How can i create multiple fargate profiles with single namespace and different labels using terraform? - kubernetes

Question

I am trying to create fargate profiles for EKS using terraform, the requirement is to create multiple fargate profiles bound to single namespace but different label.
I have defined the selector variable as below :
variable "selectors" {
description = "description"
type = list(object({
namespace = string
labels = any
}))
default = []
}
and the fargate module block as below :
resource "aws_eks_fargate_profile" "eks_fargate_profile" {
for_each = {for namespace in var.selectors: namespace.namespace => namespace}
cluster_name = var.cluster_name
fargate_profile_name = format("%s-%s","fargate",each.value.namespace)
pod_execution_role_arn = aws_iam_role.eks_fargate_role.arn
subnet_ids = var.vpc_subnets
selector {
namespace = each.value.namespace
labels = each.value.labels
}
and calling the module as below :
selectors = [
{
namespace = "ns"
labels = {
Application = "fargate-1"
}
},
{
namespace = "ns"
labels = {
Application = "fargate-2"
}
}
]
When i try to run terraform plan, i am getting below error :
Two different items produced the key "jenkinsbuild" in this 'for' expression. If duplicates are expected, use the ellipsis (...) after the value expression to enable grouping by key.
I tried giving (...) at the end of the for loop, this time i am getting another error as below :
each.value is tuple with 1 element
│
│ This value does not have any attributes.
I also defined selectors variable type as any, as well tried type casting the output to string(namespace) and object(labels), but no luck.
So could you please help me in achieving the same, It seems i am close but i am missing something here.
Thanks and Regards,
Sandeep.

Answer 1

In Terraform, when using for_each, the keys must be unique. If you do not have unique keys, then use count:
resource "aws_eks_fargate_profile" "eks_fargate_profile" {
count = length(var.selectors)
selector {
namespace = var.selectors[count.index].namespace
labels = var.selectors[count.index].labels
}
...
}