name', String),
)
metadata.create_all(engine) 

class User(declarative_base()):
    __tablename__ = 'users'
    
    id = Column(Integer, primary_key=True)
    name = Column(String)
    
    def __init__(self, name):
        self.name = name

Session = sessionmaker(bind=engine)
session = Session()

user1 = User("anurag")
session.add(user1)
session.commit()

# uncommenting next line throws exception 'TypeError: 'User' object is not iterable'
#print dict(user1)
# this one also throws 'TypeError: 'User' object is not iterable'
for u in session.query(User).all():
    print dict(u)

在我的系统输出上运行这段代码:

Traceback (most recent call last):
  File "untitled-1.py", line 37, in <module>
    print dict(u)
TypeError: 'User' object is not iterable

当前回答

我对使用(太多?)字典的看法:

def serialize(_query):
#d = dictionary written to per row
#D = dictionary d is written to each time, then reset
#Master = dictionary of dictionaries; the id Key (int, unique from database) from D is used as the Key for the dictionary D entry in Master
Master = {}
D = {}
x = 0
for u in _query:
    d = u.__dict__
    D = {}
    for n in d.keys():
        if n != '_sa_instance_state':
            D[n] = d[n]
    x = d['id']
    Master[x] = D
return Master

使用flask(包括jsonify)和flask_sqlalchemy将输出打印为JSON。

使用jsonify(serialize())调用该函数。

与我迄今为止尝试过的所有SQLAlchemy查询一起工作(运行SQLite3)

2016-06-08 15:13:18

其他回答

有了这段代码，您还可以添加到您的查询“过滤器”或“连接”，这工作!

query = session.query(User)
def query_to_dict(query):
        def _create_dict(r):
            return {c.get('name'): getattr(r, c.get('name')) for c in query.column_descriptions}

    return [_create_dict(r) for r in query]

2017-11-05 09:39:17

class User(object):
    def to_dict(self):
        return dict([(k, getattr(self, k)) for k in self.__dict__.keys() if not k.startswith("_")])

这应该有用。

2010-02-10 14:07:02

参考Alex Brasetvik的答案，你可以用一行代码来解决这个问题

row_as_dict = [dict(row) for row in resultproxy]

在Alex Brasetvik的回答的评论部分，SQLAlchemy的创建者zzzeek表示这是解决这个问题的“正确方法”。

2017-07-03 10:26:39

我是一个新晋的Python程序员，遇到了使用join表获取JSON的问题。使用这里的答案中的信息，我构建了一个函数，将合理的结果返回到JSON，其中包括表名，避免使用别名或字段冲突。

简单地传递会话查询的结果:

test = Session()。查询(VMInfo、客户). join(客户).order_by (VMInfo.vm_name) .limit (50) .offset (10)

json = sqlAl2json（test）

def sqlAl2json(self, result):
    arr = []
    for rs in result.all():
        proc = []
        try:
            iterator = iter(rs)
        except TypeError:
            proc.append(rs)
        else:
            for t in rs:
                proc.append(t)

        dict = {}
        for p in proc:
            tname = type(p).__name__
            for d in dir(p):
                if d.startswith('_') | d.startswith('metadata'):
                    pass
                else:
                    key = '%s_%s' %(tname, d)
                    dict[key] = getattr(p, d)
        arr.append(dict)
    return json.dumps(arr)

2015-11-24 15:45:49