在SQLAlchemy列类型和python数据类型之间轻松转换？

Question

我正在寻找一种简单的python方法来比较SQLAlchemy中的列类型和基类型.例如,如果我的列类型是任何长度的VARCHAR,我想将其作为字符串读取.

我可以读取列类型没关系,但我不确定一种简单的方法来验证它的基本类型...如果我可以使用"if isinstance(mycolumn,int)"这样的东西会很好 - 但我是新的到python,不知道这将如何工作.

这是我到目前为止所拥有的:

from sqlalchemy import MetaData
from sqlalchemy import create_engine, Column, Table
engine = create_engine('mysql+mysqldb://user:pass@localhost:3306/mydb', pool_recycle=3600)
meta = MetaData()
meta.bind = engine
meta.reflect()
datatable = meta.tables['my_data_table']
[c.type for c in datatable.columns]

输出:

[INTEGER(display_width=11), DATE(), VARCHAR(length=127), DOUBLE(precision=None, scale=None, asdecimal=True)]

我的最终目的是双重的,首先是因为当我将它加载到我的jQuery jqGrid中时,我希望根据类型格式化输出.第二,我正在慢慢地将非规范化数据表转换为规范化结构,并且希望确保我保持我的类型一致 - (以确保我在前一个表中的数字存储为数字而不是字符串...... )

Answer 1

只需使用python_type所有AQLAlchemy类型中可用的属性,例如:

[c.type.python_type for c in datatable.columns]

Answer 2

Python 类型到 SQL 类型：

我在使用默认 sql 类型动态创建 SQL 表的问题上苦苦挣扎。我最终得到了以下方便的函数，用于我所有的 python 类型到 sql 类型的转换需求。从 sql 类型到 python 类型是微不足道的，将在下一节中解释。

import sqlalchemy
import numpy as np

import datetime
import decimal

_type_py2sql_dict = {
 int: sqlalchemy.sql.sqltypes.BigInteger,
 str: sqlalchemy.sql.sqltypes.Unicode,
 float: sqlalchemy.sql.sqltypes.Float,
 decimal.Decimal: sqlalchemy.sql.sqltypes.Numeric,
 datetime.datetime: sqlalchemy.sql.sqltypes.DateTime,
 bytes: sqlalchemy.sql.sqltypes.LargeBinary,
 bool: sqlalchemy.sql.sqltypes.Boolean,
 datetime.date: sqlalchemy.sql.sqltypes.Date,
 datetime.time: sqlalchemy.sql.sqltypes.Time,
 datetime.timedelta: sqlalchemy.sql.sqltypes.Interval,
 list: sqlalchemy.sql.sqltypes.ARRAY,
 dict: sqlalchemy.sql.sqltypes.JSON
}

def type_py2sql(pytype):
    '''Return the closest sql type for a given python type'''
    if pytype in _type_py2sql_dict:
        return _type_py2sql_dict[pytype]
    else:
        raise NotImplementedError(
            "You may add custom `sqltype` to `"+str(pytype)+"` assignment in `_type_py2sql_dict`.")

def type_np2py(dtype=None, arr=None):
    '''Return the closest python type for a given numpy dtype'''

    if ((dtype is None and arr is None) or
        (dtype is not None and arr is not None)):
        raise ValueError(
            "Provide either keyword argument `dtype` or `arr`: a numpy dtype or a numpy array.")

    if dtype is None:
        dtype = arr.dtype

    #1) Make a single-entry numpy array of the same dtype
    #2) force the array into a python 'object' dtype
    #3) the array entry should now be the closest python type
    single_entry = np.empty([1], dtype=dtype).astype(object)

    return type(single_entry[0])

def type_np2sql(dtype=None, arr=None):
    '''Return the closest sql type for a given numpy dtype'''
    return type_py2sql(type_np2py(dtype=dtype, arr=arr))

一些用例：

>>> sqlalchemy.Column(type_py2sql(int))
Column(None, BigInteger(), table=None)

>>> type_py2sql(type('hello'))
sqlalchemy.sql.sqltypes.Unicode

>>> type_np2sql(arr=np.array([1.,2.,3.]))
sqlalchemy.sql.sqltypes.Float

我如何选择我的转换集：

我所做的是将所有 sql 类型映射到它们等效的 python 类型。然后我打印了哪种 python 类型对应于哪些 sql-types，并为每个 python 类型选择了最好的 sql-type。这是我用来生成此映射的代码：

#********** SQL to Python: one to one **********
type_sql2py_dict = {}
for key in sqlalchemy.types.__dict__['__all__']:
    sqltype = getattr(sqlalchemy.types, key)

    if 'python_type' in dir(sqltype) and not sqltype.__name__.startswith('Type'):
        try:
            typeinst = sqltype()
        except TypeError as e: #List/array wants inner-type
            typeinst = sqltype(None)

        try:
            type_sql2py_dict[sqltype] = typeinst.python_type
        except NotImplementedError:
            pass

#********** Python to SQL: one to many **********
type_py2sql_dict = {}
for key, val in type_sql2py_dict.items():
    if not val in type_py2sql_dict:
        type_py2sql_dict[val] = [key]
    else:
        type_py2sql_dict[val].append(key)

这是type_py2sql_dictsqlalchemy 1.3.5 版下的输出：

{int: [sqlalchemy.sql.sqltypes.INTEGER,
  sqlalchemy.sql.sqltypes.BIGINT,
  sqlalchemy.sql.sqltypes.SMALLINT,
  sqlalchemy.sql.sqltypes.Integer,
  sqlalchemy.sql.sqltypes.SmallInteger,
  sqlalchemy.sql.sqltypes.BigInteger],
 str: [sqlalchemy.sql.sqltypes.CHAR,
  sqlalchemy.sql.sqltypes.VARCHAR,
  sqlalchemy.sql.sqltypes.NCHAR,
  sqlalchemy.sql.sqltypes.NVARCHAR,
  sqlalchemy.sql.sqltypes.TEXT,
  sqlalchemy.sql.sqltypes.Text,
  sqlalchemy.sql.sqltypes.CLOB,
  sqlalchemy.sql.sqltypes.String,
  sqlalchemy.sql.sqltypes.Unicode,
  sqlalchemy.sql.sqltypes.UnicodeText,
  sqlalchemy.sql.sqltypes.Enum],
 float: [sqlalchemy.sql.sqltypes.FLOAT,
  sqlalchemy.sql.sqltypes.REAL,
  sqlalchemy.sql.sqltypes.Float],
 decimal.Decimal: [sqlalchemy.sql.sqltypes.NUMERIC,
  sqlalchemy.sql.sqltypes.DECIMAL,
  sqlalchemy.sql.sqltypes.Numeric],
 datetime.datetime: [sqlalchemy.sql.sqltypes.TIMESTAMP,
  sqlalchemy.sql.sqltypes.DATETIME,
  sqlalchemy.sql.sqltypes.DateTime],
 bytes: [sqlalchemy.sql.sqltypes.BLOB,
  sqlalchemy.sql.sqltypes.BINARY,
  sqlalchemy.sql.sqltypes.VARBINARY,
  sqlalchemy.sql.sqltypes.LargeBinary,
  sqlalchemy.sql.sqltypes.Binary],
 bool: [sqlalchemy.sql.sqltypes.BOOLEAN, sqlalchemy.sql.sqltypes.Boolean],
 datetime.date: [sqlalchemy.sql.sqltypes.DATE, sqlalchemy.sql.sqltypes.Date],
 datetime.time: [sqlalchemy.sql.sqltypes.TIME, sqlalchemy.sql.sqltypes.Time],
 datetime.timedelta: [sqlalchemy.sql.sqltypes.Interval],
 list: [sqlalchemy.sql.sqltypes.ARRAY],
 dict: [sqlalchemy.sql.sqltypes.JSON]}

Answer 3

一种解决方案是手动进行转换 - 例如，这有效：

def convert(self, saType):
    type = "Unknown"
    if isinstance(saType,sqlalchemy.types.INTEGER):
        type = "Integer"
    elif isinstance(saType,sqlalchemy.types.VARCHAR):
        type = "String"
    elif isinstance(saType,sqlalchemy.types.DATE):
        type = "Date"
    elif isinstance(saType,sqlalchemy.dialects.mysql.base._FloatType):
        type = "Double"
    return type

不确定这是否是一种正常的 Python 做事方式......我仍然像一个 Java 程序员一样思考。