主页 / coding / 问题 / 77903327
Accepted
Kar
Asked: 2024-01-30 08:34:58 +0800 CST

Terraform:Snowflake:动态架构//表/数据库

  • 772

需求：需要给var中的值赋值。选择数据库名称、架构名称和表名称

我想将每个 table_name 授予 share_name

问题 ：

│   19:   dynamic "schema_name" {
│
│ Blocks of type "schema_name" are not expected here.

代码：下面是我写的但无法解决问题

resource "snowflake_table_grant" "grant" {
  for_each = var.select

  database_name = each.key

  dynamic "schema_name" {
    for_each = each.value
    content {
      schema_name = schema_name.key

      dynamic "table_name" {
        for_each = schema_name.value
        content {
          table_name = table_name.value

          privilege         = "SELECT"
          roles             = ["role1"]
          shares            = ["share1"]
          on_future         = false
          with_grant_option = false
        }
      }
    }
  }
}

变量.tf

variable "select" {
  type    = map(map(list(string)))
  default = {
    database1 = {
      schema1 = ["table1", "table2"],
      schema2 = ["table3", "table4"],
    },
  }
}
terraform

1 个回答

  1. Best Answer

    你必须展平你的变量：

    
locals {

  select_flat = merge(flatten([
    for database_name, schemas in var.select: [
      for schema_name, tables in schemas: {
          for table_name in tables: 
            "${database_name}-${schema_name}-${table_name}" => {
                 "database_name" = database_name
                 "schema_name" = schema_name
                 "table_name" = table_name
             }                 
      }
    ] 
  ])...) # do NOT remove the dots
}

    然后

    resource "snowflake_table_grant" "grant" {
    for_each      = local.select_flat

    database_name = each.value.database_name
    schema_name   = each.value.database_name
    table_name    = each.value.table_name  

    privilege         = "SELECT"
    roles             = ["role1"]
    shares            = ["share1"]
    on_future         = false
    with_grant_option = false
}
    • 0

相关问题