常规建表语句：
CREATE TABLE `test_table` (
`id` int(11) NOT NULL,
`name` char(64) NOT NULL,
`password` char(64) NOT NULL,
PRIMARY KEY (`name`,`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='test';
解析脚本代码：
# coding:utf-8 import re
def table_design_transfer(table_design):
type_map = {"varchar":"String","datetime":"DateTime","bigint":"BigInteger","smallint":"SmallInteger","tinyint":"SmallInteger","text":"Text","int":"Integer","double":"Float","char":"String","set":"Enum"}
l = table_design.split("\n") # 表设计行拆分
length = len(l)
s = []
primary_key_l = []
for i in range(length): # 遍历表设计行
j = l[length-1-i].strip().split(" ") # 倒序遍历，并按空格切分
if len(j)>2: # 只关注行长度超过2的元素
column = j[0].replace("`","")
i_type = j[1]
if column == "PRIMARY":
primary_key_l = re.sub(r'`|\(|\)','',j[2]).split(",") # 拿到主键key
continue
elif column == "CREATE": # 获取表名
table_name = j[2].replace("`","")
s.append(" "+'__tablename__ = "%s"' % table_name)
s.append("class %s(Base):" % table_name)
continue
elif column in ("UNIQUE",")","KEY"): # 非表列名，跳过
continue
if i_type in type_map.keys(): # 类型存在映射表中
i_type = i_type.replace(i_type,type_map[i_type])+"()"
elif "(" in i_type and i_type.split("(")[0] in type_map.keys(): # 类型有长度声明，提取类型字段，找到映射表映射value，并替换
old_type = i_type.split("(")[0]
new_type = type_map[i_type.split("(")[0]]
i_type = i_type.replace(old_type,new_type)
else:
print "Catch any case not in type_map:%s" % i_type if column in primary_key_l: # 列名存在主键数组中
i_type = i_type + ", primary_key=True"
s.append(" "+column + " = Column(" + i_type + ")")
for i in s[::-1]: # 反序输出
print i
输出结果：
class test_table(Base):
__tablename__ = "test_table"
id = Column(Integer(11), primary_key=True)
name = Column(String(64), primary_key=True)
password = Column(String(64))
sqlalchemy库官方文档：http://docs.sqlalchemy.org/en/latest/contents.html
sqlalchemy库官方文档（中文）：http://www.cnblogs.com/iwangzc/p/4112078.html（感谢作者的分享）
faker库官方文档：https://faker.readthedocs.io/en/master/locales/zh_CN.html
faker库博客：https://www.jianshu.com/p/6bd6869631d9