Terraform 使用'；时会导致意外结果的回路地形；展平'；和'；元素'；功能

我试图通过每隔一个元素将一个数组解析为两个数组

示例： [“托马斯1”、“托马斯2”、“托马斯3”、“托马斯4”、“托马斯5”] 拆分为两个新阵列： [“托马斯1”、“托马斯3”、“托马斯5”] 和 [“托马斯2”、“托马斯4”]

也许有更好的方法可以做到这一点，但在Terraform中，当我使用下面的代码时，我的第二个新数组得到了一个非常奇怪的结果。我完全弄不懂这是怎么可能的

locals {
  hostnames = ["thomas1", "thomas2", "thomas3", "thomas4", "thomas5"]
  clusterA_range = range(0,length(local.hostnames),2)
  clusterB_range = range(1,length(local.hostnames),2)
  clusterA = flatten ( [ for i in local.clusterA_range : element(local.hostnames,(element(local.clusterA_range,i))) ] )
  clusterB = flatten ( [ for x in local.clusterB_range : element(local.hostnames,(element(local.clusterB_range,x))) ] )
}

output "hostnames" {
  value = local.hostnames
}
output "clusterA" {
  value = local.clusterA
}
output "clusterB" {
  value = local.clusterB
}
output "clusterA_range" {
  value = local.clusterA_range
}
output "clusterB_range" {
  value = local.clusterB_range
}

其结果是：

$ terraform apply

An execution plan has been generated and is shown below.
Resource actions are indicated with the following symbols:

Terraform will perform the following actions:

Plan: 0 to add, 0 to change, 0 to destroy.

Do you want to perform these actions?
  Terraform will perform the actions described above.
  Only 'yes' will be accepted to approve.

  Enter a value: yes


Apply complete! Resources: 0 added, 0 changed, 0 destroyed.

Outputs:

clusterA = [
  "thomas1",
  "thomas5",
  "thomas3",
]
clusterA_range = [
  0,
  2,
  4,
]
clusterB = [
  "thomas4",
  "thomas4",
]
clusterB_range = [
  1,
  3,
]
hostnames = [
  "thomas1",
  "thomas2",
  "thomas3",
  "thomas4",
  "thomas5",
]

为什么clusterB[“thomas4”，“thomas4”]

---

Terraform版本v0.13.0-dev发生这种情况是因为您正在迭代范围的值，然后使用这些值再次选择范围值。如果要保持结构，必须在以下范围内迭代索引：

locals {
  hostnames = ["thomas1", "thomas2", "thomas3", "thomas4", "thomas5"]
  
  clusterA_range = range(0,length(local.hostnames), 2)
  clusterB_range = range(1,length(local.hostnames), 2)
  
  clusterA = flatten ( [ 
      for idxA, i in local.clusterA_range : 
          element(local.hostnames,(element(local.clusterA_range, idxA))) ] )
 
  clusterB = flatten ( 
        [ for idxB, x in local.clusterB_range : 
        element(local.hostnames,(element(local.clusterB_range, idxB))) ] )
}

但是，对于您的特定示例，您可以简化选择：

  clusterA = [for  i in local.clusterA_range :  element(local.hostnames, i) ]
  
  clusterB = [ for x in local.clusterB_range :  element(local.hostnames, x) ]

---

啊！现在这很有道理。非常感谢你。您的简化解决方案更加优雅@汤马斯：没问题。很高兴它成功了：-）这不是对你问题的直接回答，但我只是想指出，你在这里的预期目标似乎有点类似，以防这有助于简化解决方案。