OperationalError：250003：无法获取响应。绞刑？方法：POST

问题描述

我正在尝试使用我的登录凭据连接到Snowflake。我使用以下代码：

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，收到以下错误：

OperationalError：250003：无法获取响应。绞刑？方法：post，url：https://hm53485.us-east-2.aws.snowflakecomputing.com:443/session/v1/login-request?request_id=fcfdd77a-11ff-4956-9ed8-bcc332c5989a&databaseName=S3_DB&schemaName=PUBLIC&warehouse=COMPUTE_WH&request_guid=b9fdb5c9-81cb-4ecb-8d20-abef44249bbf

我确信我所有的包都是最新的。我使用的是python 3.6.4和最新的Snowflake_Connector_python。

我当前位于AWS中的us-East-2位置。

有人能帮我一下吗？

---

解决方案

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()