如何。迭代terraform中的list（map（list））

debugcn 发表于 Dev

露西

我想在s3中创建文件（密钥）。通过以下代码，我可以在s3中成功创建文件-

locals {
  rules = [
  {
    user = "user-1",
    roles = "test"
  }
}

resource "aws_s3_bucket_object" "this" {

  for_each    = zipmap(local.rules.*.user, local.rules.*)
   bucket      = "test-09129"
   acl         = "private"
   source      = "/dev/null"
   key         = "${each.value.roles}/${each.key}"
}

s3中的文件-创建了test / user-1

但是角色可以是多个，所以我创建了角色列表。

locals {
  rules = [
  {
    user = "user-1",
    roles = ["test"]
  },
  {
    user = "user-2",
    roles = ["test", "dev", "prd"]
  }]
}

现在，我想基于上面的列表创建文件。如何根据用户和角色迭代列表（角色）以创建文件？欲望输出-

file in s3 -
test/user-1 
test/user-2 
dev/user-2
prd/user-2

马辛

您可以拼合你rules到helper_rules（假设我理解正确的话你想要的结果）：


locals {
  rules = [
  {
    user = "user-1",
    roles = ["test"]
  },
  {
    user = "user-2",
    roles = ["test", "dev", "prd"]
  }]
  
  
  helper_rules = merge([
      for user, rules in zipmap(local.rules.*.user, local.rules.*.roles):
         { for rule in rules:
           "${user}-${rule}" => {user = user, rule = rule}
         }
  ]...)
}

这将导致helper_rules：

{
  "user-1-test" = {
    "rule" = "test"
    "user" = "user-1"
  }
  "user-2-dev" = {
    "rule" = "dev"
    "user" = "user-2"
  }
  "user-2-prd" = {
    "rule" = "prd"
    "user" = "user-2"
  }
  "user-2-test" = {
    "rule" = "test"
    "user" = "user-2"
  }
}

这样，您的对象可能是：

resource "aws_s3_bucket_object" "this" {

  for_each     = local.helper_rules
   bucket      = "test-09129"
   acl         = "private"
   source      = "/dev/null"
   key         = "${each.value.role}/${each.value.user}"
}

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