Json响应pyodbc的更好方式
Json响应pyodbc的更好方式
提问于 2021-04-17 00:48:16
这是有效的，但我必须做100个反应中的30个。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()```回答 1
Stack Overflow用户
回答已采纳
发布于 2021-04-17 06:52:23
假设json.loads()提供了一个简单的dict对象列表，那么这正是SQLAlchemy可以直接使用的格式：
# https://stackoverflow.com/q/67129218/2144390
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')]页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/67129218
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