Json 响应 pyodbc 更好的方式

Question

这有效，但我必须对 100 个猫科动物做出 30 个响应。有没有更好的方法？

            for record in data:
            record["lastName"] = record["lastName"].replace("'","''")
            record["lastName"] = record["lastName"].replace("'","''")
            cursor.execute("Insert Into emp_temp (employeeId, firstName, lastName) values ('" + record["employeeId"] +"','"+ record["firstName"] + "','"+ record["lastName"] +"')")
            cursor.commit()
        cursor.close()
        conn.close()```

Answer 1

假设 json.loads() 为您提供了一个简单的字典对象列表，那么这正是 SQLAlchemy 可以直接使用的格式：

# 
import json

import sqlalchemy as sa

response_text = '''\
[{"employeeId": 1, "firstName": "Gord", "lastName": "Thompson"},
 {"employeeId": 2, "firstName": "Bob", "lastName": "Loblaw"}]'''
data = json.loads(response_text)
print(type(data))  # <class 'list'>
print(type(data[0]))  # <class 'dict'>

engine = sa.create_engine("mssql+pyodbc://@mssqlLocal64")

emp_temp = sa.Table("emp_temp", sa.MetaData(), autoload_with=engine)

with engine.begin() as conn:
    conn.execute(emp_temp.insert(), data)

# check results
with engine.begin() as conn:
    results = conn.execute(sa.text("SELECT * FROM emp_temp")).fetchall()
    print(results)
    # [(1, 'Gord', 'Thompson'), (2, 'Bob', 'Loblaw')]

Json 响应 pyodbc 更好的方式

Json Response to pyodbc Better way

python

json

pyodbc

python-requests