Terraform-输出索引和项目

Question

是否有方法修改此输出代码以包含带名称的索引号？下面的当前代码创建了一个索引为"0"的输出，但理想情况下，每个名称都应该在其自己的索引(0、1、2等(中

terraform {
required_providers {
azurerm = {
source  = "hashicorp/azurerm"
version = "3.1.0"
}
}
}
provider "azurerm" {
features {
}
}
variable "private_link_scopes" {
description = "A list of Azure private link scopes."
type = list(object({
name                = string
resource_group_name = string
}))
}
locals {
rgs_map = {
for n in var.private_link_scopes : n.resource_group_name => {
name = n.resource_group_name
}
}
}
data "azurerm_resource_group" "rg" {
for_each = local.rgs_map
name     = each.value.name
}
output "list" {
value = { for index, item in [var.private_link_scopes] : index => item[*].name }
}

Answer 1

这里似乎发生了很多结构变化。这可能是由于其他包袱或依赖它的原因，但我认为这可以简化。我使用局部变量来代替变量，但我希望这会有所帮助。

我不确定那里的splat操作员是否是你想要的。这与获取该item值中所有项目的名称属性的列表相同，但每个item只有一个，所以这看起来很奇怪。

locals {
scopes = [
{
name    = "name_a"
rg_name = "rg_a"
},
{
name    = "name_b"
rg_name = "rg_b"
}
]
}
data "azurerm_resource_group" "rg" {
# I don't see a reason to generate the intermediary map
# or to build another object in each map value. This is more
# simple. for_each with an object by name: rg_name
value = { for scope in local.scopes : scope.name => scope.rg_name }
name  = each.value
}
output "list" {
# I don't see a need to use a splat expression here.
# You can just build a map with the index and the name
# here directly.
value = { for i, v in local.scopes : i => v.name }
}

事实上，如果您不需要按名称键入资源组资源，则可以进一步简化为：

data "azurerm_resource_group" "rg" {
for_each = toset([for scope in local.scopes : scope.rg_name])
name     = each.value
}