Python 操作错误：250003:未能获取响应。绞刑？方法：邮寄

我正在尝试使用我的登录凭据连接到snowflake。我正在使用以下代码：

snowflake.connector.connect(
    user="<my_user_name>",
    password="<my_password>",
    account="<my_account_name_with_region_and_cloud>"
    )

当我尝试运行上述代码时，出现以下错误：

操作错误：250003:未能获取响应。绞刑？方法：post，url:

我确信我所有的包裹都是最新的。我正在使用Python3.6.4和最新的snowflake_connector_python

我目前在aws的us-east-2位置

---

有人能帮我解决这个问题吗？？？

我正在使用sqlalchemy，您可以通过pip安装它：

pip install SQLAlchemy

以下是我笔记本开头的内容：

import snowflake.connector
import pandas as pd
from sqlalchemy import create_engine
from snowflake.sqlalchemy import URL

url = URL(
    account = 'xxxxxxxx.east-us-2.azure',
    user = 'xxxxxxxx',
    password = 'xxxxxxxx',
    database = 'xxxxxxxx',
    schema = 'xxxxxxxx',
    warehouse = 'xxxxxxxx',
    role='xxxxxxxx'
)

engine = create_engine(url)
connection = engine.connect()

query = '''
    select 1 AS VAL;
'''

df = pd.read_sql(query, connection)

df

---

如果有人帮了你，你应该选择一个答案。我们实际上需要区域和云，否则我会出错。我必须把FCXXXX.us-east-2.aws

Just Give your account name in the account .We dont need the region and full URL.
Please check below .

----------------------------------------------------------------------
import snowflake.connector


PASSWORD = '*******'
USER = '<USERNAME>'
ACCOUNT = 'SFCSUPPORT'
WAREHOUSE = '<WHNAME>'
DATABASE = '<DBNAME>'
SCHEMA = 'PUBLIC'

print("Connecting...")
# -- (> ------------------- SECTION=connect_to_snowflake --------------------
con = snowflake.connector.connect(
  user=USER,
  password=PASSWORD,
  account=ACCOUNT,
  warehouse=WAREHOUSE,
  database=DATABASE,
  schema=SCHEMA
)


con.cursor().execute("USE WAREHOUSE " + WAREHOUSE)
con.cursor().execute("USE DATABASE " + DATABASE)
#con.cursor().execute("USE SCHEMA INFORMATION_SCHEMA")


try:
    result = con.cursor().execute("Select * from <TABLE>")
    result_list = result.fetchall()
    print(result_list)

finally:
    con.cursor().close()
con.cursor().close()