Python 3.x 如何从“中选择值”;“存储集成”;在雪花里?
我已经通过执行下面的命令创建了存储集成Python 3.x 如何从“中选择值”;“存储集成”;在雪花里?,python-3.x,snowflake-cloud-data-platform,snowflake-schema,Python 3.x,Snowflake Cloud Data Platform,Snowflake Schema,我已经通过执行下面的命令创建了存储集成 create or replace storage integration stager type = external_stage storage_provider = s3 enabled = true storage_aws_role_arn = 'arn:aws:iam::24545426:role/test' storage_allowed_locations = ('s3:
create or replace storage integration stager
type = external_stage
storage_provider = s3
enabled = true
storage_aws_role_arn = 'arn:aws:iam::24545426:role/test'
storage_allowed_locations = ('s3://testb/')
之后,我执行了DESC集成stager
我得到了这样的结果
现在我想选择stager的存储\u AWS\u外部\u ID
和属性值
如何使用query或python选择STORAGE\u AWS\u EXTERNAL\u ID
和属性值?您可以使用RESULT\u SCAN函数处理另一个查询的结果(使用SQL):
对于python:
import snowflake.connector
ctx = snowflake.connector.connect(
...
)
cs = ctx.cursor()
try:
sql = "DESC INTEGRATION stager"
cs.execute( sql )
for (c_property, c_type, c_value, c_default) in cs:
if c_property == "STORAGE_AWS_EXTERNAL_ID":
print('{0}, {1}'.format(c_property, c_value))
finally:
cs.close()
这是上述Python脚本的结果:
STORAGE_AWS_EXTERNAL_ID, WX65722_SFCRole=2_GpY+ZF0b41Nu3d2ZDFYPfCUbBxk=
感谢您的快速回复,在执行上述操作后,我得到了“property property\u value property\u value property\u value property\u value property\u value property\u value property\u value property\u value”如果值不打印我们得到的列标题,则不会进入内部Sai,让我分享整个脚本-我使用snowflake.connector包感谢您的帮助:)