在 Azure 数据工厂中创建 SnowFlake 链接服务时出现索引超出范围错误
Index Out of Range Error When Creating SnowFlake Linked Service in Azure Data Factory
-
azure
-
azure-data-factory
-
snowflake-cloud-data-platform
-
azure-data-factory-2
-
azure-data-factory-pipeline
我正在传递所需的凭据和参数,但出现错误
The value of the property 'index' is invalid: 'Index was out of range.
Must be non-negative and less than the size of the collection.
Parameter name: index'. Index was out of range. Must be non-negative
and less than the size of the collection. Parameter name: index
Activity ID: 36a4265d-3607-4472-8641-332f5656661d.
似乎 UI 没有正确生成链接服务。使用 Microsoft Docs Example JSON 我在尝试创建链接服务时收到相同的索引错误。如果我从连接字符串中删除密码并将其添加为单独的 属性 我能够成功生成链接服务。
Microsoft 文档示例(不起作用)
{
"name": "SnowflakeLinkedService",
"properties": {
"type": "Snowflake",
"typeProperties": {
"connectionString": "jdbc:snowflake://<accountname>.snowflakecomputing.com/?user=<username>&password=<password>&db=<database>&warehouse=<warehouse>&role=<myRole>"
},
"connectVia": {
"referenceName": "<name of Integration Runtime>",
"type": "IntegrationRuntimeReference"
}
}
}
工作示例
{
"name": "SnowflakeLinkedService",
"properties": {
"type": "Snowflake",
"typeProperties": {
"connectionString": "jdbc:snowflake://<accountname>.snowflakecomputing.com/?user=<username>&db=<database>&warehouse=<warehouse>",
"password": {
"type": "SecureString",
"value": "<password>"
}
},
"connectVia": {
"referenceName": "<name of Integration Runtime>",
"type": "IntegrationRuntimeReference"
}
}
}
我遇到了同样的问题,密码中包含一个 ',这就是问题所在。更改了没有符号的密码,效果很好
azure
azure-data-factory
snowflake-cloud-data-platform
azure-data-factory-2
azure-data-factory-pipeline
我正在传递所需的凭据和参数,但出现错误
The value of the property 'index' is invalid: 'Index was out of range. Must be non-negative and less than the size of the collection. Parameter name: index'. Index was out of range. Must be non-negative and less than the size of the collection. Parameter name: index Activity ID: 36a4265d-3607-4472-8641-332f5656661d.
似乎 UI 没有正确生成链接服务。使用 Microsoft Docs Example JSON 我在尝试创建链接服务时收到相同的索引错误。如果我从连接字符串中删除密码并将其添加为单独的 属性 我能够成功生成链接服务。
Microsoft 文档示例(不起作用)
{
"name": "SnowflakeLinkedService",
"properties": {
"type": "Snowflake",
"typeProperties": {
"connectionString": "jdbc:snowflake://<accountname>.snowflakecomputing.com/?user=<username>&password=<password>&db=<database>&warehouse=<warehouse>&role=<myRole>"
},
"connectVia": {
"referenceName": "<name of Integration Runtime>",
"type": "IntegrationRuntimeReference"
}
}
}
工作示例
{
"name": "SnowflakeLinkedService",
"properties": {
"type": "Snowflake",
"typeProperties": {
"connectionString": "jdbc:snowflake://<accountname>.snowflakecomputing.com/?user=<username>&db=<database>&warehouse=<warehouse>",
"password": {
"type": "SecureString",
"value": "<password>"
}
},
"connectVia": {
"referenceName": "<name of Integration Runtime>",
"type": "IntegrationRuntimeReference"
}
}
}
我遇到了同样的问题,密码中包含一个 ',这就是问题所在。更改了没有符号的密码,效果很好